Chapter 10. Reduction-Oxidation: Electrochemistry
Faculty Resource and Organizational Guide (FROG)
Table of Contents
Materials for Chapter 10 Activities ..............................................................................................5
Reagents for Chapter 10 Activities: .............................................................................................6
Introductory Notes .........................................................................................................................8
Consider This 10.1. What atoms are oxidized and reduced in reactions (10.1) and (10.2)? ......8
Section 10.1. Electrolysis ..............................................................................................................9
Learning Objectives for Section 10.1 ..........................................................................................9
Investigate This 10.2. What happens when electric current passes through water? ...................9
Consider This 10.3. What chemical reactions does an electric current cause?.........................11
Consider This 10.4. What happens to the Mg2+(aq) and SO42-(aq)? ........................................11
Consider This 10.5. How do you analyze the gas from water electrolysis? .............................13
Investigate This 10.7. What other reactions can occur in an electrolytic cell? .........................15
Consider This 10.8. What are the products of electrolysis of a CuSO4 solution? ....................16
Section 10.2. Electric Current from Chemical Reactions .......................................................17
Learning Objectives for Section 10.2 ........................................................................................17
Investigate This 10.14. How do you get electricity from a chemical reaction?........................17
Consider This 10.15. How do you get electricity from a chemical reaction? ...........................19
Consider This 10.16. How are reduction half reactions combined? .........................................20
Consider This 10.17. What are the net charges on the solutions in Figure 10.4? .....................21
Section 10.3. Cell Potentials ........................................................................................................21
Learning Objectives for Section 10.3 ........................................................................................21
Investigate This 10.21. What are cell voltages for different half-cell combinations? ..............22
Consider This 10.22. Are cell voltages from different half-cell combinations related? ...........23
Consider This 10.23. What are the cell potentials and reactions in Investigate This 10.21? ...24
Section 10.4. Half-Cell Potentials: Reduction Potentials .........................................................25
Learning Objectives for Section 10.4 ........................................................................................25
Investigate This 10.26. How do metals react with acids? .........................................................25
Consider This 10.27. What are the reactions of metals with acids? .........................................26
Section 10.5. Work from Electrochemical Cells: Free Energy ...............................................27
Learning Objectives for Section 10.5 ........................................................................................27
Investigate This 10.31. How can we get work from a galvanic cell? .......................................28
Consider This 10.32. What reaction in the zinc-copper galvanic cell produces work? ............30
Consider This 10.35. Why are there so many sizes of batteries? .............................................31
Section 10.6. Concentration Dependence of Cell Potentials: The Nernst Equation .............31
Learning Objectives for Section 10.6: .......................................................................................31
Investigate This 10.38. Does the silver-copper cell potential depend on [Ag+]? .....................32
Consider This 10.39. How does the silver-copper cell potential depend [Ag+]? ......................34
Investigate This 10.44. Does the silver-quinhydrone cell potential depend on pH? ................35
Consider This 10.45. How does the silver-quinhydrone cell potential depend on pH?............36
April 2005
ACS Chemistry FROG
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Reduction-Oxidation: Electrochemistry
Chapter 10
Section 10.7. Reduction Potentials and the Nernst Equation..................................................38
Learning Objectives for Section 10.7: .......................................................................................38
Investigate This 10.53. Can pH change the direction of a cell reaction? ..................................38
Consider This 10.54. How does pH change the direction of a cell reaction? ...........................39
Section 10.8. Coupled Redox Reactions ....................................................................................40
Learning Objectives for Section 10.8 ........................................................................................40
Investigate This 10.56. Does copper ion react with sugars? .....................................................41
Consider This 10.57. What is the reaction of copper ion with glucose? ..................................42
Investigate This 10.61. Can you characterize the Blue-Bottle reaction? ..................................43
Consider This 10.62. Can you interpret the Blue-Bottle reaction? ...........................................44
Consider This 10.63. Can you explain the timing of the Blue-Bottle reaction? .......................45
Consider This 10.68. What sort of redox coupling occurs in biological systems? ...................47
Section 10.12. Extension — Cell Potentials and Non-Redox Equilibria .................................48
Investigate This 10.73. Does addition of ethylenediamine affect ECu2+, Cu? .............................48
Consider This 10.74. Why does addition of ethylenediamine affect ECu2+,Cu ? ........................49
Consider This 10.75. Do the copper concentration cells behave as expected?.........................50
Consider This 10.77. How does addition of ethylenediamine affect [Cu2+(aq)]? ....................51
Solutions for Chapter 10 Check This Activities .........................................................................52
Check This 10.6. Reactions in an electrophoresis apparatus. ...................................................52
Check This 10.9. The oxidation reaction in Investigate This 10.7. ..........................................52
Check This 10.11. Electrochemical stoichiometry. ..................................................................52
Check This 10.13. Electrolytic purification of copper metal. ...................................................53
Check This 10.18. Ion migrations in a salt bridge. ...................................................................53
Check This 10.20. Interpreting the line notation for a copper–zinc cell...................................54
Check This 10.24. Combining cell potentials and interpreting cell reactions. .........................54
Check This 10.25. Standard reduction potentials. ....................................................................55
Check This 10.28. What are the signs of reduction potentials for other metals? .....................55
Check This 10.30. The lead-acid battery. .................................................................................55
Check This 10.34. Electrical work. ...........................................................................................56
Check This 10.37. Amount of work available from an electrochemical cell. ..........................56
Check This 10.40. Nernst equation at 25 C (298 K). ..............................................................57
Check This 10.42. Nernst equation applied to silver–copper cells. ..........................................57
Check This 10.43. Three problems from the Web Companion. ...............................................58
Check This 10.46. Deriving equation (10.74)...........................................................................59
Check This 10.47. pH and cell potentials for silver-quinhydrone cells. ...................................59
Check This 10.49. Conditions for developing photographic film. ...........................................60
Check This 10.50. Simplifying equation (10.79). .....................................................................61
Check This 10.52. Hydronium ion reduction potential as a function of pH. ............................61
Check This 10.55. Predictions you can make about biological redox reactions.......................62
Check This 10.58. Alcohols in Benedict’s reagent. ..................................................................63
Check This 10.60. Reduction potential for copper(II) in Benedict’s reagent. ..........................63
Check This 10.64. Methylene blue, MB+, reduction in basic solution. ....................................63
Check This 10.65. MB+–glucose redox reaction in basic solution. ..........................................63
Check This 10.67. Oxidation of methylene blue by oxygen in the Blue-Bottle reaction. ........63
Check This 10.69. Reduction potentials in a glucose-air fuel cell............................................64
2
ACS Chemistry FROG
Chapter 10
Check This 10.70.
Check This 10.71.
Check This 10.72.
Check This 10.76.
Check This 10.79.
Reduction-Oxidation: Electrochemistry
Intermediate potentials in the biofuel cell. .................................................65
Role of NAD+-NADH in the glucose oxidation pathway. .........................66
Net reaction for oxygen reduction. .............................................................66
Cell potentials in Investigate This 10.73. ...................................................67
K for formation of Cu(en)22+(aq). ...............................................................67
Solutions for Chapter 10 End-of-Chapter Problems..................................................................69
Problem 10.1. .............................................................................................................................69
Problem 10.2. .............................................................................................................................69
Problem 10.3. .............................................................................................................................70
Problem 10.4. .............................................................................................................................70
Problem 10.5. .............................................................................................................................71
Problem 10.6. .............................................................................................................................71
Problem 10.7. .............................................................................................................................72
Problem 10.8. .............................................................................................................................72
Problem 10.9. .............................................................................................................................73
Problem 10.10. ...........................................................................................................................73
Problem 10.11. ...........................................................................................................................74
Problem 10.12. ...........................................................................................................................74
Problem 10.13. ...........................................................................................................................75
Problem 10.14. ...........................................................................................................................75
Problem 10.15. ...........................................................................................................................76
Problem 10.16. ...........................................................................................................................76
Problem 10.17. ...........................................................................................................................77
Problem 10.18. ...........................................................................................................................77
Problem 10.19. ...........................................................................................................................78
Problem 10.20. ...........................................................................................................................78
Problem 10.21. ...........................................................................................................................78
Problem 10.22. ...........................................................................................................................79
Problem 10.23. ...........................................................................................................................79
Problem 10.24. ...........................................................................................................................79
Problem 10.25. ...........................................................................................................................80
Problem 10.26. ...........................................................................................................................81
Problem 10.27. ...........................................................................................................................81
Problem 10.28. ...........................................................................................................................82
Problem 10.29. ...........................................................................................................................82
Problem 10.30. ...........................................................................................................................82
Problem 10.31. ...........................................................................................................................83
Problem 10.32. ...........................................................................................................................83
Problem 10.33. ...........................................................................................................................84
Problem 10.34. ...........................................................................................................................85
Problem 10.35. ...........................................................................................................................85
Problem 10.36. ...........................................................................................................................86
Problem 10.37. ...........................................................................................................................86
Problem 10.38. ...........................................................................................................................86
Problem 10.39. ...........................................................................................................................87
Problem 10.40. ...........................................................................................................................88
ACS Chemistry FROG
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Reduction-Oxidation: Electrochemistry
Chapter 10
Problem 10.41. ...........................................................................................................................88
Problem 10.42. ...........................................................................................................................89
Problem 10.43. ...........................................................................................................................90
Problem 10.44. ...........................................................................................................................90
Problem 10.45. ...........................................................................................................................92
Problem 10.46. ...........................................................................................................................92
Problem 10.47. ...........................................................................................................................93
Problem 10.48. ...........................................................................................................................94
Problem 10.49. ...........................................................................................................................94
Problem 10.50. ...........................................................................................................................95
Problem 10.51. ...........................................................................................................................96
Problem 10.52. ...........................................................................................................................96
Problem 10.53. ...........................................................................................................................97
Problem 10.54. ...........................................................................................................................97
Problem 10.55. ...........................................................................................................................98
Problem 10.56. ...........................................................................................................................99
Problem 10.57. ...........................................................................................................................99
Problem 10.58. .........................................................................................................................100
Problem 10.59. .........................................................................................................................101
Problem 10.60. .........................................................................................................................101
Problem 10.61. .........................................................................................................................102
Problem 10.62. .........................................................................................................................102
Problem 10.63. .........................................................................................................................103
Problem 10.64. .........................................................................................................................104
Problem 10.65. .........................................................................................................................105
Problem 10.66. .........................................................................................................................105
Problem 10.67. .........................................................................................................................107
Problem 10.68. .........................................................................................................................107
Problem 10.69. .........................................................................................................................108
Problem 10.70. .........................................................................................................................109
Problem 10.71. .........................................................................................................................110
Problem 10.72. .........................................................................................................................111
Problem 10.73. .........................................................................................................................113
Problem 10.74. .........................................................................................................................113
Problem 10.75. .........................................................................................................................114
Problem 10.76. .........................................................................................................................115
Problem 10.77. .........................................................................................................................115
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ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry
Materials for Chapter 10 Activities
Activity
Materials
Total Quantity
10.2; 10.7
6-well plate
1
10.2; 10.7
Golf pencils (or broken pencil)
sharpened at both ends
2
10.2
9-V battery
1
10.2
9-V battery snap-on connector
with alligator clips attached to
the wires
1 set
10.2; 10.7
Toothpick or small stirring rod
1
10.7
6-V battery
1
10.7
Wires with alligator clips at
both ends
2
10.14; 10.21; 10.31; 10.38;
10.44; 10.53; 10.73
Digital multimeter with leads
1
10.14; 10.21; 10.38; 10.44;
10.53; 10.73
24-well plate
1
10.14; 10.26; 10.38,
Copper wire (18-22 gauge)
2 cm long
10.14; 10.21; 10.26; 10.38;
10.44; 10.73
Silver wire (18-22 gauge)
2 cm long
10.14; 10.21; 10.31; 10.38;
10.44; 10.73
1  5-cm rectangle of filter
paper
1
10.31
1  8-cm piece of copper sheet. 1
10.21; 10.31,
1  8-cm piece of zinc sheet.
1
10.21; 10.26; 10.44; 10.53;
10.73
Platinum wire (18-22 gauge)
1-2
10.26
Iron wire (18-22 gauge) or thin
strip
2 cm long
10.26
Magnesium wire (18-22
gauge) or thin strip
2 cm long
10.26
Zinc wire (18-22 gauge) or
thin strip
2 cm long
10.56
Test tubes
2
10.61
Large, capped vial
1
ACS Chemistry FROG
5
Reduction-Oxidation: Electrochemistry
Chapter 10
Reagents for Chapter 10 Activities:
Activity
6
Reagents
Total Quantity
10.2
Distilled water
10.2
Magnesium sulfate crystals,
MgSO4·7H2O
1-2 small crystals
10.2
Universal acid-base
indicator
Several drops
10.7; 10.14; 10.38; 10.73
0.10 M aqueous copper
sulfate, CuSO4·5H2O
2.50 g in 100 mL water
10.14; 10.38
0.10 M aqueous silver
nitrate, AgNO3
1.70 g in 100 mL water
10.14;, 10.21; 10.26; 10.31;
10.38; 10.53; 10.73
10% aqueous solution of
potassium nitrate, KNO3
10 g in 90 mL of water
10.21
0.10 M aqueous zinc
sulfate, ZnSO4
2.87 g in 100 mL of water
or dilute 1 M 1:10
10.26; 10.53
1 M hydrochloric acid, HCl
Dilute from stock
10.26
1 M sulfuric acid, H2SO4
Dilute from stock
10.31
1 M aqueous copper sulfate,
CuSO4·5H2O
2.5 g in 10 mL of solution
10.31
1 M aqueous zinc sulfate,
ZnSO4
2.9 g in 10 mL of solution
10.44
pH 3 acid-base buffer
Use commercial solution
10.44; 10.53
pH 7 acid-base buffer
Use commercial solution
10.44; 10.53
Quinhydrone solid.
Few crystals
10.38
0.01 M aqueous silver
nitrate, AgNO3
Dilute from 0.10 M AgNO3
10.38
1.0 M aqueous silver
nitrate, AgNO3
1.7 g in 10.0 mL aqueous
solution
10.53
0.10 M aqueous potassium
ferricyanide, K3Fe(CN)6
3.29 g per 100 mL
10.53
0.10 M aqueous potassium
ferrocyanide, K4Fe(CN)6
3.68 g per 100 mL
10.56
Glucose
Two small crystals
10.56
Benedict's reagent
Use commercial solution
10.56
Ethanol
Two drops
ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry
10.61
0.28 M glucose
5.0 g per 100 mL water
10.61
1.0 M KOH
5.6 g per 100 mL water
10.61
10-3 M methylene blue
0.04 g per 100 mL water
10.73
0.01 M aqueous copper
sulfate, CuSO4·5H2O
Dilute from 0.01 M CuSO4
10.73
Ethylenediamine (1,2diaminoethane),
H2NCH2CH2NH2
Drops
ACS Chemistry FROG
7
Reduction-Oxidation: Electrochemistry
Chapter 10
Introductory Notes
The “storyline” in this chapter involves the reduction-oxidation chemistry that is part of aerobic
metabolism in living organisms. Most of the redox concepts are developed in relatively familiar
chemical systems. The redox reactions of carbon-containing molecules may be less familiar, but
some understanding of them is essential for further study in biochemistry and a good deal of
biology at the molecular/cellular level. Least familiar to chemists are the “curved arrow”
diagrams used extensively in biochemistry to keep track of metabolic intermediates in reaction
pathways without having to write multiple separate chemical reactions to account for the
participation of other species in the reactions. We introduce a variant of such diagrams in Section
10.8 to illustrate coupled reactions. For many students, this might be the most important section
in the chapter. Reduction-oxidation reactions were introduced as one of the three large classes of
reactions in Chapter 6, and this background is built upon in this chapter, beginning with Consider
This 10.1 in the introductory material, which is designed to remind students what they are
supposed to have learned about reduction and oxidation and where to go to review, if necessary.
Consider This 10.1. What atoms are oxidized and reduced in reactions (10.1) and (10.2)?
Goal:
Decide which atoms in reactions (10.1) and (10.2) are oxidized and which reduced.
Classroom options:
 Allow about 3 minutes for students, working in small groups, to answer the questions and
then, have the groups share them with the class.
 This activity could be conducted as an open class discussion to introduce this chapter, if your
students are comfortable possibly exposing what they have forgotten from the introduction to
redox in Chapter 6.
Time for activity:
 5-10 minutes.
Instructor notes:
 Show or refer students to reactions (10.1) and (10.2) as they undertake this activity.
Students should reason and conclude:
(a) The C atoms bonded to the alcohol functional groups are oxidized (from an oxidation
number of +1 to +2 – see textbook pages 404 and 407) and the two Fe3+ ions are reduced
(from an oxidation number of +3 to +2 in Fe2+).
(b) The C atom of the carbonyl group is oxidized (from an oxidation number of +2 to +3)
and the O atom represented by one-half a molecule of oxygen is reduced (from an oxidation
number of 0 to –1).
Follow-up activities:
 Investigate This 10.2. What happens when electric current passes through water?
 Consider This 10.3. What chemical reactions does an electric current cause?
 Consider This 10.4. What happens to the Mg2+(aq) and SO42-(aq)?
 Consider This 10.5. How do you analyze the gases from water electrolysis?
 Check This 10.6. Reactions in an electrophoresis apparatus.
8
ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry
Section 10.1. Electrolysis
Learning Objectives for Section 10.1
 Use evidence from experimental observations to write probable half reactions for the
reductions and oxidations taking place in an electrolytic cell.
 Describe and draw molecular level diagrams of the processes going on and the flow of charge
in an electrochemical cell.
 Use the Faraday and relationships among time, electric current, cell potential, and cell reaction
stoichiometry to calculate the amounts of products from electrolysis or the amount of work
available from the reactants in a galvanic cell.
 Use the known cell reaction to identify the anode and cathode of an electrochemical cell.
Investigate This 10.2. What happens when electric current passes through water?
Goal:
Observe what happens when electrodes attached to the terminals of a battery are immersed in
water containing a dissolved ionic salt and universal acid-base indicator.
Set-up time:
 10-15 minutes.
Time for activity:

5 minutes
Materials:
 6-well plate (alternatively, a small petri dish, small watch glass or 12-well plate can be used).
 Golf pencils sharpened at both ends or any pencil that can be sharpened at both ends. Platinum
or nichrome wire or other inert conductor could also be used for electrodes, but the pencils are
convenient and inexpensive.
 One 9-V battery.
 Alligator clips connected to the wires of a 9-V battery snap-on attachment (can be purchased
at electronic hobby shops such as Radio Shack).
 Toothpick or small stirring rod.
Reagents:
 Distilled water.
 Magnesium sulfate crystals, MgSO4•7H2O.
 Universal acid-base indicator.
Procedure:
 Conduct this as a class investigation with student volunteers and the rest of the class working
in small groups to discuss and analyze the results.
SAFETY NOTE
Wear your safety goggles.

(a) Half fill one well of a 6-well plate with distilled water and add (and dissolve) one or two
crystals of magnesium sulfate, MgSO4·7H2O (Epsom salt) and several drops of universal acidbase indicator. The color of the indicator ranges from red in acidic solutions through green in
neutral solutions to blue in basic solutions. Use an overhead projector to show the well on a
screen and note and record the color of the solution. Use thin carbon rods (pencil lead) as
electrodes. Use wires with alligator clips to connect one end of each electrode to a 9-volt
battery. Record which electrode is connected to the positive and which to the negative
terminal of the battery. Immerse the other ends of the electrodes in the solution and hold them
ACS Chemistry FROG
9
Reduction-Oxidation: Electrochemistry
Chapter 10
about two centimeters apart in the solution for 30-45 seconds and then remove them. Note and
record any evidence for chemical reactions that you see occurring.
 (b) After you remove the electrodes from the solution, stir the solution gently and record any
evidence for chemical reactions that you see occurring.
Anticipated results:
 (a) This activity seems to work best if the ionic concentration is relatively low. The original
color of the solution with indicator is greenish, about pH 6 or so. Short lengths of broken lead
pencils (or “golf pencils” available in office supply stores) sharpened at both ends make good
electrodes since they are sturdy and the wood prevents breaking. Bubbles will be produced at
both electrodes in the solution. The solution color near the cathode (electrode attached to the
negative terminal of the battery) will become blue or purple, indicating that the solution
becomes basic in the region near this electrode. The solution color near the anode (electrode
attached to the positive terminal of the battery) will become yellow or pink, indicating that the
solution becomes acidic in this region. Carry out the electrolysis long enough for these
changes to become apparent and noted by students, but do not prolong the electrolysis. Colors
will begin to appear within a several seconds.
NOTE: It’s likely that it is hydronium ion that is reduced at the cathode to produce bubbles of
hydrogen gas. The solution around the electrode becomes somewhat basic as water molecules
react to maintain the water dissociation equilibrium (Le Chatelier’s principle). The hydronium
ion continues to be reduced and the hydroxide ion concentration builds up. To avoid going
through this chain of reasoning in the textbook, we simply write the net half reaction (10.3) and
say that water is reduced. Oxidation of water occurs at the anode and produces oxygen gas
(bubbles) and hydronium ion in solution, half reaction (10.4).

(b) It is most effective to use a toothpick or other small stirrer to intermix the blue and pink
areas of the solution after electrolysis and watch the color return to the original greenish hue.
Without a control, it looks like the color returns to the original green, but it actually does not.
The electrolysis usually seems to produce a slight excess of hydronium ion over hydroxide, so
their reaction afterwards is not quite stoichiometric (although we assume that it is for the
discussion in the text). If you repeat the electrolysis two or three more times, it becomes quite
evident that the solution moves toward a more yellow (acidic) color. Thus, you should only
repeat the experiment on the same solution once, if the class wishes to observe the reactions
again.
Disposal:
 Wash the solution down the drain with lots of water.
Follow-up discussion:
 Use Consider This 10.3 to initiate discussion of the results of this activity.
Follow-up activities:
 Consider This 10.3. What chemical reactions does an electric current cause?
 Consider This 10.4. What happens to the Mg2+(aq) and SO42-(aq)?
 Consider This 10.5. How do you analyze the gases from water electrolysis?
 Check This 10.6. Reactions in an electrophoresis apparatus.
 End of chapter problems 10.1 and 10.2.
10
ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry
Consider This 10.3. What chemical reactions does an electric current cause?
Goal:
Reasoning from the results in Investigate This 10.2, suggest likely reaction products from the
chemical reactions when an electric current passes through the solution.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to work on this activity and then
have the groups share with the class, summarizing answers on the chalkboard or an overhead
transparency.
 This activity could be conducted as an open class discussion.
Time for activity:
 5-10 minutes.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.2.
Students should reason and conclude:
(a) Small bubbles formed at both electrodes and the color of the solution near each electrode
changed. Formation of a gas and the change in color of the solution at both electrodes means
that a reaction took place at both electrodes. One product is likely to be the hydroxide ion,
OH–(aq), which could be the cause of the blue color of the acid-base universal indicator in
the region near the cathode (electrode connected to the negative terminal of the battery).
Another product is likely to be the hydronium ion, H3O+(aq), which could be the cause of the
pink color of the universal indicator in the region near the anode (electrode connected to the
positive terminal of the battery).
(b) When the pink and blue regions of the solution, produced by passing electric current
through it, are mixed, the solution returns to its original color (the color before the electric
current passed through it). This change could be caused by reaction between the OH–(aq) and
H3O+(aq) [which we postulated in part (a)] to form H2O. These possibilities are internally
consistent and make sense of the observations we have.
Follow-up discussion:
 Use this discussion as a lead-in to further discussion of the electrolysis of H2O, including a
consideration of the identity of the gas(es) produced and thence to equations (10.3) and (10.4).
Follow-up activities:
 Consider This 10.4. What happens to the Mg2+(aq) and SO42-(aq)?
 Consider This 10.5. How do you analyze the gases from water electrolysis?
 Check This 10.6. Reactions in an electrophoresis apparatus.
 End of chapter problems 10.1 and 10.2.
Consider This 10.4. What happens to the Mg2+(aq) and SO42-(aq)?
Goal:
Use reasoning based on avoidance of charge separation to explain in words and pictures what
happens to the Mg2+(aq) and SO42-(aq) in Investigate This 10.2.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to work on this activity and then
have the groups share their pictures with the class on the chalkboard or an overhead
transparency.
 This activity could be conducted as an open class discussion, but loses the impact of seeing
several different representations of what is going on at the molecular level.
ACS Chemistry FROG
11
Reduction-Oxidation: Electrochemistry
Chapter 10
Time for activity:
 15-25 minutes including discussion of the implications of the reasoning demonstrated in the
sketches students make.
Instructor notes:
 Show or refer students to Figure 10.1 as they carry out this activity.
Students should reason and conclude:
(a) The movement of positive and negative ions shown in Figure 10.1 explains how charge
separation could be avoided when water is oxidized and reduced by an electric current. The
positive cations are shown moving toward the negatively charged anode where hydroxide
anions are produced by half reaction (10.3). The negative anions are shown moving toward
the positively charged cathode where hydronium cations are produced by half reaction (10.4).
Thus, charge separation is avoided and the movement of the cations and anions [Mg2+(aq)
and SO42–(aq) in Investigate This 10.2] is responsible for charge flow through the solution.
The ionic compound is needed in this experiment, in order to have a large flow of charge
through the liquid, so reaction occurs rapidly. The flow of charge through pure water, which
contains only a tiny concentration of ions, is so slow that the oxidation and reduction
reactions of water would not be observable in a reasonable time.
(b) A crude sketch that tries to catch the essence of the processes occurring in the electrolysis
of a dilute aqueous magnesium sulfate solution is:
This sketch shows the overall stoichiometry of the electrolysis (discussed in the text just
succeeding this activity), which produces equivalent numbers of hydronium cations, H+(aq),
and hydroxide anions, OH–(aq). Four electrons leave the electrode where water is oxidized
(loses electrons) in the solution and flow in the external circuit to the positive terminal of the
battery. Four electrons from the negative terminal of the battery flow in the external circuit to
the electrode where water is reduced (gains electrons) in the solution. Thus, the number of
electrons produced by oxidation is the same as the number used by reduction in the overall
process. As hydroxide anion builds up around the left-hand electrode in the diagram,
magnesium cations move toward the electrode and the charge around the electrode is
balanced. Similarly, sulfate anions move toward the right-hand electrode and balance the
charge produced by the hydronium cations formed there. This movement of the ions in the
12
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solution completes the electric circuit within the solution. The stoichiometry shown is
required to balance the hydroxide and hydronium ions produced by the electrolysis. The
gaseous products, hydrogen and oxygen gas, are shown in bubbles to indicate their presence
as gases and their stoichiometry is indicated by the number of bubbles of each. This
stoichiometry, captured symbolically in equations (10.5) and (10.7), is the key to the analysis
in Consider This 10.5.
Follow-up discussion:
 Discuss how the chemical reactions observed in Investigate This 10.2 require a flow of
electrons in the electrodes and wires outside the solution and a flow/movement of charged
particles--not electrons--within the solution.
 Emphasize that the charge carriers are not the same in all parts of the circuit represented in
Figure 10.1 and the correct sketches from part (b) of this activity.
 Emphasize the stoichiometry of the overall electrolysis reaction that must produce the same
number of electrons in the oxidation half reaction as are used in the reduction half reaction.
Ideally, this stoichiometry should be part of the sketches from part (b).
Follow-up activities:
 Consider This 10.5. How do you analyze the gases from water electrolysis?
 Check This 10.6. Reactions in an electrophoresis apparatus.
 End of chapter problems 10.1 and 10.2.
Consider This 10.5. How do you analyze the gas from water electrolysis?
Goal:
Give the reasoning, based on the experimental results shown for a simple electrolysis cell and the
stoichiometry of the reaction, for which gas is which as well as which electrode is positive and
which negative.
Classroom options:
 Allow about 3 minutes for students, working in small groups, to complete this activity and
then have the groups share their conclusions with the class.
 This activity could be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, although this loses the immediacy of the analysis based on the previous and ongoing discussion.
Time for activity:
 5-10 minutes that might be an extension of the discussion on Consider This 10.4 and the ideas
that follow it in the text. This is a good activity for assessing the implications and reasoning
demonstrated in the sketches students made in Consider This 10l.4(b).
Instructor notes:
 Sow or refer students to equation (10.7) and the figure of the electrolysis cell from Consider
This 10.5.
 The objective here is for students to observe that the left-hand electrode is producing more gas
than the right-hand electrode. Since twice as much hydrogen as oxygen is produced by water
electrolysis, we infer that hydrogen is produced by the left-hand electrode reaction, that is, the
reduction of water is occurring at this electrode, so it is the negative electrode. Discussion of
this system might lead to a suggestion that the identity of the gases be tested (as we did for the
peroxide decomposition reaction in Chapter 7, Investigate This 7.36). At this point, a standard
Hoffman electrolysis apparatus could be shown or demonstrated or a picture of one, like this
ACS Chemistry FROG
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Reduction-Oxidation: Electrochemistry
Chapter 10
one from Pauling, College Chemistry, 3rd ed. (Freeman, San Francisco, 1964), Fig. 6-2, page
150, could be shown to make it easier to see how the gases could be accessed for testing.
Students should reason and conclude:
The diagram shows that more gas is formed at the left electrode. From equations (10.7) and
(10.9), we see that the ratio of H2(g) to O2(g) for the electrolysis reaction is 2:1. Thus, the
reduction of H in H2O is occurring at the left electrode and the oxidation of O in H2O is
occurring at the right electrode. The left electrode is attached to the negative terminal of the
current source. It donates the electrons for the reduction. The right electrode is connected to
the positive terminal of the current source. It accepts electrons after oxidation occurs. The
signs of the electrodes in the solution are shown in this diagram:
Follow-up discussion:
 Use the discussion as an opportunity to introduce and define anode and cathode (as easier
terms to use than “electrode at which oxidation occurs”, for example.
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Follow-up activities:
 Check This 10.6. Reactions in an electrophoresis apparatus.
 Investigate This 10.7. What other reactions can occur in an electrolytic cell?
 Consider This 10.8. What are the products of electrolysis of a CuSO4 solution?
 End of chapter problems 10.1 and 10.2.
Investigate This 10.7. What other reactions can occur in an electrolytic cell?
Goal:
Observe the results when an aqueous 0.1 M CuSO4 solution is electrolyzed with carbon rod
electrodes.
Set-up time:
 5-10 minutes, assuming that the aqueous 0.1 M CuSO4 solution and wires with alligator clips
at both ends have been prepared previously.
Time for activity:
 10-15 minutes (including discussion).
Materials:
 6-well plate (alternatively, a small Petri dish, small watch glass or 12-well plate can be used).
 Golf pencils sharpened at both ends for electrodes.
 6-V battery. This can be a lantern battery, or you can use a battery holder (available in stores
like Radio Shack) and three or four 1.5-V batteries. (The reaction is slower at 4.5 volts, but
still takes only a few minutes.)
 Wires with alligator clips at both ends.
 Toothpick or small stirring rod.
Reagents:
 0.1 M aqueous CuSO4 solution.
Procedure:
 Conduct this as a class investigation with student volunteers to carry out the procedure and
report on what cannot be observed in a projected image of the reaction system. The rest of the
class works in small groups to discuss and analyze the results.
SAFETY NOTE
Wear your safety goggles.

Fill one well of a 6-well plate about one-third full of 0.1 M aqueous copper sulfate, CuSO4,
solution. This activity can also be done by simply using several drops of the copper sulfate
solution to make a 2-3 cm diameter puddle of liquid on a transparency on the overhead stage
as in the illustration that is shown for Investigate This 10.2. Then hold the electrodes (the
pencils sharpened at both ends, as in Investigate This 10.2) in opposite sides of the puddle.
 Use short pieces of pencils (or golf pencils) sharpened at both ends as electrodes and wires
with alligator clips to connect them to a 6-V battery.
 Record which electrode is attached to the positive and which to the negative terminal of the
battery.
 Immerse the electrodes in the solution and hold them about 2 cm apart for about 45 seconds
(or until a noticeable layer of copper has formed on the cathode) and then remove them.
 Note and record any evidence for chemical reactions that you observe.
Anticipated results:
 However this activity is done, copper should deposit as a “copper-colored” layer on the black
pencil lead (graphite) cathode and some bubbles (of oxygen) should be formed at the anode,
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Reduction-Oxidation: Electrochemistry
Chapter 10
but not the cathode. The copper deposit can be reported by volunteers or shown by video. The
electrode can be passed around the class. Using the pencil sharpened at both ends makes this a
striking result, since the graphite at one end is still black (and can be used for writing) and at
the other end is copper colored. If the electrolysis is carried out for a couple of minutes, the
copper coating is thick enough – and usually strong enough – to make this end of the pencil
useless for writing.
NOTE: Do not use a higher than 6-V volt battery. At higher voltages, the electrolysis is so rapid
that the copper layer is quite fragile and easy to wipe off. Also, confounding the interpretation,
gas will be produced so rapidly at the anode that it causes the pencil lead to disintegrate and
produce lots of dark particles in the solution. This also occurs at the lower voltage, but at a great
deal lower rate, so the particles are not apparent if the electrolysis is carried out only long enough
to observe the formation of bubbles and deposition of a visible layer of copper. An even lower
voltage will work in this investigation, but the reaction becomes slower and may take too much
time for a good class investigation. An alternative to the carbon electrodes would be platinum
wires: the cathode would become copper plated and bubbles would form at the anode with no
problems of disintegration messing up the observations. The copper can easily be removed from
the platinum by rinsing in nitric acid.
Disposal:
Place the used copper solution in a properly labeled waste container in accord with local
regulations. The copper ion can be precipitated as a hydroxide or sulfide and disposed of as a
solid, so as to make the volume of waste smaller.
Follow-up discussion:
 Use Consider This 10.8 to initiate discussion of the results of this activity.
Follow-up activities:
 Check This 10.9. The oxidation reaction in Investigate This 10.7.
 Worked Example 10.10. Stoichiometry of nickel electrodeposition.
 Check This 10.11 – Electrochemical stoichiometry.
 Worked Example 10.12. Electrolytic production of aluminum metal.
 Check This 10.13. Electrolytic purification of copper metal.
 End of chapter problems (on the stoichiometry of electrodeposition reactions) 10.3 through
10.8.
Consider This 10.8. What are the products of electrolysis of a CuSO4 solution?
Goal:
From the results of Investigate This 10.7, draw conclusions about the products of electrolysis of
the CuSO4 solution.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to draw their conclusions and then
have the groups share them with the class.
 This activity could be conducted as an open class discussion.
Time for activity:
 About 10 minutes.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.7.
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Students should reason and conclude:
(a) A copper-colored coating forms on the cathode (electrode attached to the negative
terminal of the battery). We assume that, in a solution containing Cu2+(aq), this is a coating
of copper metal so that Cu2+(aq) has been reduced: Cu2+(aq) + 2e-  Cu(s).
(b) Bubbles appeared at the anode (electrode attached to the negative terminal of the battery).
A gas must be produced by oxidation of some species in the solution. By analogy with the
electrolysis of water, we hypothesize that water is being oxidized at the anode to produce
oxygen gas: 2H2O(l)  O2(g) + 4H2(g) + 4e-.
(c) As Cu2+(aq) is removed at the cathode, more Cu2+(aq) migrates/diffuses toward the
cathode where its concentration is lower. The oxidation of water at the anode produces
hydronium ion in the vicinity of the electrode and sulfate ion, SO42–(aq) migrates toward the
anode to maintain electrical neutrality in the vicinity of the electrode. The movement of the
Cu2+(aq) SO42–(aq) ions are analogous to the migration of the Mg2+(aq) SO42–(aq) ions
shown in the sketch above for Consider This 10.4. In this case, the metal ion is the reactant at
the cathode and little or no hydrogen gas is produced. The reaction at the anode is the same
as the one shown in the sketch. The movement of the ions in the solution carries charge from
one electrode to the other and completes the electric circuit in the solution. Electrons must
flow in the external circuit (from the anode to the battery and from the battery to the cathode)
to complete the circuit and produce the results observed in Investigate This 10.2.
Follow-up discussion:
 Use the discussion of these reactions to introduce electrodeposition and electroplating and
quantification of electrolysis via the mass of metal electrodeposited and its relationship to the
number of moles of electrons that flow in the external circuit.
Follow-up activities:
 Check This 10.9. The oxidation reaction in Investigate This 10.7.
 Worked Example 10.10. Stoichiometry of nickel electrodeposition.
 Check This 10.11. Electrochemical stoichiometry.
 Worked Example 10.12. Electrolytic production of aluminum metal.
 Check This 10.13 Electrolytic purification of copper metal.
 End of chapter problems (on the stoichiometry of electrodeposition reactions) 10.3 through
10.8.
Section 10.2. Electric Current from Chemical Reactions
Learning Objectives for Section 10.2
 Describe and draw molecular level diagrams of the processes going on and the flow of charge
in an electrochemical cell.
 Show how to connect two metal-metal ion half cells to make a galvanic cell and explain the
role of each component of the cell.
 Use the known cell reaction to identify the anode and cathode of an electrochemical cell.
 Translate a physical cell set up to the conventional line notation for cells and vice versa.
Investigate This 10.14. How do you get electricity from a chemical reaction?
Goal:
Discover the requirements for a set up to obtain a current flow from a reduction-oxidation
reaction.
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Reduction-Oxidation: Electrochemistry
Chapter 10
Set-up time:
 15 minutes, assuming the reagent solutions have been prepared.
Time for activity:
 5-15 minutes, including discussion initiated by Consider This 10.15.
Materials:
 Digital multimeter (should be able to read milliamps as well as millivolts) with leads that can
be attached to the wire electrodes. If the readout can be projected or the measurement made
with probes whose output can be projected, the activity is enhanced.
 24-well plate (comparable Petri dishes or small beakers can be used, but the size of the salt
bridge will have to be altered to accommodate different containers).
 Copper wire (18-22 gauge). Before use, clean the copper wire by dipping it in 1 M HCl for
several seconds and then rinsing well with water.
 Silver wire (18-22 gauge). Before use, clean the silver wire with a fine emery cloth or silver
polish.
 1  5-cm rectangle of filter paper.
Reagents:
 0.1 M aqueous silver nitrate, AgNO3.
 0.1 M aqueous copper sulfate, CuSO4·5H2O.
 10% aqueous solution of potassium nitrate, KNO3.
Procedure:
 Conduct this activity as a class investigation with student volunteers to carry out the
procedure and the rest of the class working in small groups to discuss and analyze the results.
SAFETY NOTES
Wear your safety goggles.
Solutions of Ag+(aq) will stain clothing and skin. Wear
plastic gloves when handling these solutions.

(a) Do Investigate This 6.62 (if not done previously) or review the notes and discussion from
that activity.
 (b) Fill one well of a 24-well plate about two-thirds full of 0.10 M aqueous silver nitrate,
AgNO3, solution. Place a silver wire in the solution with one end sticking out of the well. Fill
an adjacent well about two-thirds full of 0.10 M aqueous copper sulfate, CuSO4, solution and
place a copper wire in the solution with one end sticking out of the well. Connect one lead
from a digital multimeter to one of the wires and the other lead to the other wire. Set the
multimeter to read current (amperes – one of the more sensitive scales) and record the current
reading from the meter.
 (c) Roll a 1  5-cm rectangle of filter paper into a tight, 5-cm-long cylinder and fold it into a
square “U” shape. Thoroughly wet the paper with a 10% aqueous solution of potassium
nitrate, KNO3. Place one leg of the “U” in the silver nitrate well and the other leg in the
copper sulfate well and again record the current reading from the meter.
Anticipated results:
 (a) Investigate This 6.62 shows that copper metal reacts with silver ion in aqueous solution to
produce a solution containing copper ion and solid silver metal plating out on the copper
metal. The reverse reaction of silver metal with copper ion does not occur. These results will
be used to help students rationalize why there should be a current detected from a silvercopper cell and to predict the spontaneous direction of the reaction in the cell.
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
(b) Without the salt bridge, the current should be zero or very close to zero, as there is no
connection between the silver and copper wells and no way for current to flow.
 (c) With the salt bridge, there will be a measurable, although not very large current, now that
there is a pathway for ions to move between the silver and copper wells, thus completing the
circuit in the solutions. The exact current reading is unimportant and will depend on the
details of the way the cell is set up. The objective is for students to observe that electrons are
flowing between the electrodes when the salt bridge is included in the set up.
Disposal:
Place the used metal ion solutions in a properly labeled waste container in accord with local
regulations. The silver ion can be precipitated as its chloride and the copper ion as a hydroxide or
sulfide and disposed of as solids, so as to make the volume of waste smaller.
Follow-up discussion:
 Use Consider This 10.15 to initiate discussion of the results of this activity.
Follow-up activities:
 Consider This 10.16. How are reduction half reactions combined?
 Consider This 10.17. What are the net charges on the solutions in Figure 10.4?
 Check This 10.18. How can you detect ion migrations in a salt bridge?
Consider This 10.15. How do you get electricity from a chemical reaction?
Goal:
Reason from the Investigate This 10.14 results what chemistry is going on in the copper-silver
electrochemical cell and what may be happening at the molecular level in the cell.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to work on their analysis of the
results from Investigate This 10.14 and then have the groups share their analyses with the
class.
 This activity could be conducted as an open class discussion.
Time for activity:
 5-10 minutes included as an extension of Investigate This 10.14.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.14. If necessary, review
the results from Investigate This 6.62.
Students should reason and conclude:
(a) No current was detected in Investigate This 10.14(b), but current was observed in part (c).
In part (b), there was no pathway for the ions to flow between the separated solutions, so no
reduction-oxidation reaction could occur in the solutions and, hence, there could be no flow
of electrons in the external part of the circuit where the meter is located. The electrical circuit
could not be completed. When the salt bridge was added in part (c), a pathway was created
for ions to move between the silver and copper ion solutions, redox reactions could occur, the
circuit was complete, and electrons could flow in the external circuit. An electric cell was
created.
(b) The results from Investigate This 6.62 were that Cu(s) was oxidized and Ag+(aq) was
reduced. The reaction that occurred in this investigation was:
Cu(s) + 2Ag+(aq)  Cu2+(aq) + Ag(s)
In Investigate This 6.62, the reduction-oxidation reaction occurred in the same solution. Cu
metal was placed directly into an aqueous solution of Ag+(aq). In Investigate This 10.14,
Cu(s)/ Cu2+(aq) and Ag(s)/Ag+(aq) were in separate wells, so the direct reaction observed in
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Reduction-Oxidation: Electrochemistry
Chapter 10
Investigate This 6.62 could not occur. However, when a salt bridge was added (part (c)), ions
could migrate between the solutions and the reduction-oxidation reaction could occur.
Reaction occurs when the electrodes are connected by the external part of the circuit through
which electrons can flow from the copper electrode (where copper metal is oxidized) through
the meter to the silver electrode (where silver cations are reduced).
Follow-up discussion:
 Use the discussion to introduce the standard electrochemical convention: half reactions
written as reduction reactions, as in equations (10.20) and (10.21).
Follow-up activities:
 Consider This 10.16. How are reduction half reactions combined?
 Consider This 10.17. What are the net charges on the solutions in Figure 10.4?
 Check This 10.18. How can you detect ion migrations in a salt bridge?
Consider This 10.16. How are reduction half reactions combined?
Goal:
Show how to combine two reduction half reactions to give an overall reaction and note the
cancellation of the electrons between the two half reactions.
Time for activity:
 5-10 minutes including discussion of the visualization of the reaction in Figure 10.4.
Classroom options:
 Allow about three minutes for students, working in small groups, to complete this activity and
then have the groups share their results with the class.
 This activity could be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class section.
Instructor notes:
Students should reason and conclude:
(a) To obtain reaction (10.22), two silver ions and silver metal atoms are required, so half
reaction (10.20) has to be multiplied by two. Half-reaction (10.21) shows copper(II) ion as
the reactant, but it is a product in reaction(10.22), so we will have to subtract half reaction
(10.21) from half reaction (10.20) to get the correct overall reaction.
2Ag+(aq) + 2e–  2Ag(s)
–[Cu2+(aq) + 2e–  Cu(s)]
2Ag+(aq) + Cu(s)  2Ag(s) + Cu2+(aq)
(b) The electrons cancel out in the combination of the half-reaction equations in part (a).
There are never free electrons in the overall chemical reaction. In a reduction-oxidation
reaction the number of electrons involved in the reduction must equal the number of
electrons involved in the oxidation. This is the same outcome as found in equation (10.5) in
Section 10.1 where the electrons cancelled out in the combination of two half reactions to
yield an overall reaction.
Follow-up discussion:
 Show or refer students to Figure 10.4, as you use the discussion to lead to this visualization of
the cell set-up in Investigate This 10.14(b) and thence to Consider This 10.17.
Follow-up activities:
 Consider This 10.17. What are the net charges on the solutions in Figure 10.4?
 Check This 10.18. How can you detect ion migrations in a salt bridge?
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Consider This 10.17. What are the net charges on the solutions in Figure 10.4?
Goal:
Determine the net charges on the solutions represented in Figure 10.4.
Classroom options:
 This activity can be conducted as an open class discussion following the analysis in Consider
This 10.16.
 This activity could also be assigned as a homework problem and then discussed at the next
class session.
Time for activity:
 5-10 minutes.
Instructor notes:
 Show or refer students to Figure 10.4 as this activity is conducted.
Students should reason and conclude:
(a) A copper atom from the copper electrode loses two electrons to the copper metal and
enters the solution as Cu2+(aq). Thus, the copper atom is oxidized to Cu2+(aq). The two
electrons travel through the external part of the circuit to the silver electrode. (This is a part
of the process that is not represented by reaction equation (10.22). The electrons do not
appear in the reaction equation, because they are not in the solution.) Now, the silver metal
electrode has two extra electrons that are used to reduce two Ag+(aq) to Ag(s) on the surface
of the electrode.
(b) The silver well in Figure 10.4(a) contains 7 Ag+(aq) and 7 NO3–(aq) ions. The net charge
in the well is 0 (= 7 – 7). The silver well in Figure 10.4(b) contains 5 Ag+(aq) and 7 NO3–(aq)
ions. The net charge in the well is –2 (= 5 – 7). The copper well in Figure 10.4(a) contains 7
Cu2+(aq) and 7 SO42–(aq) ions. The net charge in the well is 0 (= 14 – 14). The copper well in
Figure 10.4(b) contains 8 Cu2+(aq) and 7 SO42–(aq) ions. The net charge in the well is +2
(= 16 – 14). Figure 10.4(a) represents the two wells containing electrically neutral ionic
solutions before any connection is made between them. After the external electrical
connection is made, Figure 10.4(b), the process described in part (a) would add a copper(II)
ion to the copper well and would remove two silver ions from the silver well. This would
give the net charges we found for Figure 10.4(b).
Follow-up discussion:
 Discuss incomplete electrical circuit that has a build up of charges.
 Introduce/discuss why a salt bridge is important.
 Discuss Figure 10.5 (Does electron transfer occur with a salt bridge to complete the circuit?).
Follow-up activities:
 Check This 10.18. How can you detect ion migrations in a salt bridge?
 End of chapter problems 10.9 through 10.16.
Section 10.3. Cell Potentials
Learning Objectives for Section 10.3
 Use the known cell reaction to identify the anode and cathode of an electrochemical cell.
 Use the known sign for a galvanic cell potential to identify the direction of the cell reaction
and the anode and cathode of the cell.
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Investigate This 10.21. What are cell voltages for different half-cell combinations?
Goal:
Measure the cell voltages for different half-cell combinations.
Set-up time:
 10-15 minutes, assuming reagents have been prepared.
Time for activity:
 10-15 minutes (including discussion).
Materials:
 Digital multimeter with leads that can be attached to the wire electrodes. If the readout can be
projected or the measurement made with probes whose output can be projected, the activity is
enhanced.
 Three 1  5-cm rectangle of filter paper for salt bridges.
 24-well plate (comparable Petri dishes or small beakers can be used, but the size of the salt
bridges will have to be altered to accommodate different containers).
 2-cm length of (18-22 gauge) silver wire.
 2-cm piece of (18-22 gauge) copper wire.
 Thin 2-cm strip of zinc.
Reagents:
 0.10 M aqueous silver nitrate, AgNO3
 0.10 M aqueous copper sulfate, CuSO4·5H2O
 0.10 M aqueous zinc sulfate, ZnSO4
 10% aqueous potassium nitrate, KNO3
Procedure:
 Conduct this activity as a class investigation with student volunteers to take the data and
report the results, if projected readings are not possible. The rest of the class should work in
small groups to discuss and analyze the results.
SAFETY NOTES
Wear your safety goggles.
Solutions of Ag+(aq) will stain clothing and skin. Wear
plastic gloves when handling these solutions.

As shown in the diagram, fill one of the central wells of a 24-well plate, about half full of 10%
KNO3 solution. Fill a well adjacent to the KNO3 well about two-thirds full of 0.10 M aqueous
silver nitrate, AgNO3, solution and insert a silver wire as an electrode. Fill two other adjacent
wells about two-thirds full of 0.10 M aqueous zinc sulfate, ZnSO4, and 0.10 M aqueous
copper sulfate, CuSO4, respectively. As electrodes, place a strip of zinc metal in the zinc ion
solution and a copper wire in the copper ion solution.
 Use filter paper soaked in 10% KNO3 solution to make salt bridge
connections between the KNO3 well and each of the other three wells.
This interconnects all three half cells through the central KNO3 well.
NOTE: The directions here are slightly different than in the text. We
suggest that the KNO3 well should be filled only half full and the other
wells about two-thirds full. If any siphoning of solutions occurs through
the salt bridges, it should, therefore, go in the direction of the KNO3 well.
If the siphoning were into the half-cell wells, it would dilute the solutions
and perhaps throw off the results in ways that would be hard to explain.
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
(a) Connect the input leads from a digital voltmeter to the Cu and Zn electrodes. Note the
meter reading, including its units and sign, and which lead is connected to the Cu electrode
and which to the Zn. Record the cell voltage to the nearest millivolt. Reverse the connections
and record the same information as previously.
 (b) Repeat the entire procedure in part (a) for the other two pairs of half cells, Cu with Ag,
and Zn with Ag.
Anticipated results:
 A representative set of results is given in this table:
Red lead
(cathode)
Black lead
(anode)
Voltage,
volts
Zn
Cu
–1.019
Cu
Zn
1.020
Cu
Ag
–0.424
Ag
Cu
0.427
Zn
Ag
–1.455
Ag
Zn
1.442
NOTE: These results are internally consistent, but not all are consistent with standard reduction
potentials. Using standard reduction potentials and the Nernst equation to calculate the voltage
for the Ag/Cu cell with 0.10 M solutions gives E = 0.432 V, which compares very favorably with
the measured value here. A similar calculation for the Ag/Zn cell gives E = 1.532 V, which is
about 0.08 V higher than the measured value. If this is due to a problem with the Zn half cell,
then we would expect the measured voltage for the Cu/Zn cell to be about 0.08 V too low and,
indeed, this is exactly what we observe. The calculated voltage is 1.100 V, which is just 0.08 V
higher than the measured value. Zinc half cells that act ideally are difficult to make, because the
relatively reactive zinc metal often has surface contamination that changes the reduction
potential of the half cell. Use the cleanest and shiniest zinc sheet you can find for the electrode
material in this activity or clean what you use as thoroughly as possible.
Disposal:
 Dispose of the used reagents according to local ordinances in an appropriately labeled
container for metal ion waste. All three metal ions could be precipitated as their sulfide salts
(by addition of a solution of Na2S) to reduce the volume of waste.
Follow-up discussion:
 Use Consider This 10.22 to initiate discussion of the results of this activity.
Follow-up activities:
 Consider This 10.23. What are the cell potentials and reactions in Investigate This 10.21?
 Check This 10.24. Combining cell potentials and interpreting cell reactions.
 End of chapter problems 10.17 through 10.20.
Consider This 10.22. Are cell voltages from different half-cell combinations related?
Goal:
Determine if cell voltages from different half-cell combinations are related.
ACS Chemistry FROG
23
Reduction-Oxidation: Electrochemistry
Chapter 10
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer these questions and then
have groups share their answers with the class using the chalkboard or an overhead
transparency.
 This activity could be conducted as an open class discussion.
Time for activity:
 About 10 minutes.
Instructor notes:
 Try to be sure that the class agrees on the results from Investigate This 10.21 as they begin
this activity.
Students should reason and conclude:
(a) The absolute values for the two readings for a given pair of half cells are essentially
identical; they have the same magnitude, but are opposite in sign. The difference between the
two readings for any pair is that the connections to the multimeter were reversed. Reversing
the connections reverses the sign of the voltage reading.
(b) Any two of the positive voltages measured in Investigate This 10.21 can be combined to
give the third voltage. For example, the Ag/Zn voltage, 1.442 V, minus the Ag/Cu voltage,
0.427 V, is equal to the Cu/Zn voltage, 1.020 V [≈ (1.442 V) – (0.427 V)]. This same pattern
holds for the negative readings.
Follow-up discussion:
 Use this discussion to lead into a discussion of the convention for the sign of cell potentials
and its correlation with the spontaneous reaction in the cell.
Follow-up activities:
 Consider This 10.23. What are the cell potentials and reactions in Investigate This 10.21?
 Check This 10.24. Combining cell potentials and interpreting cell reactions.
 End of chapter problems 10.17 through 10.20.
Consider This 10.23. What are the cell potentials and reactions in Investigate This 10.21?
Goal:
Use the measured cell potentials and conventions for cell potentials to write the overall reactions
for the cells in Investigate This 10.21.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to answer these questions and
then have groups share their answers with the class using the chalkboard or an overhead
transparency.
 This activity can be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class section, either in small groups initially or open class discussion.
Time for activity:
 10-15 minutes to give students a chance to use and reinforce the convention relating the sign
of the cell potential and the direction of cell reactions.
Instructor notes:
 Try to be sure that the class agrees on the results from Investigate This 10.21 and the
conclusions from Consider This 10.22 as they begin this activity.
 Show or refer students to Figure 10.6 as this activity is carried out.
24
ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry
Students should reason and conclude:
(a) In Investigate This 10.21, when the red (cathode) lead was connected to the copper
electrode and the black (anode) lead to the zinc electrode, the measured voltage was positive.
This result is consistent with the illustration in Figure 10.6, which shows the same
connections and a positive voltage reading on the meter. The result indicates that Zn metal is
being oxidized at the anode and Cu2+(aq) is being reduced at the cathode. The experimental
voltage, 1.020 V, is relatively close to +1.10 V. [See the note accompanying the data in
Investigate This 10.21 for the rationale as to why these values are not closer.]
(b) The cell represented in Check This 10.20 shows the zinc electrode as the anode and the
copper electrode as the cathode with the half cell connected by a salt bridge, “||”. This is
exactly the set up shown in Figure 10.6. The cell reaction for both the cell in Figure 10.6 and
the zinc-copper cell in Investigate This 10.21 [see part (a)] is:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
(c) The data from Investigate This 10.21 give the cell potential for the copper-silver cell as
0.43 V and the cell potential for the zinc-silver cell as 1.44 V. In the copper-silver cell, the
copper electrode is the anode, so copper metal is being oxidized, silver ions are being
reduced, and the cell reaction is:
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
In the zinc-silver cell, the zinc electrode is the anode, so zinc metal is being oxidized, silver
ions are being reduced, and the cell reaction is:
Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s)
Follow-up discussion:
 Use this discussion as a lead in to show how cell potentials and redox reactions can be
combined to get new reactions and their reduction potentials. Remind students about the
patterns they found in Consider This 10.22, which they will revisit in Check This 10.24, from
the perspective of equation (10.32).
Follow-up activities:
 Check This 10.24. Combining cell potentials and interpreting cell reactions.
 End of chapter problems 10.17 through 10.20.
Section 10.4. Half-Cell Potentials: Reduction Potentials
Learning Objectives for Section 10.4
 Use the known cell reaction to identify the anode and cathode of an electrochemical cell.
 Use the known sign for a galvanic cell potential to identify the direction of the cell reaction
and the anode and cathode of the cell.
 Determine an unknown cell potential for a cell reaction by combining cell reactions with
known cell potentials to give the desired cell reaction and its cell potential.
 Use a table of standard reduction potentials to predict the direction of any redox reaction (for
which data are given) and determine its standard cell potential.
Investigate This 10.26. How do metals react with acids?
Goal:
Determine which metals react with acids by observing the metal in the acid solution.
ACS Chemistry FROG
25
Reduction-Oxidation: Electrochemistry
Chapter 10
Set-up time:
 10-15 minutes, assuming solutions have been prepared.
Time for activity:
 10-15 minutes, including discussion starting with Consider This 10.27.
Materials:
 24-well plate.
Reagents:
 1 M hydrochloric acid, HCl.
 1 M sulfuric acid, H2SO4.
 Ten metal samples (as 2-cm lengths of wire or thin strips), two each of Ag, Cu, Fe, Mg, and
Zn.
Procedure:
 Conduct this activity as a class investigation with student volunteers and have the other
students work in small groups to discuss and analyze the results. These reactions produce gas
bubbles that are readily observed when projected via an overhead projector.
SAFETY NOTE
Wear your safety goggles.

Fill each of five wells in one row of a 24-well plastic plate about two-thirds full of 1 M
hydrochloric acid, HCl.
 Repeat with 1 M sulfuric acid, H2SO4, in five wells of another row.
 Place a different one of the five metal wires or strips in each HCl well. Record which metal is
in which well.
 Observe what happens and note any indication of reaction between a metal and the acid.
 Repeat with the five metals in H2SO4.
Anticipated results:
 Bubbles form when Fe, Mg, and Zn metals are placed in contact with the acids. Sometimes
the bubbles form slowly on Zn, so students have to be alert. Neither acid shows any indication
of reaction with Ag and Cu. In the cases where bubbles form, the solid metal piece becomes
smaller. This is most evident with Mg, which reacts most vigorously.
Disposal:
 The Fe, Zn, and Mg solutions can be disposed down the drain. The used Ag and Cu solutions
should be disposed in a properly labeled waste container or converted to solid sulfide salts,
and disposed of as solids. If possible, recycle the metal pieces.
Follow-up discussion:
 Use Consider This 10.27 to initiate discussion of the results of this activity.
Follow-up activities:
 Consider This 10.28. What are the signs of reduction potentials for other metals?
 Worked Example 10.29. The aluminum-air cell.
 Check This 10.30. The lead-acid battery.
Consider This 10.27. What are the reactions of metals with acids?
Goal:
Write the reactions for those metals that reacted with acids in Investigate This 10.26.
26
ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to analyze the results and write their
equations and then have the groups share them with the class on the chalkboard or an
overhead transparency.
 This activity could be conducted as an open class discussion.
Time for activity:
 About 10 minutes as a part of Investigate This 10.26.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.26 as this activity
begins.
Students should reason and conclude:
(a) The evidence for reactions between the metals and the acids is the observation of bubbles
forming in the Mg, Zn, and Fe wells with both acids. The other two metals, Ag and Cu, did
not produce bubbles in either acid. The Mg, Zn, and Fe react the same way with acids, to
produce a gas, although Mg reacted more vigorously in both acids. Since both acids produced
the same results, we conclude that it is the common species in both acids, the hydronium ion,
that is responsible for the reaction with metals. Some metals react with hydronium ion and
others do not.
(b) Hydrogen, H2(g), is likely to be the gas that is produced by the reaction of a metal with
hydronium ion. This is a reduction reaction, so the metal must be oxidized. The pieces of
metal that reacted became smaller during the reaction, which is consistent with their being
oxidized and going into solution as their cations. The balanced reactions that fit this model
are:
Mg(s) + 2H+(aq)  Mg2+(aq) + H2(g)
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
2Fe(s) + 6H+(aq)  2Fe3+(aq) + 3H2(g)
Follow-up discussion:
 Relate these results to the direction of the reactions with respect to the reaction in the standard
hydrogen electrode (SHE). Discuss the sign of the cell potentials, if the SHE is assigned as the
anode of the cells set up with these reactive metals and their cations as the other half cell.
Follow-up activities:
 Consider This 10.28. What are the signs of reduction potentials for other metals?
 Worked Example 10.29. The aluminum-air cell.
 Check This 10.30. The lead-acid battery.
 End of chapter problems 10.21 through 10.33.
Section 10.5. Work from Electrochemical Cells: Free Energy
Learning Objectives for Section 10.5
 Use the known cell reaction to identify the anode and cathode of an electrochemical cell.
 Use the Faraday and relationships among time, electric current, cell potential, and cell reaction
stoichiometry to calculate the amounts of products from electrolysis or the amount of work
available from the reactants in a galvanic cell.
 Use the known sign for a galvanic cell potential to identify the direction of the cell reaction
and the anode and cathode of the cell.
ACS Chemistry FROG
27
Reduction-Oxidation: Electrochemistry
Chapter 10

Determine an unknown cell potential for a cell reaction by combining cell reactions with
known cell potentials to give the desired cell reaction and its cell potential.
 Determine free energy change for a cell reaction from cell potential and visa versa.
Investigate This 10.31. How can we get work from a galvanic cell?
NOTE: This activity can’t work as written. Although small motors that will operate on about one
volt are available, most require currents on the order of several tens of milliamps. A cell (or even
series of cells) like the one described in this activity has such a high internal resistance (from the
tiny salt bridge) that a current of a few milliamps is all it can provide. The zinc-copper Daniell
(or gravity) cell illustrated in end-of-chapter problem 10.74 avoided this problem by having no
salt bridge; there is direct contact between the two solutions that are separated by their density
difference. It’s probably not worthwhile to construct such a cell or use another cell set up that has
a smaller internal resistance unless you already have one available. You might wish to do the
activity with a standard commercial battery. For example, a 1.581 V alkaline AAA battery used
to power two different small motors provided 0.130 amp for one motor and 0.079 amp for the
other.
Goal:
Students investigate how we can get work from a galvanic cell.
Set-up time:
 15 minutes, assuming that solutions have been prepared previously.
Time for activity:
 10-15 minutes including discussion and analysis in Consider This 10.32.
Materials:
 Digital multimeter (should be able to read milliamps as well as millivolts) with leads that can
be attached to the wire electrodes. If the readout can be projected or the measurement made
with probes whose output can be projected, the activity is enhanced.
 6-well plate or two small Petri dishes.
 1  8-cm piece of clean copper sheet.
 1  8-cm piece of clean zinc sheet.
NOTE: Use copper and zinc sheets thick enough to stay bent into shape. You want electrodes
that extend across the diameter of the wells and will stay flat in the well when you connect clip
leads to them. The easiest way to accomplish this is to bend the electrodes into a shape that will
fit over the edge of the well and provide a place to attach the clip lead:
Nothing this elaborate is required for part (a), but will be useful for part (b). It is sometimes
difficult to find a motor that will run on the voltage supplied by a copper-zinc cell. If so, you can
increase the voltage using a copper-magnesium cell (use 1 M MgSO4 solution and about 25 cm
of magnesium ribbon doubled back on itself twice for the electrode), a copper-silver cell (use
1 M AgNO3 solution and silver wire or thin sheet, if available, for the electrode), or a silvermagnesium cell for the largest voltage. Alternatively, you can make two cells and connect them
in series to obtain enough voltage. If you do this, measure the voltage of the combination in part
(a), since you will need the total voltage of the cell for the later calculation of the work available
from the cell or cells in Check This 10.34.
 2  5-cm rectangle of filter paper for salt bridge.
28
ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry

Small electric motor with an attached fan blade. (Little motors that run on 1-2 volt are usually
available from model and novelty stores. The fan blade can be made of cardboard or it could
be a disk with radial stripes so that motion in obvious).
Reagents:
 1 M aqueous copper sulfate, CuSO4·5H2O, solution.
 1 M aqueous zinc sulfate, ZnSO4, solution.
Procedure:
 Conduct this as a class investigation with student volunteers to carry out the procedure and
provide meter readings for the class (if they are not projected), with the rest of the class
working in small groups to discuss and analyze the results.
SAFETY NOTE
Wear your safety goggles.

(a) Fill one well of a six-well plate about half full of aqueous 1.0 M zinc nitrate, ZnSO4,
solution. Bend a 1  8-cm piece of zinc sheet into an “L” (or the shape shown above), so that
the longer leg can rest flat on the bottom of the well with the shorter leg up and out of the
well. Place the sheet in the well with the ZnSO4 solution. Repeat in an adjacent well, using a
solution of aqueous 1.0 M copper nitrate, CuSO4, solution and a 1  8-cm piece of copper
sheet. Finish making a galvanic cell by connecting the half cells with a filter-paper salt bridge
that has been soaked in 10% aqueous potassium nitrate, KNO3 solution. Connect the leads
from a multimeter to the copper and zinc electrodes. Note the electrode to which each lead is
attached. Turn the multimeter to its voltage scale and record the reading, including its sign and
units, to the nearest millivolt. Reverse the connections from the multimeter and again record
the reading, including its sign and units.
 (b) Disconnect one lead from the multimeter and then connect a small electric motor with an
attached fan blade between this lead and the electrode, as shown schematically in Figure
10.10. Turn the multimeter to a current scale and record the current reading in the circuit when
the fan is going. Usually the multimeter will have to be set on an appropriate current scale
before the motor will run. The current will probably be in the range from 100 to 200 milliamp.
Anticipated results:
 (a) The voltage of a copper-zinc cell with 1 M solutions of the cations will be near 1.1 V. A
representative voltage in such one set up was 1.083 V, with the anode lead connected to the
zinc electrode.
 (b) Will not work, see the note above.
NOTE: The blue color of the copper(II) solution may be seen to fade a bit, if the cell is run for
long enough to reduce a substantial amount of the cation. (Simply connect the electrodes with an
external wire to short the cell and observe this result.) If this happens, you can refer to this
phenomenon in later discussions of batteries running down.
Disposal:
 Discard used copper solution in a properly labeled waste container. The used zinc solution can
be disposed down the drain with plenty of water.
Follow-up discussion:
 Use Consider This 10.32 to initiate discussion of the results of this activity.
Follow-up activities:
 Worked Example 10.33. Electrical work.
 Check This 10.34. Electrical work.
ACS Chemistry FROG
29
Reduction-Oxidation: Electrochemistry




Chapter 10
Consider This 10.35. Why are there so many sizes of batteries?
Worked Example 10.36. Amount of work available from an electrochemical cell
Check This 10.37. Amount of work available from an electrochemical cell
End of chapter problems 10.34 through 10.39
Consider This 10.32. What reaction in the zinc-copper galvanic cell produces work?
NOTE: Although the cell in Investigate This 10.31 cannot be used to run a motor, the copperzinc cell has been introduced and extensively discussed previously in the chapter, so the
questions can be answered from what is already known, in particular Figure 10.6. The
differences between the cell shown in Figure 10.6 and the one in Figure 10.10 are the
concentrations of the metal ion solutions. Since, in both cases, the concentrations of the two
solutions are the same, the cell potentials are the same, although the argument for this conclusion
requires use of the Nernst equation, which is introduced in the next section.
Goal:
Based on the results from Investigate This 10.31, determine the reaction in the zinc-copper
galvanic cell that produces work.
Classroom options:
 Allow about 3 minutes for students, working in small groups, to complete their analysis and
then have groups share their analyses with the class.
 This activity could be conducted as an open class discussion.
Time for activity:
 About 10 minutes integrated with Investigate This 10.31.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.31 as this activity is
done.
Students should reason and conclude:
(a) The cell can do useful work in the surroundings. The fan rotates, showing that the electric
motor, powered by the cell, is able to move objects.
(b) The voltage reading on the meter is positive (no sign shown) when the anode lead (black)
is connected to the zinc electrode. Therefore, the Zn/Zn2+ half cell is the anode and this is
where oxidation occurs. The Cu/Cu2+ half cell is the cathode and this is where reduction
occurs.
(c) Based on the reasoning in part (b), the reduction half reaction for copper proceeds as
written in equation (10.21), but the half reaction for zinc is oxidation, so equation (10.43) has
to be reversed. that is subtracted from equation (10.21) to give the overall reaction in the cell:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
[Also, the blue solution in the Cu/ Cu2+(aq) well may have been observed to fade to a lighter
blue. This observation shows that the Cu2+(aq) is being used up, and is more evidence that
the correct overall equation involves reduction of theCu2+(aq) to Cu(s.]
Follow-up discussion:
 Use this discussion as a lead-in to definitions of electrical potential difference, volts, voltage
and electrical work.
Follow-up activities:
 Worked Example 10.33. Electrical work.
 Check This 10.34. Electrical work.
 Consider This 10.35. Why are there so many sizes of batteries?
30
ACS Chemistry FROG
Chapter 10



Reduction-Oxidation: Electrochemistry
Worked Example 10.36. Amount of work available from an electrochemical cell
Check This 10.37. Amount of work available from an electrochemical cell
End of chapter problems 10.34 through 10.39
Consider This 10.35. Why are there so many sizes of batteries?
Goal:
Conclude that different size 1.5-V batteries will produce different amounts of work because they
have different amounts of reactants.
Classroom options:
 This activity could be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, but may lose some impact.
Time for activity:
 5-10 minutes depending on how quickly the class comes to understand the limitation that
amount of reactant places on the total amount of work that can be produced.
Instructor notes:
 Discussion of electromotive force (emf) and the equation for the electrical work available per
mole of reaction should lead naturally to this activity.
Students should reason and conclude:
The amount of work available from a battery (cell) is a function of the cell potential, but this
factor is the same, 1.5 V, for all the batteries under consideration. The other factor is nF, the
number of moles of electrons per mole of cell reaction. The cell reaction is the same in all
these batteries, but the number of moles of the reaction that are available is limited in each
battery by the amount of reactants it contains. The larger the battery, the more moles of
reactants there are and, hence, the more moles of reaction and more work available.
Follow-up activities:
 Worked Example 10.36. Amount of work available from an electrochemical cell
 Check This 10.37. Amount of work available from an electrochemical cell
 End of chapter problems 10.34 through 10.39
Section 10.6. Concentration Dependence of Cell Potentials: The Nernst
Equation
Learning Objectives for Section 10.6:
 Determine an unknown cell potential for a cell reaction by combining cell reactions with
known cell potentials to give the desired cell reaction and its cell potential.
 Use the known cell reaction to identify the anode and cathode of an electrochemical cell.
 Use a table of standard reduction potentials to predict the direction of any redox reaction (for
which data are given) and determine its standard cell potential.
 Describe how an electrode senses the redox half reaction in a half cell in which the reduced
and oxidized species are both present as dissolved ions and/or molecules.
 Determine free energy change for a cell reaction from cell potential and vice versa.
 Apply Le Chatelier’s principle to predict the direction of change of cell potentials as
concentrations in the half cells are changed.
 Use the Nernst equation, which relates the cell potential, the standard cell potential, and the
reaction quotient for the cell reaction, to determine any one of these quantities, if the other
two are known.
ACS Chemistry FROG
31
Reduction-Oxidation: Electrochemistry
Chapter 10

Use the standard cell potential to determine the equilibrium constant for a cell reaction and
visa versa.
 Use the Nernst equation to convert standard reduction potentials to reduction potentials under
non-standard conditions and vice versa.
Investigate This 10.38. Does the silver-copper cell potential depend on [Ag+]?
Goal:
Measure the silver-copper cell potential for different [Ag+(aq)].
Set-up time:
 10-15 minutes, assuming reagents have been prepared.
Time for activity:
 10-20 minutes, incorporating discussion in Consider This 10.39.
Materials:
 Digital multimeter with leads that can be attached to the wire electrodes. If the readout can be
projected or the measurement made with probes whose output can be projected, the activity is
enhanced.
 Four 1  5-cm rectangle of filter paper for salt bridges.
 24-well plate (comparable Petri dishes or small beakers can be used).
 2-cm piece of clean silver wire. Clean metal electrodes will make the measurements as close
as possible to the “book” values in this simple set up.
 2-cm piece of clean copper wire.
Reagents:
 1.00 M aqueous silver nitrate, AgNO3, solution.
 0.100 M aqueous silver nitrate, AgNO3, solution.
 0.0100 M aqueous silver nitrate, AgNO3, solution.
NOTE: The silver ion solutions for this activity should be carefully prepared shortly before use,
in order to observe the linear dependence of the cell emf on the logarithm of the silver ion
concentration. Make a 1.00 M solution using a volumetric flask; twenty-five milliliters of this
stock solution will be enough. Dilute aliquots (pipet) of this solution by a factor of 10 and 100 in
volumetric flasks.
 10% aqueous potassium nitrate solution
 0.10 M copper sulfate, CuSO4. The linearity of the potentials does not depend on the
concentration of this solution, but agreement with the values calculated from standard
reduction potentials does. Students may question experimental results that are not reasonably
in agreement with such calculated values.
Procedure:
 Conduct this as a class investigation with volunteers to carry out the procedure and report the
results, if they are not projected. The class should work in small groups to discuss and analyze
the results.
SAFETY NOTES
Wear your safety goggles.
Solutions of Ag+(aq) will stain clothing and skin. Wear
plastic gloves when handling these solutions.
32
ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry

Measurements should be taken quickly (without undue rush) and as soon as feasible after
adding the salt bridges, to prevent significant dilution by siphoning of solutions from one well
to another.
 As shown in the diagram, fill one of the central wells of a 24-well
plate about half full of 10% KNO3 solution.
 Fill three other wells adjacent to the KNO3 well about two-thirds
full of 1.00 M, 0.100 M, and 0.0100 M aqueous silver nitrate,
AgNO3, solutions, respectively.
 Place a clean silver wire in the 0.0100 M silver solution as an
electrode.
 Fill a fourth adjacent well about two-thirds full of 0.10 M
aqueous copper sulfate, CuSO4, solution, and place a clean
copper wire in the solution as an electrode.
 Use filter paper soaked in 10% KNO3 solution to make salt bridge connections between the
KNO3 well and each of the other wells.
 (a) Connect the input leads from a digital voltmeter to the electrodes so that the meter reading
is positive. Note which lead is connected to the silver electrode and which to the copper.
Record the cell potential to the nearest millivolt.
 (b) Remove the silver electrode from the 0.0100 M silver solution, wipe it dry with a paper
towel, place it in the 0.100 M silver solution, and measure the new cell potential.
 (c) Remove the silver electrode from the 0.100 M silver solution, wipe it dry with a paper
towel, place it in the 1.00 M silver solution, and measure the new cell potential.
Anticipated results:
 For all the cells, the potential readings are positive when the anode lead (black) is connected
to the copper electrode, that is, copper metal is oxidized to copper(II) cation in the cell
reactions. The calculated results for ideal cells and the experimental results for a
representative set of measurements are:

[Ag+(aq)]
log[Ag+(aq)]
E (calc), V
E (expt’l), V
0.0100
–2.00
0.373
0.365
0.100
–1.00
0.432
0.416
1.00
0.00
0.492
0.500
–0.301
0.474
0.457
The experimental values are all lower than the calculated values, which suggests a systematic
error in the experiment. Incorrect solution concentrations (too high for [Cu2+(aq)] or too low
for [Ag+(aq)]) or non-ideal cell conditions or electrodes are all possible problems. Care needs
to be exercised in making up the solutions, cleaning the electrodes, and preparing the cells.
The two higher concentration cells give potentials that are 16 and 17 mV too low. If these
results are used to calculate E for this system, the result is 16 mV too low. If all three
experimental values are used, about the same result is obtained (see Check This 10.42),
although the data do not obey the Nernst equation quite as well.
ACS Chemistry FROG
33
Reduction-Oxidation: Electrochemistry
Chapter 10
Disposal:
 The used Ag and Cu solutions should be disposed in a properly labeled waste container or
converted to solid sulfide salts, and disposed of as solids. The KNO3 solution can be washed
down the drain.
Follow-up discussion:
 Use Consider This 10.39 to initiate discussion of the results of this activity.
Follow-up activities:
 Check This 10.40. Nernst equation at 25 C (298 K).
 Worked Example 10.41. Nernst equation applied to a copper-cadmium cell.
 Check This 10.42. Nernst equation applied to silver-copper cells.
 Check This 10.43. Three problems from the Web Companion.
 Investigate This 10.44. Does the silver-quinhydrone cell potential depend on pH?
 Consider This 10.45. How does the silver-quinhydrone cell potential depend on pH?
 End of chapter problems 10.40 and 10.41.
Consider This 10.39. How does the silver-copper cell potential depend [Ag+]?
Goal:
Based on the observations from Investigate This 10.38, conclude that the silver-copper cell
potential dependence on [Ag+(aq)] is consistent with Le Chatelier’s principle.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to make their predictions and then
share them and their reasoning with the class.
 This activity could be conducted as an open class discussion.
Time for activity:
 About 10 minutes as an extension of Investigate This 10.38.
Instructor notes:
 Try to sure the class agrees on the results from Investigate This 10.38 as this activity is done.
Students should reason and conclude:
(a) The copper electrode in each cell is the anode. A representative set of results for a slightly
different set of concentrations is given in the table under “Anticipated results” in Investigate
This 10.38 above.
(b) Since copper metal is oxidized in the cells, the cell reaction is:
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
Increasing the concentration of silver ion is a disturbance to the system and Le Chatelier's
principle says that the system will react by trying to reduce its concentration, that is, by going
in the direction written. The prediction is that the greater driving force to go in the direction
written will result in a higher electrical potential for the cell. As the concentration of the
silver ion increases, the cell potential should also increase.
(c) The results from Investigate This 10.38 are consistent with the prediction in part (b).
Follow-up discussion:
 Use the discussion of these results to lead into further discussion of the reaction quotient, Q
and the Nernst equation.
Follow-up activities:
 Check This 10.40. Nernst equation at 25 C (298 K).
 Worked Example 10.41. Nernst equation applied to a copper-cadmium cell.
 Check This 10.42. Nernst equation applied to silver-copper cells.
34
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Chapter 10




Reduction-Oxidation: Electrochemistry
Check This 10.43. Three problems from the Web Companion.
Investigate This 10.44. Does the silver-quinhydrone cell potential depend on pH?
Consider This 10.45. How does the silver-quinhydrone cell potential depend on pH?
End of chapter problems 10.40 and 10.41.
Investigate This 10.44. Does the silver-quinhydrone cell potential depend on pH?
Goal:
Measure the silver-quinhydrone cell potential with solutions of different pH in the quinhydrone
half cell.
Set-up time:
 10-15 minutes, assuming reagents are previously prepared.
Time for activity:
 15-20 minutes, including discussion incorporating Consider This 10.45.
Materials:
 Digital multimeter with leads that can be attached to the wire electrodes. If the readout can be
projected or the measurement made with probes whose output can be projected, the activity is
enhanced.
 Three 1  5-cm rectangle of filter paper for salt bridges.
 24-well plate (comparable Petri dishes or small beakers can be used).
 2-cm piece of clean silver wire. Clean metal electrodes will make the measurements as close
as possible to the “book” values in this simple set up.
 2-cm piece of clean platinum wire.
Reagents:
 0.100 M silver nitrate, AgNO3. The cell potentials will have the expected sign, even if this
solution is not exactly this molarity, but in order to come close to calculated values, the
solution should be made with care and the concentration accurately known.
 pH 3.00 acid-base buffer solution
 pH 7.00 acid-base buffer solution
 10% aqueous potassium nitrate solution
 Quinhydrone solid.
Procedure:
 Conduct this as a class investigation with student volunteers to carry out the procedure and
report the voltages, if they cannot be projected. The class should work in small groups to
discuss and analyze the results.
SAFETY NOTES
Wear your safety goggles.
Solutions of Ag+(aq) will stain clothing and skin. Wear
plastic gloves when handling these solutions.

(a) Fill one of the central wells of a 24-well plate, as shown in the diagram, about half full of
10% KNO3 solution. Fill an adjacent well about two-thirds full of 0.100 M aqueous silver
nitrate, AgNO3, solution and insert a silver wire as an electrode. Fill two other adjacent wells
about two-thirds full of pH 7.00 and pH 3.00 acid-base buffer solutions, respectively. To each
buffer well add a tiny amount of solid quinhydrone. Place a platinum wire in the pH 7.00 well
as an electrode and to stir the solution.
ACS Chemistry FROG
35
Reduction-Oxidation: Electrochemistry
Chapter 10

Use filter paper soaked in 10% KNO3 solution to make salt bridge
connections between the KNO3 well and each of the other wells. This
set up interconnects all three half cells through the central KNO3 well.
Connect the input leads from a digital voltmeter to the electrodes so
that the meter reading is positive. Note which lead is connected to the
silver electrode and which to the platinum. Record the cell potential to
the nearest millivolt.
 (b) Remove the platinum wire from the pH 7.00 quinhydrone well,
wipe it off, and place it in the pH 3.00 quinhydrone well. Stir the
solution in the well and again connect the digital voltmeter. Note which
lead is connected to which electrode and record the cell potential.
Anticipated results:
 A set of results obtained by one instructor is given in this table:
pH
anode
E (calc), V
E(expt’l), V
3.00
platinum
0.218
0.183
7.00
platinum
0.454
0.367

Clearly, there is a problem with these experimental results. Although we do not expect exact
agreement with the values calculated from the reduction potentials in Appendix C of the text,
we expect better agreement and better internal consistency than here. Before carrying out this
activity in the classroom, check that your reagents give results that are consistent with the
theoretical values.
Disposal:
 The used Ag solution should be disposed in a properly labeled waste container or converted to
the solid sulfide or chloride salt, and disposed of as solid metal waste. The buffer solutions
can be washed down the drain.
Follow-up discussion:
 Use Consider This 10.45 to initiate discussion of the results of this activity.
Follow-up activities:
 Check This 10.46. Deriving equation (10.74).
 Check This 10.47. pH and cell potentials for silver-quinhydrone cells.
 Worked Example 10.48. Equilibrium constant for reaction of silver ion with hydroquinone.
 Check This 10.49. Conditions for developing photographic film.
 End of Chapter problems 10.42 through 10.53.
Consider This 10.45. How does the silver-quinhydrone cell potential depend on pH?
Goal:
Analyze the results from Investigate This 10.44 to show that the silver-quinhydrone cell potential
depends on pH as would be predicted by Le Chatelier’s principle.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to answer the questions and
make their predictions and then share them and their reasoning with the class.
 This activity could be conducted as an open class discussion, or some parts could be assigned
as homework and discussed at the next class session.
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Time for activity:
 About 10 minutes as an extension of Investigate This 10.44.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.44 as this activity is
done.
Students should reason and conclude:
(a) Because pH 5.00 is midway between pH 3.00 and 7.00, we might expect the cell potential
for a cell with a quinhydrone half cell prepared with pH 5.00 buffer to be midway between
the potentials of the cells measured in Investigate This 10.44 at pH 3.00 and 7.00. A linear
dependence on pH can be rationalized, because the Nernst equation shows that the cell
potential is proportional to the logarithm of the concentrations in the cells and the pH is the
logarithm of a concentration. Thus, it would not be surprising if the cell potential were
directly proportional to the pH. If this is true, the experimental values in the table in
“Anticipated results” for Investigate This 10.44 above predict that a cell with a quinhydrone
half cell prepared with a pH 5.00 buffer would have a cell potential of 0.275 V {= [(0.183 V)
+ (0.367 V)]/2}. The ideal value is0.336 V {= [(0.218 V) + (0.454 V)]/2}.
(b) When the platinum wire electrode is connected to the anode lead of the multimeter, the
voltage reading is positive for both cells we made. The platinum wire in the quinhydrone half
cell is the anode for both the pH 3.00 and the pH 7.00 cells, so it would also be the anode if
the half cell were prepared with a pH 5.00 buffer.
(c) Since the platinum electrode is the anode, oxidation is occurring in the quinhydrone half
cell; hydroquinone is being oxidized to quinone, the reverse of reaction equation (10.70). The
line notation and cell reaction for this cell are:
Pt | c M QH2(aq); c M Q(aq); pH 3, 5 or 7 || 0.100 M Ag+(aq) | Ag(s)
2Ag+(aq) + QH2(aq)  2Ag(s) + Q(aq) + 2H+(aq)
Note that the concentrations of hydroquinone and quinone are shown as the same in the cell
notation, because reaction equation (10.69) shows that quinhydrone dissolves to give equal
numbers of each molecule.
(d) Changing the concentration of hydronium ion in the equilibrium represented in part (c) is
a disturbance to which the system reacts by minimizing the disturbance. An increase in pH is
a decrease in [H+(aq)]. If [H+(aq)] is decreased, the system responds by trying to make more
of it, that is, by proceeding in the direction shown in part (c). This driving force to go in the
direction written will increase the cell potential and this is what is observed experimentally;
the cell potential increased with pH. Le Chatelier's principle does explain the direction of the
variation of the cell potential with pH.
Follow-up discussion:
 Use this discussion to lead into a quantitative analysis of the dependence of the silverquinhydrone cell potential on pH.
Follow-up activities:
 Check This 10.46. Deriving equation (10.74).
 Check This 10.47. pH and cell potentials for silver-quinhydrone cells.
 Worked Example 10.48. Equilibrium constant for reaction of silver ion with hydroquinone.
 Check This 10.49. Conditions for developing photographic film.
 End of Chapter problems 10.42 through 10.53.
ACS Chemistry FROG
37
Reduction-Oxidation: Electrochemistry
Chapter 10
Section 10.7. Reduction Potentials and the Nernst Equation
Learning Objectives for Section 10.7:
 Determine an unknown cell potential for a cell reaction by combining cell reactions with
known cell potentials to give the desired cell reaction and its cell potential.
 Apply Le Chatelier’s principle to predict the direction of change of cell potentials as
concentrations in the half cells are changed.
 Determine free energy change for a cell reaction from cell potential and vice versa.
 Use the standard cell potential to determine the equilibrium constant for a cell reaction and
vice versa.
 Use a table of standard reduction potentials to predict the direction of any redox reaction (for
which data are given) and determine its standard cell potential.
 Use the Nernst equation, which relates the cell potential, the standard cell potential, and the
reaction quotient for the cell reaction, to determine any one of these quantities, if the other
two are known.
 Use the Nernst equation to convert standard reduction potentials to reduction potentials under
non-standard conditions and vice versa.
 Use a table of standard reduction potentials at pH 7 to predict the direction and standard cell
potential at pH 7 for redox reactions in biological systems.
Investigate This 10.53. Can pH change the direction of a cell reaction?
Goal:
Measure the cell potential and its sign when the pH is changed in one half cell of a cell.
Set-up time:
 10-15 minutes, assuming the reagents are prepared previously.
Time for activity:
 15-20 minutes, including discussion based on Consider This 10.54.
Materials:
 Digital multimeter with leads that can be attached to the wire electrodes. If the readout can be
projected or the measurement made with probes whose output can be projected, the activity is
enhanced.
 Three 1  5-cm rectangles of filter paper for salt bridges.
 24-well plate (comparable Petri dishes or small beakers can be used).
 Two 2-cm pieces of platinum wire for electrodes. One will remain in the Fe(CN)63–-Fe(CN)64–
half cell and the other will be in quinhydrone half cell being measured.
Reagents:
 0.10 M aqueous potassium ferricyanide, K3Fe(CN)6
 0.10 M aqueous potassium ferrocyanide K4Fe(CN)6
 10% aqueous potassium nitrate, KNO3
 Quinhydrone
 1 M aqueous hydrochloric acid, HCl
 pH 7 buffer
Procedure:
 Conduct this as a class investigation with volunteers and work in small groups to discuss and
analyze the results.
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Chapter 10
Reduction-Oxidation: Electrochemistry
SAFETY NOTE
Wear your safety goggles.

Use three wells grouped around one containing 10% aqueous KNO3 solution, as diagrammed
in Investigate This 10.44, to prepare three half cells. Prepare the wells with (1) a solution that
is 0.10 M in aqueous potassium ferricyanide, K3Fe(CN)6, and 0.10 M in aqueous potassium
ferrocyanide, K2Fe(CN)6, with a platinum wire electrode, (2) a quinhydrone solution in 1 M
HCl (pH  0) with a platinum wire electrode, and (3) a quinhydrone solution in a pH = 7
buffer solution with a platinum wire electrode. Use filter paper salt bridges soaked in 10%
aqueous KNO3 solution to connect each half cell to the KNO3 well.
 Connect one lead of a digital voltmeter to the electrode in the Fe(CN)63–-Fe(CN)64– solution
and the other lead to the electrode in the pH  0 quinhydrone half cell. Record which electrode
is the anode and the cell potential to the nearest millivolt.
 Switch the lead (or move the electrode, after wiping it off) from the pH  0 quinhydrone half
cell to the pH = 7 quinhydrone half cell and repeat the recording.
Anticipated results:
 The results of representative measurements are:
Quinhydrone
half cell pH
Anode connection
Cell
potential, V
≈0
Fe(CN)63–Fe(CN)64–
0.192
7
Fe(CN)63–Fe(CN)64–
–0.135

The platinum wire electrode in the quinhydrone half cell is the actual anode of the cell when
the solution in the quinhydrone half cell is pH 7.
 These values are not the “theoretical” values from standard reduction potentials (although they
are in the right direction), but the ferri-ferrocyanide system is far from ideal. The difference
between the two potentials should be about 0.42 V. The experimental difference, 0.33 V, is
not terrific agreement, but the objective here is to see the change in the sign of the potential as
the pH changes, and this is accomplished.
Disposal:
 Dispose of the used ferri-ferrocyanide solution in an appropriately labeled waste container
according to local ordinances. The other solutions can be rinsed down the drain with lots of
water.
Follow-up discussion:
 Use Consider This 10.54 to initiate discussion of the results of this activity.
Follow-up activities:
 Check This 10.55. Predictions you can make about biological redox reactions.
 End of chapter problem 10.54 through 10.61.
Consider This 10.54. How does pH change the direction of a cell reaction?
Goal:
Conclude that a pH change in one half cell can change the direction of a cell reaction.
ACS Chemistry FROG
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Reduction-Oxidation: Electrochemistry
Chapter 10
Classroom options:
 This activity can be conducted as an open class discussion.
 Allow 3-5 minutes for students, working in small groups, to answer these questions. Then,
the instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency.
Time for activity:
 About 10 minutes as an extension of Investigate This 10.53.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.58.
Students should reason and conclude:
(a) The cell notation and cell reactions for the two cells constructed in Investigate This 10.53
are
Pt(s) | 0.10 M Fe(CN)63–(aq); 0.10 M Fe(CN)64–(aq) || c M QH2(aq); c M Q(aq); pH 0 | Pt(s)
2Fe(CN)64–(aq) + Q(aq) + 2H+(aq)  2Fe(CN)63–(aq) + QH2(aq)
Pt(s) | c M QH2(aq); c M Q(aq); pH 7 || 0.10 M Fe(CN)63–(aq); 0.10 M Fe(CN)64–(aq) | Pt(s)
2Fe(CN)63–(aq) + QH2(aq)  2Fe(CN)64–(aq) + Q(aq) + 2H+(aq)
(b) The ferri-ferrocyanide half cell is the anode in the cell with the pH ≈ 0 quinhydrone halfcell solution. The cell reaction is the ferrocyanide ion, Fe(CN)64–(aq), reducing the quinone
to hydroquinone, which requires the presence of hydronium ion, H+(aq), in the solution with
hydroquinone. Decreasing the [H+(aq)] (increasing the pH of the quinhydrone solution) is a
disturbance to the system, which responds by going in the direction to try to increase the
[H+(aq)], that is, the reverse of the reaction in the quinhydrone half-cell solution at pH ≈ 0. If
the hydronium concentration is decreased enough (pH is increased enough), the reaction
might be expected to become spontaneous in the reverse direction, that is, hydroquinone
reducing the ferricyanide ion, Fe(CN)63–(aq), to ferrocyanide ion, Fe(CN)64–(aq). This is the
direction of the cell reaction observed with the quinhydrone half-cell solution at pH ≈ 7, for
which the quinhydrone electrode is the anode.
Follow-up discussion:
 Use the discussion as an introduction to reduction potentials under biological conditions
(usually near pH 7), E°´, and Check This 10.55.
Follow-up activities:
 Check This 10.55. Predictions you can make about biological redox reactions.
 End of chapter problem 10.54 through 10.61.
Section 10.8. Coupled Redox Reactions
Learning Objectives for Section 10.8
 Determine an unknown cell potential for a cell reaction by combining cell reactions with
known cell potentials to give the desired cell reaction and its cell potential.
 Use a table of standard reduction potentials to predict the direction of any redox reaction (for
which data are given) and determine its standard cell potential.
 Determine free energy change for a cell reaction from a cell potential and vice versa.
 Apply Le Chatelier’s principle to predict the direction of change of cell potentials as
concentrations in the half cells are changed [Sections 10.6, 10.7, 10.8, and 10.10].
40
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Chapter 10







Reduction-Oxidation: Electrochemistry
Use the Nernst equation, which relates the cell potential, the standard cell potential, and the
reaction quotient for the cell reaction, to determine any one of these quantities, if the other
two are known.
Use the standard cell potential to determine the equilibrium constant for a cell reaction and
vice versa.
Use the Nernst equation to convert standard reduction potentials to reduction potentials under
non-standard conditions and vice versa.
Use a table of standard reduction potentials at pH 7 to predict the direction and standard cell
potential at pH 7 for redox reactions in biological systems.
Combine free energy changes for redox reactions, complexation equilibria, and solubility
equilibria to get the free energy change and reduction potential for the net redox reactions that
occur in systems with such interrelated reactions.
Convert a stepwise series of reactions to a diagram that shows the coupling of reactions and
vice versa.
Use standard reduction potentials to determine the probable sequence of a series of coupled
redox reactions and vice versa.
Investigate This 10.56. Does copper ion react with sugars?
Goal:
Observe that a blue solution containing Cu(II) cation reacts with glucose, but not ethanol, to
form a muddy brown opaque suspension upon heating.
Set-up time:
 10-15 minutes, assuming reagents are prepared previously.
Time for activity:
 15-20 minutes including discussion based on Consider This 10.57.
Materials:
 2 small test tubes.
 Boiling water bath. A beaker of water on a hot plate is completely adequate.
Reagents:
 10 mL Benedict's reagent. Benedict’s reagent (for qualitative determination) is available from
chemical suppliers and is inexpensive. It is hardly worth the trouble, but the reagent can be
made by dissolving 17.3 g sodium citrate and 10.0 g sodium carbonate in about 80 mL of
water with heating and stirring. Filter, if the solution is not clear, and bring the volume to 85
mL. Dissolve 1.73 g of copper sulfate pentahydrate in 10 mL of water and add this solution to
the citrate-carbonate solution slowly with constant stirring. Bring the final volume to 100 mL.
 2 drops ethanol.
 1 or 2 small crystals of glucose. Different sugars (used in this activity were galactose, glucose,
lactose, maltose, and sucrose).
Procedure:
 Conduct as a class investigation with student volunteers and work in small groups to discuss
and analyze the results.
SAFETY NOTE
Wear your safety goggles.
Add 5 mL of Benedict’s reagent to each of two small, clean, labeled test tubes. To one of the
test tubes, add 2 drops of ethanol.
 To the other test tube add one or two small crystals of glucose.

ACS Chemistry FROG
41
Reduction-Oxidation: Electrochemistry
Chapter 10

Swirl to mix and dissolve the alcohol and sugars. Note the color and appearance of each
mixture.
 Place the test tubes in a boiling water bath for about five minutes. Note the color and
appearance of the mixtures after heating.
NOTE: In an earlier version of this activity, which some instructors may continue to use,
additional reducing and non-reducing sugars were also included. Among these were galactose,
lactose, maltose, and sucrose. If more sugars are included, you must be prepared to go beyond
the material in the present text to explain reducing and non-reducing sugars. At this point in the
text “storyline,” this is irrelevant information and is a side issue that is a distraction from the
central idea of the aldehyde group as a reducing agent, which is why the activity has been
stripped to its bare essentials.
Anticipated results:
 The mixture of ethanol and Benedict’s reagent is clear blue (no change in the reagent color)
and remains so upon heating.
 The mixture of glucose and Benedict’s reagent starts out clear blue. If left at room temperature
for about five minutes, the solution turns green. When placed in the boiling water bath, the
mixture turns opaque and reddish-brown (indicating the formation of solid copper(I) oxide,
Cu2O).
Disposal:
 Dispose of the used copper ion solutions in an appropriately labeled waste container or
precipitate the copper ions as insoluble sulfide salts and dispose of as solid metal ion waste.
Follow-up discussion:
 Use Consider This 10.57 to initiate discussion of the results of this activity.
Follow-up activities:
 Check This 10.58. Alcohols in Benedict's reagent
 Worked Example 10.59. Standard reduction potential for copper (II) in Benedict's solution.
 Check This 10.60. Reduction potential for copper(II) in Benedict's reagent.
 Investigate This 10.61. Can you characterize the Blue-Bottle reaction?
 End of chapter problem 10.62.
Consider This 10.57. What is the reaction of copper ion with glucose?
Goal:
Conclude, based on the changes observed in Investigate This 10.56, that copper(II) is reduced to
copper(I) in a reaction with glucose (but not ethanol).
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer these questions and then
share their conclusions with the class.
 This activity could be conducted as an open class discussion.
Time for activity:
 About 10 minutes as an extension of Investigate This 10.56.
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.56 as this activity is
done.
Students should reason and conclude:
42
ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry
(a) The Benedict’s reagent solution containing copper(II) cation and glucose changed from a
clear blue to a cloudy red-brown precipitate upon heating. The Benedict’s reagent solution
containing copper(II) cation and ethanol did not change from a clear blue, even upon heating.
(b) The appearance of the red-brown precipitate indicates that Cu2O(s) formed in a reaction
between the blue citrate complex of Cu2+(aq) and glucose. Cu2+ is reduced to Cu+. Since this
change did not occur when ethanol was tested with the reagent, the functional group in the
glucose molecule responsible for the reduction is probably not an alcohol group.
Follow-up discussion:
 Use the discussion in this activity to introduce the oxidation of the aldehyde group in glucose
to a carboxylic acid group. Go on to consider the oxidizing Cu2+–Cu+ redox system.
Follow-up activities:
 Check This 10.58. Alcohols in Benedict's reagent
 Worked Example 10.59. Standard reduction potential for copper (II) in Benedict's solution.
 Check This 10.60. Reduction potential for copper(II) in Benedict's reagent.
 Investigate This 10.61. Can you characterize the Blue-Bottle reaction?
 End of chapter problem 10.62.
Investigate This 10.61. Can you characterize the Blue-Bottle reaction?
Goal:
Characterize the Blue-Bottle reaction in terms of the patterns of behavior observed after shaking
the reaction mixture and variations on this treatment.
Set-up time:
 20-30 minutes. The Blue-Bottle mixture needs to be made and parceled out among the
reaction vessels as near the time they will be used as possible. The time required here assumes
that the reagent solutions have been previously prepared.
Time for activity:
 15-20 minutes.
Materials:
 Enough large, capped vials so each group will have one.
Reagents:
 0.28 M glucose solution.
 1.0 M KOH.
 10-3 M solution of methylene blue: 0.04 g in 100 mL water.
NOTE: Mix equal portions of the glucose and hydroxide solutions for the Blue-Bottle reaction
mixture. Prepare enough of the mixture to be able to fill the reaction vials/containers about onethird to one-half full. Add to the mixture 1 mL of the methylene blue solution per 100 mL of the
mixture and mix thoroughly. On standing, the solution should become clear and colorless.
Occasionally it may appear very slightly on the yellowish side. Dispense for group use in capped
containers that can be shaken without the danger of spilling or leakage. Paper towels should be
provided in case there are any spills. The caps have to be removable, so students can observe the
effect of admitting fresh air to the container before shaking. After a few hours, the solution will
begin to turn yellow. It will still work, but not as reproducibly, and the color change is not as
dramatic. Glucose reacts in basic solution to form the colored product(s).
Procedure:
 Students should work in small groups on this investigation and discuss their results and
hypotheses with the entire class as a part of Consider This 10.62..
ACS Chemistry FROG
43
Reduction-Oxidation: Electrochemistry
Chapter 10
SAFETY NOTES
Wear your safety goggles.
Solutions of hydroxide are caustic and can harm skin and
clothing. Take care not to spill any of the solution, but
work over a paper towel to catch any drops that may
escape.

Shake the vial two or three times. What, if any, changes do you observe? Watch to see
whether anything further happens.
 When the solution has returned to its original appearance, shake the vial again and see if your
observations are reproducible.
 Try variations. Uncap the vial and recap it before shaking. Shake the vial more than two or
three times.
 Keep a record of the conditions and observations for all your trials.
NOTE: No directions are given here or in the text for timing the disappearance of the blue color,
because that need should arise from the students as part of their group discussion and analysis.
Anticipated Results:
 After shaking, the solution is blue, but within about a minute or less, the solution should
suddenly turn colorless. Repeated shakings and exposure to air (oxygen) will cause the
solution again to turn blue and then fade suddenly to colorless. The more times the solution is
shaken, the longer the blue color persists.
Disposal:
 Discard the used mixtures down the drain with copious amounts of water.
Follow-up discussion:
 Use Consider This 10.62 to initiate class discussion of the outcomes of this activity from each
group.
Follow-up activities:
 Consider This 10.63. Can you explain the timing of the Blue-Bottle reaction?
 Check This 10.64. Methylene blue, MB+, reduction in basic solution.
 Check This 10.65. MB+–glucose redox reaction in basic solution.
 Worked Example 10.66. Oxidation of methylene blue by oxygen.
 Check This 10.67. Oxidation of methylene blue by oxygen in the Blue-Bottle reaction
 Consider This 10.68. What sort of redox coupling occurs in biological systems?
 End of chapter problems 10.63 through 10.66 and 10.74 through 10.77.
Consider This 10.62. Can you interpret the Blue-Bottle reaction?
Goal:
Attempt an interpretation of the pattern of observations on the Blue-Bottle reaction from
Investigate This 10.61, given the redox and corresponding color properties of methylene blue.
Classroom options:
 This is the time for groups to share their observations and the patterns they have detected in
the Blue-Bottle reaction from Investigate This 10.61. Lists on the chalkboard or overhead
transparencies will facilitate the discussion as the class grapples together with the
interpretation of these data.
Time for activity:
 15-20 minutes, which may include beginning Consider This 10.63.
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Chapter 10
Reduction-Oxidation: Electrochemistry
Instructor notes:
 Try to get the class to reach a consensus about the factors that affect the reaction and the
direction and magnitude of the effects.
Students should reason and conclude:
(a) The blue color is essentially the same and persists for about the same length of time (for
any particular group of students) if the amount of shaking is the same. The time required for
the blue color to disappear increases as the amount of shaking increases, although the depth
of the blue color appears to be the same for any amount of shaking. Timing is more
consistent if the reaction container is opened between trials. The disappearance of the blue
color is sudden and dramatic. [This is a classic clock reaction in which the disappearance of a
reactant—oxygen in this case—is signaled by a sudden change in color of the reaction
mixture.]
(b) Since the only blue species in the mixture is the oxidized form of methylene blue, MB+, it
must be MB+ that is formed when the solution is shaken. Some species in the reaction
mixture must reduce the methylene blue to its colorless form, MBH. We have already learned
that glucose can reduce copper(II) ion in basic solution, so it is likely that glucose is also the
reducing agent in this system as well. What species might be responsible for oxidizing the
MBH back to MB+ when the solution is shaken? Perhaps molecular oxygen, dissolved from
the air when the solution is shaken, is this oxidizing agent. This hypothesis is consistent with
the blue color persisting longer with more shaking. If more oxygen is dissolved, more MB+
can be formed and the blue color could last longer. The things this model does not explain
are the sudden disappearance of the blue color, rather than a slow fading over time, and why
the blue color should be about the same whether the mixture is shaken a few or many times.
We are faced with an incomplete model that needs further refinement.
Follow-up discussion:
 Use this discussion and analysis to lead to the questions in Consider This 10.63 and to the
timing and constant blue color puzzles.
Follow-up activities:
 Consider This 10.63. Can you explain the timing of the Blue-Bottle reaction?
 Check This 10.64. Methylene blue, MB+, reduction in basic solution.
 Check This 10.65. MB+–glucose redox reaction in basic solution.
 Worked Example 10.66. Oxidation of methylene blue by oxygen.
 Check This 10.67. Oxidation of methylene blue by oxygen in the Blue-Bottle reaction
 Consider This 10.68. What sort of redox coupling occurs in biological systems?
 End of chapter problems 10.63 through 10.66 and 10.74 through 10.77.
Consider This 10.63. Can you explain the timing of the Blue-Bottle reaction?
Goal:
Attempt an explanation for the relationship of the persistence of color to amount of shaking in
the Blue-Bottle reaction and for the sudden disappearance of the color that is the same under all
conditions.
Classroom options:
 It is probably best to couple this activity with the discussion immediately following Consider
This 10.62, perhaps by allowing 3-5 minutes for students, working in their groups, to answer
the questions and then share their answers with the class. perhaps using the chalkboard or an
overhead transparency to illustrate their reasoning.
 This activity could be conducted as an open class discussion.
ACS Chemistry FROG
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Reduction-Oxidation: Electrochemistry
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Time for activity:
 15-20 minutes, depending on how long it takes the class to be comfortable with the reaction
model.
Instructor notes:
 Try to be sure the class has reached a consensus on the patterns observed in Investigate This
10.61 and the factors they need to account for to explain the timing of the Blue-Bottle
reaction.
Students should reason and conclude:
(a) The blue color must persist in the solution as long as there is molecular oxygen present to
produce MB+ from MBH. The color should fade away as glucose reduces MB+. To remain
blue, the oxidation must be faster than the reduction. When the oxygen has been used up, the
solution quickly fades to colorless.
(b) Shaking mixes the solution with air and allows more oxygen to dissolve into the solution.
There is very little methylene blue in solution, so it does not take much oxygen for the rapid
oxidation to return MBH to MB+.
(c) In this reaction system, the methylene blue never gets used up as it simply cycles between
its oxidized and reduced forms. Each shaking adds only a tiny amount of oxygen to the
reaction system, so only a tiny amount of glucose gets used up to reduce the MB+ produced
by the reaction of MBH with oxygen. Since there is a substantial concentration of glucose in
the reaction mixture, only a small amount of it gets used up by the relatively small amount of
oxygen that can be added, even with many repetitions of the shaking. Thus, the system is
able to repeat the cycle over and over.
(d) The more oxygen that dissolves, the longer the cycle of oxidation and reduction of
methylene blue can go on before the oxygen is used up. Hence, the blue color persists longer.
When the oxygen concentration gets low enough, it can’t oxidize the MBH fast enough to
compete with the MB+ reduction by glucose and the blue solution color fades very rapidly.
Since the other concentrations don’t vary appreciably, the low concentration of oxygen that
triggers the fading of the color is the same under all conditions in this activity, so the rapid
fading is essentially the same in all trials.
Follow-up discussion:
 Carry on this discussion until you feel that the class is comfortable with the model that
explains the Blue-Bottle reaction. Remind them that rates of reaction are the subject of the
next chapter, so the Blue-Bottle reaction model is an introduction to this topic. The cycling of
methylene blue in this system provides a lead-in to discussion of coupling of redox reactions
and the use of the curved arrow diagrams exemplified for this simple case by Figure 10.12.
Follow-up activities:
 Check This 10.64. Methylene blue, MB+, reduction in basic solution.
 Check This 10.65. MB+–glucose redox reaction in basic solution.
 Worked Example 10.66. Oxidation of methylene blue by oxygen.
 Check This 10.67. Oxidation of methylene blue by oxygen in the Blue-Bottle reaction
 Consider This 10.68. What sort of redox coupling occurs in biological systems?
 End of chapter problems 10.63 through 10.66 and 10.74 through 10.77.
46
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Consider This 10.68. What sort of redox coupling occurs in biological systems?
Goal:
Show that the coupling species cancel out when the reactions they couple are combined to give
the net reaction and that the coupling species show up in a cycle in the curved-arrow diagram of
the reactions.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to find the net reaction and answer
these questions and then share their conclusions with the class on the chalkboard on overhead
transparency.
 This activity could be conducted as an open class discussion or in a recitation session.
Time for activity:
 About 10 minutes.
Instructor notes:
 Try to be sure, before starting the activity, that students understand that the net reaction must
convert all the carbons in glucose to carbon dioxide and ethanol.
Students should reason and conclude:
(a) In order to combine—add—reactions (10.99) and (10.100) to give the net reaction for
conversion of glucose to carbon dioxide and ethanol, reaction (10.100) must be multiplied by
two to account for the two pyruvate ions formed in reaction (10.99):
C6H12O6 + 2NAD+  2C3H3O3– + 2NADH + 4H+
(10.99)
2C3H3O3– + 2NADH + 4H+  2CO2 + 2CH3CH2OH + 2NAD+
2  (10.100)
C6H12O6  2CO2 + 2CH3CH2OH
NAD + and NADH do not appear in the net reaction, because, for every NADH produced in
reaction (10.99), an NADH is used up in reaction (10.100). The net result is that the
nicotinamide dinucleotide cycles between NADH and NAD+ in the two reactions.
(b) The coupling species is the nicotinamide dinucleotide that cycles between its reduced and
oxidized forms, NADH and NAD+, as shown in this coupling diagram:
2ADP + 2Pi + 2H+
C6H12 O6
2NAD+
2C3H3O2 – + 4H+
2NADH
2ATP + 2H2 O
2CO2 + 2CH3CH2OH
2C3H3O3 –+ 4H+
Follow-up discussion:
 Use the discussion in this activity to introduce the requirement that the coupling species must
have a reduction potential intermediate between the reduction potentials of the two reactions it
couples coupled reactions and go on to show how this plays out in a glucose biofuel cell.
Follow-up activities:
 Check This 10.69. Reduction potentials in a glucose-air fuel cell.
 Check This 10.70. Intermediate potentials in the biofuel cell.
 Check This 10.71. Role of NAD+–NADH in the glucose oxidation pathway.
ACS Chemistry FROG
47
Reduction-Oxidation: Electrochemistry


Chapter 10
Check This 10.72. Net reaction for oxygen reduction.
End of chapter problems 10.63 through 10.66 and 10.74 through 10.77.
Section 10.12. Extension — Cell Potentials and Non-Redox Equilibria
Investigate This 10.73. Does addition of ethylenediamine affect ECu2+, Cu?
Goal:
Find that the cell potential for a copper(II) ion concentration cell is a function of the difference in
concentration of free copper(II) cation between the two half cells. Addition of ethylenediamine
to one half cell changes the cell potential.
Set-up time:
 10-15 minutes, assuming reagents are prepared previously.
Time for activity:
 15-20 minutes including discussion based on Consider This 10.74.
Materials:
 Digital multimeter with leads that can be attached to the wire electrodes. If the measurement
can be made with probes whose output can be projected, the activity is enhanced.
 Three 1  5-cm rectangles of filter paper for salt bridges.
 24-well plate (comparable Petri dishes or small beakers can be used).
 Three 2-cm pieces of copper wire.
Reagents:
 0.1 M aqueous copper sulfate, CuSO4.
 0.01 M aqueous copper sulfate, CuSO4.
 Ethylenediamine (1,2-diaminoethane), H2NCH2CH2NH2.
 10% aqueous potassium nitrate solution
Procedure:
 Conduct this as a class investigation with student volunteers to carry out the activity and
report the cell potentials, if they are not projected, and work in small groups to discuss and
analyze the results.
SAFETY NOTE
Wear your safety goggles.

(a) Use a 24-well plate to set up a central well about half full of 10% KNO3 solution and half
cells in three adjacent wells, each with a salt bridge to the central well. For two of the half
cells, fill the wells about two-thirds full of aqueous 0.10 M copper sulfate, CuSO4, solution.
Prepare the third half cell with aqueous 0.010 M CuSO4 solution. Place a clean copper wire in
each well as an electrode. Use a digital multimeter to read the cell potential to the nearest
millivolt and determine which electrode is the anode for the three cells you can make by
combining pairs of half cells.
 (b) To one of the wells containing 0.10 M CuSO4 solution, add one drop of ethylenediamine
(1,2-diaminoethane), H2NCH2CH2NH2 (= en), and stir the mixture with the electrode to make
sure it is uniform. Record your observations. Use the digital multimeter to read the cell
potentials and determine which electrode is the anode for the two cells you can make
connecting the half cell with added ethylenediamine to each of the other half cells.
48
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Chapter 10
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Anticipated results:
 One incomplete set of results is [Values in parentheses are approximately what we would
expect on the basis of ideal systems. The experimental results are not very close to what we
would expect them to be theoretically, although in the right direction.]:

Anode connection
Cathode connection
E, V
0.10 M Cu2+
0.10 M Cu2+
(≈ 0)
0.010 M Cu2+
0.10 M Cu2+
0.050 (≈ 0.030)
0.10 M Cu2+ + en
0.010 M Cu2+
0.110 (≈ 0.48)
0.10 M Cu2+ + en
0.10 M Cu2+
(≈ 0.51)
These results suggest that the half cell with the lower concentration of free (uncomplexed)
copper(II) ion is the anode of these concentration cells.
NOTE: The cell potential for the cell with ethylenediamine in one half cell is a sensitive function
of the exact conditions in the half cell, so the expected values in the table are approximations
based on an equilibrium constant for formation of Cu(en)22+ of about 1019. Minor variations in
the concentrations can result in values that differ by tenths of volts, usually lower than shown.
Disposal:
 Dispose of CuSO4 solutions in a properly labeled waste container or precipitate the metal ion
as its sulfide or hydroxide and dispose as solid metal ion waste according to local ordinances.
Follow-up discussion:
 Use Consider This 10.74 to initiate discussion of the results of this activity.
Follow-up activities:
 Consider This 10.75. Do the copper concentration cells behave as expected?
 Check This 10.76. Cell potentials in Investigate This 10.73.
 Consider This 10.77. How does addition of ethylenediamine affect ECu2+,Cu?
 Worked Example 10.78. K for formation of Ag(NH3)2+(aq).
 Check This 10.79. K for formation of Cu(en)22+(aq).
 End of chapter problem 10.70.
Consider This 10.74. Why does addition of ethylenediamine affect ECu2+,Cu ?
Goal:
Conclude that the results from Investigate This 10.73 show that the anode of a copper(II)
concentration cell is the half cell with the lower copper(II) concentration and, hence, that the
addition of ethylenediamine lowers the free copper(II) concentration.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to reach their conclusions and then
share them with the class.
 This activity could be assigned as homework and discussed the next day in class, although
some of the immediacy is lost.
Time for activity:
 About 10 minutes as an extension of Investigate This 10.73.
ACS Chemistry FROG
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Reduction-Oxidation: Electrochemistry
Chapter 10
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.73 as they work on this
activity.
Students should reason and conclude:
(a) Connecting two half cells with the same copper(II) ion concentration produces a potential
of about zero volts. This is what might be expected, because, when the ion concentrations are
equal, there is no driving force for this reaction between half cells anode and cathode:
Cu(s, anode) + Cu2+(aq, cathode)  Cu(s, cathode) + Cu2+(aq, anode)
If the concentration of copper(II) ion in the anode half cell is lower than that in the cathode
half cell, this represents a disturbance to the system of equal concentrations. Le Chatelier’s
principle tells us that the cell will respond by increasing the lower concentration at the
expense of the higher. That is, the favored direction of the reaction above will be as it is
written, which means that the lower concentration half cell will be the anode in the cell, just
as we observed in Investigate This 10.73.
(b) The (Cu2+) = 0.10 half cell to which ethylenediamine is added is the anode of cells
produced by coupling it to either the other (Cu2+) = 0.10 half cell or the (Cu2+) = 0.010 half
cell. The cell potentials for these two cells are a good deal higher than that for the one
between a (Cu2+) = 0.10 half cell and a (Cu2+) = 0.010 half cell, indicating that the
concentration of free copper(II) cation is a good deal smaller (hence, a larger driving force)
than 0.010 M. It seems likely that some reaction between copper ion and ethylenediamine
effectively removes the metal ion from the solution. The information from Chapter 6, Section
6.6, suggests that formation of a complex between copper(II) cation and one or more
ethylenediamine molecules may be responsible.
Follow-up discussion:
 Use this discussion to introduce concentration cells and the concept that it is the concentration
of the “free” metal ion species that determines the cell potential. This phenomenon provides a
way to study how other species interact with and change the metal ion concentration.
Follow-up activities:
 Consider This 10.75. Do the copper concentration cells behave as expected?
 Check This 10.76. Cell potentials in Investigate This 10.73.
 Consider This 10.77. How does addition of ethylenediamine affect ECu2+,Cu?
 Worked Example 10.78. K for formation of Ag(NH3)2+(aq).
 Check This 10.79. K for formation of Cu(en)22+(aq).
 End of chapter problem 10.70.
Consider This 10.75. Do the copper concentration cells behave as expected?
Goal:
Students analyze the results from Investigate This 10.73 to determine if the copper concentration
cells behave as expected.
Classroom options:
 This activity could be conducted as an open class discussion.
 The activity could also be assigned as a homework problem and then discussed at the next
class section.
Time for activity:
 5-10 minutes in class.
50
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Chapter 10
Reduction-Oxidation: Electrochemistry
Instructor notes:
 Try to be sure the class agrees on the results from Investigate This 10.73 as they do this
activity.
Students should reason and conclude:
(a) The measured cell potential for identical copper half cells should be near zero and support
the prediction that the cell potential is zero. There is no drive to transfer electrons either way.
(b) With the non-identical copper half cells, there was a measurable cell potential with the
half cell with the lower copper(II) cation concentration as the anode. The difference in
concentrations, a concentration gradient, creates a driving force toward reducing the higher
concentration in the cathodic half cell and increasing the lower concentration ion the anodic
half cell.
Follow-up activities:
 Check This 10.76. Cell potentials in Investigate This 10.73.
 Consider This 10.77. How does addition of ethylenediamine affect ECu2+,Cu?
 Worked Example 10.78. K for formation of Ag(NH3)2+(aq).
 Check This 10.79. K for formation of Cu(en)22+(aq).
 End of chapter problem 10.70.
Consider This 10.77. How does addition of ethylenediamine affect [Cu2+(aq)]?
Goal:
Conclude that the results from Investigate This 10.73 show that adding ethylenediamine
decreases [Cu2+(aq)].
Classroom options:
 This activity can be conducted as an open class discussion.
Time for activity:
 5-10 minutes in class, depending on how much is a review of what has already been
discussed.
Instructor notes:
 Use the results from Investigate This 10.73 for the analysis.
Students should reason and conclude:
(a) Based on the results of Investigate This 10.73(a), the cell with the lower concentration of
Cu2+ should be the anode and the cell with the higher concentration the cathode of a
concentration cell.
(b) When ethylenediamine is added to one of the 0.10 M Cu2+ half cells, this cell is the anode
in a cell created from an identical 0.10 M Cu2+ half cell as well as from a cell created with a
0.010 M Cu2+ half cell. Thus, addition of ethylenediamine produces a Cu2+ concentration
even less than 0.010 M..
(c) The cell created from two originally identical copper half cells, to one of which
ethylenediamine has been added, is:
Cu(s)  Cu2+ (0.10 M) +ethylenediamine  Cu2+(0.10 M)  Cu(s)
More uncomplexed copper(II) ions are present in the cathodic half cell and fewer in the
anodic half cell, just as shown in Figure 10.16(b).
Follow-up activities:
 Worked Example 10.78. K for formation of Ag(NH3)2+(aq).
 Check This 10.79. K for formation of Cu(en)22+(aq).
 End of chapter problems 10.67 through 10.73.
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Reduction-Oxidation: Electrochemistry
Chapter 10
Solutions for Chapter 10 Check This Activities
Check This 10.6. Reactions in an electrophoresis apparatus.
Electrical charge flows through the ionic solution and ions in the gel in an electrophoresis
apparatus, just as in the electrolysis cell pictured in diagram in Consider This 10.5. Reduction of
water (or of hydronium ions) occurs at the negative electrode and produces bubbles of H2(g)
according to equation (10.3). Oxidation of water occurs at the positive electrode and produces
bubbles of O2(g) according to equation (10.4).
Check This 10.9. The oxidation reaction in Investigate This 10.7.
The gas produced at the anode is oxygen gas. The oxidation reaction is
2H2O(l)  O2(g) + 4H+(aq) + 4e–
Hydronium ion is the other product of the oxidation, so you could test the solution near the anode
to see whether it is more acidic than the original solution. Because the solution is colored (blue
from the aquated copper(II) cation), an indicator added to the solution might not work, but you
could test the anodic solution and the bulk of the solution with pH indicator paper to check your
prediction.
Check This 10.11. Electrochemical stoichiometry.
(a) The electrolytic reductions that will occur in the two cells in series are:
Ni2+(aq) + 2e–  Ni(s)
2Ag+(aq) + 2e–  2Ag(s)
The silver ion reduction, reaction (10.12), has been multiplied by two here, because the same
number of electrons have to pass through electrolytic cells that are connected in series. Thus, we
see that the number of moles of silver metal produced by a given amount of current (number of
electrons) is twice the number of moles of nickel metal produced. We use the molar masses of
the two metals to convert their mole ratio to a mass ratio:
2 mol Ag  107.87 g Ag 1 mol 215.7 g Ag
2 mol Ag
=
=
= 3.675 (g Ag)·(g Ni)–1
58.71
g
Ni
1 mol Ni
58.71
g
Ni
1 mol Ni 
1 mol




(b) Calculate the number of moles of electrons that pass through the circuit and then the number
of moles of each metal that will be deposited by this many moles of electrons. Convert the
number of moles of metal deposited to grams of metal and take the ratio to compare with the
ratio in part (a).
coulombs passed = (0.106 A)(7200. s) = 763 C
 1 mol electrons 
moles of electrons = (763 C) 
= 7.91  10–3 mol
 96,485 C 
From the equations above, we see that the number of moles of nickel metal deposited is one-half
the number of moles of electrons that pass through the circuit and the number of moles of silver
metal is the same as the number of moles of electrons. Thus, we will get 3.96  10–3 mol nickel
and 7.91  10–3 mol of silver and their mass ratio is:
52
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Chapter 10
Reduction-Oxidation: Electrochemistry




7.91 10 –3 mol Ag  107.87 g Ag 1 mol 0.853 g Ag
7.91  10 –3 mol Ag
=
=
–3
–3
58.71
g
Ni
3.96  10 mol Ni
0.232 g Ni
3.96  10 mol Ni 
1 mol
mass ratio = 3.67 (g Ag)·(g Ni)–1
Since the calculation in part (a) and that in part (b) both depend on the ratio of number of moles
on each metal deposited and their molar masses, it should not be surprising to find that the ratio
in part (b) is the same, within the round-off uncertainty in the data, as that in part (a).
Check This 10.13. Electrolytic purification of copper metal.
From the mass of copper deposited in 14 days, calculate the number of moles of copper and
hence the number of moles of electrons (two moles of electrons per mole of copper). Convert
moles of electrons to coulombs and then the number of amperes of current that must pass for 14
days to provide this many coulombs.
mass Cu deposited = 163 kg
 1000 g   1 mol 
moles of Cu = (163 kg) 
= 2.57  103 mol Cu
 1 kg   63.5 g 
Convert moles of copper deposited to coulombs required:
 2 mol electrons 
2.57  103 mol Cu = (2.57  103 mol Cu) 
 1 mol Cu 
 96,485 C 


1 mol electrons 
= 4.96  108 C
 4.96  108 C   1 day   1 hour   1 min 
current = 
  24 hours   60 min   60 seconds  = 410 A

14 days
Check This 10.18. Ion migrations in a salt bridge.
(a) Drawing should show a neutral solution with charges balancing out. For example Cu2+(aq)
will cancel out two NO3-(aq) ions resulting in a neutral solution. To detect the migration of
Cu2+(aq) ions into the salt bridge, we would observe the salt bridge turning blue.
(b) Besides those listed above Check This 10.18, NO3-(aq) ions could migrate into the salt bridge
from the silver half cell as the Ag+(aq) plate out. This is unlikely since the concentration gradient
is against it. This migration would be undetectable in the salt bridge where the concentration of
NO3-(aq) ions is so high. In theory, one might detect the depletion of NO3-(aq) ions or a
concentration of K+(aq) ions smaller the concentration of Ag+(aq) that disappears. But, both
potassium and nitrate ion are very hard to analyze accurately enough for the purpose so, for
practical purposes, the nitrate migration out of the cell is undetectable.
(c) The proteins shown in Figure 5.9 are normal and sickle-cell hemoglobin at a pH where the
protein is negative (which is the most usual condition in electrophoresis). Electrophoresis gels
are made up with a buffer solution that contains a lot of relatively small ions. The
electrochemical reactions at the electrodes (the electrolysis of water) cannot occur unless there is
flow of charge carriers through the gel, which is how it acts as a salt bridge. One can imagine
that the negatively charged protein is just one of the anions that moves in the salt bridge and the
counterions that came with it from the protein sample move the other direction. Electrophoresis
ACS Chemistry FROG
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Reduction-Oxidation: Electrochemistry
Chapter 10
is not this simple, of course, and it is more the flow of the small ions that drags the protein along,
but that is not necessary to understand the direction of movement and the analogy to a the salt
bridge. Actually more than an analogy – the gel is a salt bridge. The salt bridge animation
reinforces this explanation as it shows the potassium ions and nitrate ions moving in opposite
directions. The potassium cation moves to the right toward the zinc half cell where reduction of
zinc cation is occurring and more positive charge is needed to maintain electrical neutrality. The
nitrate anion moves to the left toward the magnesium half cell where magnesium metal is being
oxidized adding positively charged magnesium cations to the solution and requiring additional
anions to maintain electrical neutrality.
Check This 10.20. Interpreting the line notation for a copper–zinc cell.
(a) The cell notation indicates that the anode reaction (half cell on the left in the this notation) is
oxidation of zinc metal:
Zn(s)  Zn2+(aq) + 2eThe cathode half cell reaction is reduction of copper(II) cation:
Cu2+(aq) + 2e-  Cu(s)
The overall cell reaction is:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s).
(b) If a piece of copper metal is placed in a solution of zinc ion, no reaction will be observed. We
know this, because the overall reaction written in part (a) is the spontaneous direction of the
reaction between zinc metal and copper(II) ion. The reverse reaction of copper metal with zinc
cation is not spontaneous. If a piece of zinc metal is placed in a solution of copper ions, a
reaction will be observed; the zinc metal will darken as finely divided copper metal deposits on
its surface and the clear blue copper(II) cation solution will become lighter blue as the cation is
reduced.
Check This 10.24. Combining cell potentials and interpreting cell reactions.
(a) Substituting measured values (in bold italics) for cell potentials from Investigate This 10.21
and Consider This 10.23 into equation (10.32) gives:
EZn|Cu + ECu|Ag = EZn|Ag = (1.02 V) + (0.43 V) = 1.45 V ≈ 1.44 V
The sum of the measured potentials for two of the half-cell combinations is equal to the third, as
the equation suggests it might be.
(b) Equation (10.30), Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s), shows the spontaneous direction for
the cell reaction with zinc and copper half cells. This equation shows that a strip of zinc metal
would be oxidized to zinc(II) cation and copper metal would begin to deposit on the surface of
the zinc metal if a strip of zinc metal is placed in a solution of copper(II) ion. There will be no
reaction if a strip of copper metal is placed in a solution of zinc(II) cation, because the reaction
of copper metal with zinc(II) cation, the reverse of this cell reaction is not spontaneous.
(c) Equation (10.31), Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s), shows the spontaneous direction
for the cell reaction with zinc and silver half cells. This equation shows that a strip of zinc metal
would be oxidized to zinc(II) cation and silver metal would begin to deposit on the surface of the
zinc metal if a strip of zinc metal is placed in a solution of silver(I) ion. There will be no reaction
if a strip of silver metal is placed in a solution of zinc(II) cation, because the reaction of silver
metal with zinc(II) cation, the reverse of this cell reaction is not spontaneous.
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(d) In Investigate This 6.62, we found that, when copper metal was placed in a solution of
silver(I) cation, silver metal needles formed on the surface of the copper metal and the solution
became blue, indicating the presence of copper(II) cation. This is the spontaneous reaction
predicted by equation (10.22), so the prediction is correct. As we showed in part (c), a similar
spontaneous reaction between zinc metal and silver(I) cation is predicted by equation (10.31).
Reactions (10.30), (10.2), and (10.31) are interrelated through cell potentials, as shown in part
(a), and combinations of cell reactions, so it’s likely that reaction (10.31) also correctly predicts
that zinc metal will reduce silver(I) cation, just as copper metal does.
Check This 10.25. Standard reduction potentials.
(a) The reaction at the cathode is reduction of copper(II) cation:
Cu2+(aq) + 2e–  Cu(s)
Since the anode of the cell is the standard hydrogen electrode, the cell potential, 0.337 V, is the
standard reduction potential for this cathode half reaction. By definition, the standard reduction
potential for any half cell is equal to the measured potential of a cell (copper, in this case) made
by coupling to the SHE to create a cell in which the SHE is taken to be the anode.
(b) If a piece of copper metal is placed in a solution containing hydronium ions, no reaction will
be observed, because, as we see from the cell [equation (10.40)] and its cell potential, the
spontaneous reaction is:
Cu2+(aq) + H2(g)  Cu(s) + 2H+(aq)
Copper metal does not react with hydronium ion under standard conditions.
Check This 10.28. What are the signs of reduction potentials for other metals?
Metals that have positive standard reduction potentials cannot be oxidized be hydronium ion
under standard conditions, so no bubbles of hydrogen will form when these metals are placed in
acidic solution. Conversely, metals that have negative standard reduction potentials can be
oxidized by hydronium ion and will produce bubbles of hydrogen gas when placed in an acidic
solution. Copper, Cu, does not react with acid, so its standard reduction potential is positive.
Iron, Fe, and magnesium, Mg, both react with acid to form bubbles of gas (hydrogen, although
you did not test to check this assumption), so their standard reduction potentials are negative.
The magnesium metal caused more vigorous bubbling than iron, which suggests that the driving
force for the reaction with magnesium is greater, that is, that the magnesium has a more negative
standard reduction potential than iron. Although this is true, it is somewhat dangerous to draw
such conclusions based on rates of reaction, because, as we have seen, reactions that have high
negative free energies, such as the oxidation of glucose, can be so slow as to be unobservable on
a reasonable time scale.
Check This 10.30. The lead-acid battery.
(a) The data in Appendix C show that the standard reduction potential for half reaction (10.49),
1.658 V, is more positive than that for reaction (10.50), –0.356 V. Thus reaction (10.49) is
combined with the reverse of reaction (10.50) to give the overall cell reaction and their standard
reduction potentials are combined in the same way to give the standard cell potential:
PbO2(s) + Pb(s) + 4H+(aq) + 2SO4(aq)  2PbSO4(s) + 2H2O
Ecell = (1.658 V) – (–0.356 V) = 2.014 V
Automobile batteries consist of several cells connected in series. If each cell provides about 2 V,
six cells in series will produce a 12-V battery. This is a true “battery,” that is, a combination of
ACS Chemistry FROG
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several cells in series. All the connections are inside the battery casing, so you cannot see the
individual cells and their connections from the outside, but the cutaway view shown in the
margin on page 692 gives some idea of the internal construction of the battery.
(b) The line notation for the cell in which the overall reaction in part (a) occurs to produce
electrical current for an automobile is:
Pb(s)  PbSO4(s) | H2SO4(aq) | PbO2(s)  PbSO4(s) | Pb(s)
When the cell discharges, the Pb(s) in the anode—the electrode that is filled with PbSO4(s)—is
oxidized to lead(II) and forms more PbSO4(s) by reaction of the lead(II) cation with sulfate anion
from the electrolyte solution. The lead(IV) in the PbO2(s) in the cathode—the electrode that is
filled with PbSO4(s) and PbO2(s)—is reduced to lead(II), which reacts with sulfate anion from
the solution to form more PbSO4(s).
(c) When the cell reaction runs in reverse, the changes at the electrodes are just the opposite of
those described in part (b). At the electrode (anode) where electrons leave the cell when it
discharges, electrons enter the cell when it is recharged and reduce lead(II) in the PbSO4(s) to
give Pb(s). At the electrode (cathode) where electrons enter the cell when it discharges, electrons
leave the cell when it is recharged as lead(II) in PbSO4(s) is oxidized to lead(IV) in PbO2(s).
Since reduction occurs at the Pb(s) electrode filled with PbSO4(s), this electrode is acting as the
cathode upon recharging.
Check This 10.34. Electrical work.
As NOTED in Investigate This 10.31, the activity is impossible, as written, so no work data are
available for the cell suggested. However, the data in the note for a AAA battery and two small
motors can be used to practice this sort of problem. The battery voltage was 1.581 V; one motor
required 0.130 A and the other 0.079 A. We’ll do the analysis for the motor that drew the larger
current and leave the analysis for the other for you to do. The current drawn by the motor is:
current = 0.130 A = 0.130 C·s–1
charge per minute = (0.130 C·s–1)(60 s) = 7.80 C
welec = (7.80 C)(1.581 V) = 12.3 J
 1 mol 
mol e– = (7.80 C) 
= 8.08  10–5 mol

96,485 C
Check This 10.37. Amount of work available from an electrochemical cell.
As NOTED in Investigate This 10.31, the activity is impossible, as written, so no work data are
available for the cell suggested. However, the data in the note for a AAA battery and two small
motors can be used to practice this sort of problem. To do the problem as written, you would
have to calculate the number of moles of electrons available from the number of moles of
copper(II) cation reduced, if half of the moles originally present are reduced and compare this
value to the number of moles required to run the motor for one minute, as in Check This 10.34.
For a commercial battery, you can look up the capacity of the battery (there are lots of Web sites
that provide this information). The capacities are usually given in milliamp-hours, mA·hr, that is,
the current the battery can produce (theoretically) in one hour. In actual use, batteries can
approximate this capacity only if a relatively small current is drawn over a long time. For a AAA
battery, the capacity is 1180 mA·hr. To make this calculation comparable to that we would do
for a battery whose stoichiometry and composition are known, we first calculate the number of
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moles of electrons the battery can produce and then compare half of this to the number of moles
required per minute by the motor analyzed in Check This 10.34. If the current is 1.18 A =
1.18 C·s–1 for one hour, the charge that passes through the circuit is:
charge = (1.18 C·s–1·hr)(60 s·min–1)(60 min·hr–1) = 4.25  103 C
Only three significant figures are used for the battery capacity, since this is only approximate for
actual use. The number of moles of electrons that would flow if half this capacity were used is:
 1 mol 
mol e– = (0.5)(4.25  103 C) 
= 2.20  10–2 mol
96,485 C 
The motor requires 8.08  10–5 mol of electrons to run for one minute, so the number of minutes
that battery can run the motor, using half the battery capacity is:


2.20  10 –2 mol
time motor can run = 
–5
–1  = 272 min = 4.5 hr
 8.08  10 mol  min 
Note that this is the same result we would obtain by dividing half the battery capacity by the
current, 130 mA, drawn by the motor.
Check This 10.40. Nernst equation at 25 C (298 K).
Start with equation (10.40) and substitute values for the constants and temperature of 298 K in
the factor that precedes the logarithmic term (note that 1 J = 1 V·C):
-1
1
2.303RT 2.302598.31451 V C  mol  K 298.15 K
= 0.0591598 V

nF
n 96,485.3 C  mol-1 
0.05916 V
log Q
n
Values of the constants (including the 2.303 ln to log conversion) are carried out to five or six
significant figures to yield the value for the Nernst equation constant that is usually shown. If the
rounded values we use for most calculations are substituted, the result is 0.05914 V.
E = Eº -
Check This 10.42. Nernst equation applied to silver–copper cells.
(a) The experimental data from Investigate This 10.38 are plotted here:
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The Nernst equation for the silver-copper cell is:
 Cu2+ (aq) 


E = E –
logQ = E –
log 

2
+
n
2
Ag
(aq)






2+
+
E = E – (0.0296 V)log(Cu (aq)) + (0.059 V)log(Ag (aq))
Rearranging this equation gives:
E = 0.059 V)log(Ag+(aq)) + [E – (0.0296 V)log(Cu2+(aq))]
This is the theoretical equation for the line on the plot. From the plot, E = (0.054 V)log(Ag+(aq))
+ 0.472 V. The experimental slope is about 10% too low, but it is strongly dependent on the third
significant figure in the experimental voltages, so a very few millivolts too low or too high can
cause the slope to vary this much. This is why great care is required in setting up the cells in
Investigate This 10.38. The intercept of the line is:
0.472 V = E – (0.0296 V)log(Cu2+(aq))
E = (0.472 V) + (0.0296 V)log(Cu2+(aq)) = (0.472 V) + (0.0296 V)log(0.10)
E = (0.472 V) – (0.030 V) = 0.442 V
The data in Table 10.1 give a calculated E = 0.462 V [= (0.799 V) – (0.337 V)], so the
experimental value is about 20 mV too low. Recall that the measured values shown in Investigate
This 10.38 were systematically too low, so the intercept of the line in the plot will be too low and
lead to a lower value for the experimental E. Since the slope of the line is so close to the
theoretical value, it is likely that the systematic error in the experimental cell values is a problem
with the reference copper half cell.
(b) As the silver ion concentration increases, the observed cell potential increases. This variation
in the cell potentials with the silver ion concentrations is consistent with Le Chatelier's principle.
See Consider This 10.39(b) above for the explanation.
0.059 V
0.059 V
Check This 10.43. Three problems from the Web Companion.
NOTE: For the these problems, the Web Companion incorrectly omits the n = 10 factor in the
Nernst equation on page 4, so the answers accepted by the Companion are incorrect. The
calculations presented here include the n = 10 factor and, of course, give different values than are
accepted as correct by the Companion. Also, after you have correctly answered the second
question about lowering the pH to 0.0, the new screen with the third question has changed the pH
in the second question to 1.0. This is some sort of programming error.
For each calculation, the appropriate concentrations are substituted into the Nernst equation and
the equation solved for E.
First calculation:
Mn 2+  Cu 2+ 

0.059 V
0.059 V
0.12 0.15
log
log + 16
E = 1.17 V –
=
1.17
V
–
2
10
0.116 0.12
10
H  MnO–4 
2
5
E = 1.11 V
Second calculation:
0.12 0.15
0.059 V

log
E = 1.17 V –
= 1.20 V
10
1.0 16 0.12
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Third calculation [We assume—as must also be done to get the result the Companion accepts—
that the change in size of the electrode is made in the original cell, not the one in which sulfuric
acid has been added to the MnO4–/Mn2+ half cell.]:
0.059 V
0.12 0.15
log
E = 1.17 V –
= 1.11 V
10
0.116 0.12
The cell potential is unchanged. The solid copper metal is not included in the reaction quotient,
because the dimensionless concentration ratio for a pure solid is taken as 1. (See Table 9.1 and
accompanying discussion, if necessary.)
Check This 10.46. Deriving equation (10.74).
Use the relationship log y x = log(y) – log(x) to rearrange the ratio in the first equation:





 H+ (aq) 
+
+
E = E – (0.059 V)·log  +
 = E – (0.059 V)[log(H (aq)) – log(Ag (aq))]
Ag
(aq)


Now rearrange this equation to give equation (10.74):
E = E + (0.059 V)·log(Ag+(aq)) – (0.059 V)·log(H+(aq))
Check This 10.47. pH and cell potentials for silver-quinhydrone cells.
(a) Equation (10.75) predicts that the cell potentials should be a linear function of pH for silverquinhydrone cells in which the silver half cell is the same, but the pH in the quinhydrone half
cell is varied. The slope of the line is predicted to be 0.059 V. The two values for cell potentials
obtained in Investigate This 10.44 can be combined with the pH values in the quinhydrone half
cells to give a slope for the change observed:
(0.387 V) – (0.183 V)
E
slope =
=
= 0.051 V
7–3
(pH)
This experimental value for the slope is about 15% lower than predicted, but, considering that it
is based on only two measurements, is not in bad agreement with the prediction. To find the
intercept of the experimental line [which we can use in part (c)], substitute the slope and one pair
of the experimental values into a linear equation model:
E = 0.387 V = (slope)pH + intercept = (0.051 V)(7) + intercept
intercept = 0.030 V
These values can also be determined graphically by plotting the points and determining the
equation of the line. The graph is:
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(b) In Consider This 10.45(d) we reasoned that an increase in pH (decrease in hydronium ion
concentration) in the quinhydrone half cell would increase the cell potential for the silverquinhydrone cell. Equation (10.75) shows that the cell potential is a linear function of pH that
increases as pH increases, so the equation is consistent with our reasoning based on
Le Chatelier’s principle.
(c) In part (a), we used the data from Investigate This 10.44 to find the intercept of the linear
equation relating cell potential and pH for the silver-quinhydrone cell. This intercept is the
constant in equation (10.75), so we can write:
constant = intercept = 0.030 V = E + (0.051 V)log(Ag+(aq))
Note that we use the experimental slope, 0.051 V, in this expression, because the linear
relationship we are analyzing is the one we determined experimentally. Rearranging this
expression and substituting the concentration of silver ion from the experiment gives:
E = (0.030 V) – (0.051 V)log(0.100) = 0.081 V
The data in Appendix C give the standard cell potential for this cell:
E(Ag+,Ag) – E(Qu,QuH2) = (0.799 V) – (0.699 V) = 0.100 V
Our experimental value is close to, but about 20% lower than, this calculated value, which is in
line with the fact that the data shown for Investigate This 10.44 are consistently lower than
values calculated from the data in Appendix C and the Nernst equation. This result again
underscores the need to take care in preparing the half cells for these measurements.
Check This 10.49. Conditions for developing photographic film.
To find the pH condition for driving the developing reaction far toward completion, we look at
the equilibrium ratio for the reaction to see how to make the concentration of silver(I) cation as
small as possible. The equilibrium constant and equilibrium constant expression are:
 Qu(aq)H+ (aq)2 

K= 
QuH2 (aq) Ag + (aq)2 

eq
(The dimensionless concentration ratio for solid silver has been omitted from the numerator,
because its value is 1.) In order to maintain the equality here, as we make the concentration of
silver(I) ion smaller in the denominator, we have to make the numerator smaller as well. An
obvious way to make the numerator smaller is to make the hydronium ion concentration smaller,
that is, to carry out the reaction (developing the film) at a high pH. If you have ever developed
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film, you probably know that developer solution feels slippery; basic (high pH) solutions feel
slippery when rubbed between your fingers. After developing, film is usually “fixed” to stop the
developing process before it goes too far and reduces too much of the silver ion. The fixing
solution is acidic (sometimes a dilute acetic acid solution is used) in order to reduce the pH and
stop the developing reaction quickly as the developer is rinsed off the film.
Check This 10.50. Simplifying equation (10.79).
The simplification requires the use of the relationship logxa = alogx:


1
0.059 V log 
+
+

E(Ag , Ag) = E°(Ag , Ag) – 
2
+
2  
Ag
(aq)
 



1
 0.059 V 
log 

E(Ag+, Ag) = E°(Ag+, Ag) – 

+

2 
 Ag (aq)
(10.79);
2


1
 0.059 V 
log 

E(Ag+, Ag) = E°(Ag+, Ag) – 2 

+

2 
 Ag (aq)


1
 0.059 V 
log 

E(Ag+, Ag) = E°(Ag+, Ag) – 

+

1 
 Ag (aq)
(10.80)
Check This 10.52. Hydronium ion reduction potential as a function of pH.
The hydronium ion is a reactant in its reduction half reaction. If its concentration is decreased
(pH is increased), Le Chatelier’s principle predicts that the reverse of the reduction reaction will
be favored, in order to try to increase the hydronium ion concentration. Thus the reduction
potential should decrease compared to the standard reduction potential, that is, go toward more
negative values. The Nernst equation for the reduction half reaction is:
 H  
 0.05916 V 
2
E(H+, H2) = E°(H+, H2) – 
log 

2 
+



2
H ( aq) 

Four significant figures in the constant in the Nernst equation, 0.05915, are required to obtain
values as precise as those given in Appendix C. We are interested in the change in E(H+, H2) as
the pH changes with all other conditions kept standard, so we substitute (H2) = 1. Also, E°(H+,
H2) = 0.000V, so we get:

1
 0.05916 V 
+
E(H , H2) = – 
 log  +

2
 H (aq)




1
 0.05916 V 
=
–
log





2
   2   H+ (aq)
2




1
1
 0.05916 V 
E(H+, H2) = – (2) 
log  +
= – (0.05916 V)log  +





2
 H (aq) 
 H (aq) 
 
Since log 1 x = – log(x), we can write:
E(H+, H2) = – (0.05916 V)[– log(H+(aq))] = – (0.05916 V)pH
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For pH 7, E(H+, H2) = – (0.05916 V)(7) = – 0.414 V
For pH 14, E(H+, H2) = – (0.05916 V)(14) = – 0.828 V
The reduction potential value we calculated for pH 7 can be compared with E°´, that is, the
reduction potential for hydronium ion reduction at pH 7, which is given in Appendix C as –
0.414 V. The values agree.
The reduction potential value at pH 14 represents the reduction when (OH–(aq)) = 1.0 M. To
convert the hydronium ion reduction to these conditions, we use the method from Section 6.10.
We add enough hydroxide ions to convert all the hydronium ions to water and then cancel
redundant waters. Applying this conversion to half-reaction equation (10.83), we add two OH–
(aq) to each side:
2H+(aq) + 2OH–(aq) + 2e–  H2(g) + 2OH–(aq)
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
The standard reduction potential for this reaction from Appendix C is 0.8281 V. Our calculated
value and this table value agree.
Check This 10.55. Predictions you can make about biological redox reactions.
(a) The data in Appendix C for the standard reduction potentials at pH 7 give:
E´(UQ,UQH2) = 0.10 V
E´(NAD+,NADH) = – 0.32 V
E´UQ|NAD = E´(UQ,UQH2) – E´(NAD+,NADH) = 0.42 V
The standard free energy for this reaction is:
Go´rxn = – nFE´UQ|NAD = 2(96,485 C·mol–1)(0.42 V) = – 81 kJ·mol–1
(b) The positive standard cell potential and negative standard free energy show that reaction
(10.84) is favored in the direction written at pH = 7. The equilibrium constant for the reaction is:
K= e
–Go 
–(–81,000 Jmol–1 )
RT
(8.314 J mol
=e
–1
–1
K )(298 K)
= 1.6  1014
(c) Reaction (10.84) is rewritten here with the unchanging side groups removed from the
molecular structures leaving just the pair of electrons that bonds the group to the rings. The
nonbonding electrons on the oxygen atoms are added for the count. The counts include all
electron pairs, both those represented by a pair of dots and those represented by a line.
O
H
H
H
O H
H+ (aq)
N
O
N
R´
NADH
UQ
NAD +
13 e– pairs
17 e– pairs
(aq)
(membrane)
(aq)
12 e– pairs
(membrane)
O H
UQH 2
18 e– pairs
There are 30 pairs of electrons on each side of the equation. When NADH is oxidized to NAD+,
it loses two electrons, which are gained by UQ as it is reduced to UQH2 (also gaining two
hydrogen atomic cores—protons). This is the two-electron change in the reaction.
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Check This 10.58. Alcohols in Benedict’s reagent.
The sample of ethanol in Investigate This 10.56 did not react with Benedict’s reagent, so we can
conclude that alcohols do not react with Benedict’s reagent.
Check This 10.60. Reduction potential for copper(II) in Benedict’s reagent.
(a) In reaction (10.90), rewritten here, hydroxide ion is a reactant. Le Chatelier’s principle
predicts that the reverse reaction, which would produce hydroxide ion, would be favored if the
concentration of hydroxide ion is reduced from its standard value. If the reverse reaction is
favored, the reduction potential would decrease from its standard value.
2Cu(cit)–(aq) + 2OH–(aq) + 2e–  Cu2O(s) + 2cit3–(aq) + H2O(l)
(b) The reaction quotient for this reaction (with electrons, solvent, and pure solid omitted) is:
Q=
cit
3–
(aq)
2
Cu(cit) (aq)  OH (aq)
–
2
–
2
(c) Use the Nernst equation, substituting appropriate values for the dimensionless concentration
ratios in Q, to find the reduction potential at pH 12, (OH–(aq)) = 10–2:

12  = 0.21 V
 0.059 V 

log


2
2
12 10 –2 


The calculated reduction potential at pH 12 is less than the standard reduction potential, as we
predicted using Le Chatelier’s principle in part (a).
 0.059 V 
E = E – 
 logQ = (0.33 V) –

2
 
Check This 10.64. Methylene blue, MB+, reduction in basic solution.
To rewrite half reaction (10.93) in terms of the hydroxide ion, we use the method from Section
6.10. We add enough hydroxide ions to convert all the hydronium ions to water and then cancel
redundant waters. Applying this conversion to half-reaction equation (10.93), we add one OH–
(aq) to each side:
MB+(aq) + H+(aq) + OH–(aq) + 2e–  MBH(aq) + OH–(aq)
MB+(aq) + H2O(l) + 2e–  MBH(aq) + OH–(aq)
The standard reduction potential for the methylene blue reduction, E°´(MB+, MBH) = 0.011 V, is
given at pH 7. The change we made from hydronium to hydroxide ion does not affect this
condition. The solution is still at pH 7, so the standard potential is still 0.011 V.
Check This 10.65. MB+–glucose redox reaction in basic solution.
Since the hydroxide ion is a reactant in reaction (10.94), increasing the concentration of
o
hydroxide would increase EM B|glucose . Reaction (10.90) is more favored as the solution becomes
more basic.
Check This 10.67. Oxidation of methylene blue by oxygen in the Blue-Bottle reaction.
The Nernst equation for the oxidation of methylene blue by oxygen, equation (10.97), is given in
Worked Example 10.66:
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Reduction-Oxidation: Electrochemistry
Chapter 10
 OH – 2 MB+ 2 
0.059
V


EO2|MB = EO2|MB – 
 log 
2 

4 
 O2 MBH 
 OH – 2   MB+  2
 0.059 V 
= EO2|MB – 

 log 


4 
 O2    MBH 
The equation was reformatted to facilitate substitution of the conditions for the Blue-Bottle
reaction specified in the problem statement:
 MB+  

 = 10
 MBH 
 20 kPa 
 = 0.20
(O2) = 
100 kPa 
(OH–) = 1
From Worked Example (10.66) we know that EO2\MB = 0.60 V at pH 14. Therefore,
EO2|MB
 12 
 0.059 V 
= (0.60 V) – 
10 2 = 0.56 V
 log 


4
 0.20 
Even though the pressure of oxygen is less than the standard pressure, oxidation is still favored.
If the ratio of oxidized to reduced methylene blue is about 1000, we get:
EO2|MB
 12 
 0.059 V 
2
= (0.60 V) – 
log 

1000  = 0.50 V



4 
 0.20 
Our conclusion remains the same: the oxidation is highly favored.
Check This 10.69. Reduction potentials in a glucose-air fuel cell.
(a) The reaction, standard reduction potential, and Nernst equation for reduction of oxygen gas
are:
O2(g) + 4H+(aq) + 4e-  2H2O(aq)
E(O2,H2O) = 1.229 V
2


H 2O(aq)

 0.059 V 
E(O2,H2O) = E(O2,H2O) – 
 log 
4

4 
 O2 (g) H+ (aq) 
Substituting the dimensionless concentration ratios for all species except hydronium ion gives:

2

1
 0.059 V 
E(O2,H2O) = E(O2,H2O) – 
 log 
4

4 
 1H+ (aq)  


1
 0.059 V 
log
= E(O2,H2O) – 

 +


4 
 H (aq)
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Chapter 10
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

1
= E(O2,H2O) – (0.059 V) log  +

 H (aq)
 
Using the relationship log 1 x = –log(x), we get:
E(O2,H2O) = E(O2,H2O) – (0.059 V)[–log(H+(aq))] = E(O2,H2O) – (0.059 V)pH
And, at pH 5:
E(O2,H2O) = 1.229 V – (0.059 V)5 = 0.934 V
The reaction, standard reduction potential, and Nernst equation for reduction of gluconate are:
RCOO–(aq) + 3H+(aq) + 2e–  RCHO(aq) + H2O(aq)
E(RCOO–,RCHO) = –0.44 V
ERCOO– , RCHO
 RCHO(aq) H O(aq) 
 2
 ;
 0.059 V 
= E(RCOO ,RCHO) – 
 log 
3
–
+

2
 RCOO (aq)H (aq)  
–
Substituting the dimensionless concentration ratios for all species except hydronium ion gives:

11 
 0.059 V 
E(RCOO–,RCHO) = E(RCOO–,RCHO) – 
log


3

2 
 1H+ (aq) 


1
 0.059 V 
log  +
E(RCOO–,RCHO) = E(RCOO–,RCHO) – 



2 
 H (aq)
3
 0.059 V 
E(RCOO–,RCHO) = (–0.44 V) – 3
 pH

2
And, at pH 5:
 0.059 V 
E(RCOO–,RCHO) = (–0.44 V) – 3
 5 = –0.88 V

2
(b) The overall cell reaction for the glucose-oxygen cell is obtained by adding the half reaction
for oxygen reduction to the reverse of the reaction for gluconate reduction (multiplied by two to
account for the number of electrons):
O2(g) + 4H+(aq) + 4e-  2H2O(aq)
E(O2,H2O) = 0.934 V
2RCHO(aq) + 2H2O(aq)  2RCOO–(aq) + 6H+(aq) + 4e– –E(RCOO–,RCHO) = 0.88 V
O2(g) + 2RCHO(aq)  2RCOO–(aq) + 2H+(aq)
ERCHO|O2 = 1.81 V
Le Chatelier's principle predicts that the cell potential would decrease if the activity of oxygen is
decreased. Because the cell potential is so large, such a decrease would have an almost negligible
effect on the driving force for the reaction, which would still be highly favored.
Check This 10.70. Intermediate potentials in the biofuel cell.
(a) The results from Check This 10.69 for the biofuel cell at pH 5 are included on this revised
version of Figure 10.14.
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Reduction-Oxidation: Electrochemistry
reduced
2RCHO
2H2O
–0.88 V
2RCOO –
oxidized
6H+
oxidized
GOx
GOx
reduced
carbon
oxidized
reduced anode
2+
Os
Os3+
e– flow in
0.3 V
0.7 V
external
3+
Os2+
Os
circuit
reduced
carbon
oxidized
cathode
Chapter 10
+
reduced 4H oxidized
LAC
O2
0.934 V
LAC
oxidized
2H2O
reduced
(b) The “ideal” potential for glucose oxidase would be midway between –0.88 V and 0.3 V (for
the osmium system) or about –0.3 V. At the other end of the system, the ideal laccase potential
would be midway between 0.7 V and 0.934 V or about 0.8 V. These midway potentials are
chosen so that the oxidation and reduction reactions in the enzyme cycles will have about the
same driving force.
(c) In Check This 10.69, we found that the overall potential drop through the cell (at pH 5) is
about 1.8 V, but only 0.4 V is captured as a drop across the electrodes (which has to do with the
way osmium ions interact in their matrices). Thus, the efficiency of the cell is:
 0.4 V 

 100% ≈ 20%.
1.8 V 
Check This 10.71. Role of NAD+-NADH in the glucose oxidation pathway.
In reaction (10.103), NADH is a reaction product and in reaction (10.104) it is a reactant. The net
outcome of these two reactions (if all the NADH produced in reaction (10.103) is used up by
reaction (10.104) is to use the electrons from glucose oxidation to reduce the oxidized form of
ubiquinone. Adding reaction (10.103) and reaction (10.104) (taken 12 times to account for all the
NADH produced), we get:
C6O12O6 + 12UQ + 6H2O  6H2O + 12UQH2
The NAD+ and NADH cancel out in the addition, since they are on opposite sides of the two
reactions. NAD+ and NADH couple the oxidation of glucose to the reduction of ubiquinone.
Electrons from the glucose oxidation are thus transferred to a molecule in the mitochondrial
membrane, which passes them on (by being reoxidized) to the cytochromes and ultimately to
oxygen.
Check This 10.72. Net reaction for oxygen reduction.
(a) The sequence of reactions, beginning with NADH, that leads to the reduction of the oxygen
molecule in mitochondria is:
UQ + NADH + H+(matrix)  UQH2 + NAD+
2Fe3+(cyt c1) + UQH2  2Fe2+(cyt c1) + UQ + 2H+(intermembrane space)
2Fe2+(cyt c1) + 2Fe3+(cyt a3)  2Fe3+(cyt c1) + 2Fe2+(cyt a3)
1/2O2 + 2Fe2+(cyt a3) + 2H+(matrix)  H2O + 2Fe3+(cyt a3)
NADH + 3H+(matrix) + 1/2O2  NAD+ + 2H+(intermembrane space) + H2O
The third reaction in this sequence represents all the redox reactions that pass electrons from one
cytochrome to another in the electron transport pathway in the inner mitochondrial membrane.
(b) The reduction potentials for the ubiquinone and cytochromes are irrelevant for the net cell
potential for the oxidation of NADH by O2 as a result of the above series of reactions. They are
simply coupling agents that are reduced in one step of the reaction sequence and oxidized in the
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Reduction-Oxidation: Electrochemistry
next and, therefore, cancel out of the net reaction. The overall reaction potential is just the
combination of the O2-H2O and NAD+-NADH standard reduction potentials (at pH 7) to give the
standard cell potential:
E(O2,H2O) – E(NAD+,NADH) = (0.816 V) – (0.32 V) = 1.14 V
Check This 10.76. Cell potentials in Investigate This 10.73.
The Investigate This 10.73 result, 0.050 V (or 50 mV), is not as close to the calculated value,
0.030 V, as we would hope, although it is small and positive, as predicted. Thus, the
experimental results qualitatively support the theoretical results calculated by the Nernst
equation, but the quantitative result based on the value reported above in Investigate This 10.73
is somewhat off. Care has to be taken in making and connecting the half cells for this sort of
investigation.
Check This 10.79. K for formation of Cu(en)22+(aq).
The equilibrium reaction and equilibrium constant expression for complex formation between
copper(II) cation and ethylenediamine is given by reaction (10.114):
Cu2+(aq) + 2en(aq)  Cu(en)22+(aq)
(Cu(en)2+
2 (aq))
K=
2+
(Cu (aq))(en(aq))2
The cell potential was 0.110 V (reported above in Investigate This 10.73) for a concentration cell
with [Cu2+(aq)] = 0.01 M in one half cell and, in the other half cell, a solution with an initial
[Cu2+(aq)] = 0.10 M to which ethylenediamine had been added. [Recall that this cell potential is
probably a good deal too low. Your result may vary considerably from this.] The half cell to
which ethylenediamine had been added was the anode of the cell, so, as you found in Consider
This 10.77, the concentration of copper(II) cation in this half cell is lower than that in the other
half cell. The Nernst equation for this cell is:
 lower (Cu 2+ ) 
 lower (Cu2+ ) 
 0.059 V 
 0.059 V 
ECu|Cu = 0.110 V = – 
= –
 log 
 log 



2 
 higher (Cu 2+ ) 
2 

0.010
Rearrange this equation and solve for the lower (Cu2+):
2 (0.110 V)
log[lower (Cu2+)] = –
+ log(0.010) = –3.73 – 2.00 = –5.73
0.059 V
lower (Cu2+) = [Cu2+(aq)] in the solution with ethylenediamine = 1.9  10–6 M
Only a tiny amount of Cu2+ is left uncomplexed in this solution. Therefore, essentially all the
copper(II) cation is complexed and we can say that [Cu(en)22+(aq)] = 0.10 M. In order to find K,
we also need the concentration of the ethylenediamine that is not complexed. We determine this
concentration by calculating the number of moles of ethylenediamine originally added to the
solution and subtracting the number of moles that are complexed. The number of moles of
ethylenediamine added to the solution is:
 0.9 g   1 mol 
mol en added = (0.05 mL) 
= 8  10-4 mol
 mL   60.0 g 
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Only one significant figure in the result is justified by the data. The number of moles of
ethylenediamine in the complex is equal to twice the number of moles of complex (in the 2 mL
of solution), since each complex ion contains two ethylenediamine molecules:
mol en complexed = 2(mol complex) = 2(0.10 M)(2  10–3 L) = 4  10–4 mol
Therefore, we find:
mol en uncomplexed = (8  10-4 mol) – (4  10-4 mol) = 4  10-4 mol
 4  10-4 mol 
[en(aq)] = 
= 0.2 M
(en(aq)) = 0.2
 2  10-3 L 
K=
(Cu(en)2 2+ ) 
(0.10)

(Cu(en)2+
2 (aq))
6
2+
2 = 
-6
2  = 1.3  10
2+
2
 (1.9  10 )(0.2) 
(Cu (aq))(en(aq)) (Cu )(en)
This calculated equilibrium constant is about 10-fold smaller than that for the silver diammine
complex calculated in Worked Example 10.78. This result would lead us to conclude that the
copper-ethylenediamine complex, Cu(en)22+(aq, is weaker (attraction is less strong between the
metal cation and the ligands) than the silver diammine complex, Ag(NH3)+(aq). However, a
check of tables of metal ion complexation constants shows that the copper-ethylenediamine
complex has an equilibrium constant for formation of about 1019. Our result is off by about 13
orders of magnitude. There are two possible sources of error that could account for this result.
The measured cell potential could be incorrect, but it is difficult to understand how it could be so
far off. It is also possible that a solution of ethylenediamine, instead of the pure liquid, was added
to half cell or that the drop of ethylenediamine was smaller than 0.05 mL. In either case, our
calculated value for (en(aq)) would be too high and lead to too small a value for K. This is yet
another reminder that care must be taken in preparing solutions and cells for these quantitative
measurements.
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Solutions for Chapter 10 End-of-Chapter Problems
Problem 10.1.
(a) We have seen (Investigate This 10.2) that electrolysis of a dilute aqueous solution of an ionic
compound (magnesium sulfate) produces a gas at both electrodes and a basic solution at the
cathode and acidic solution at the anode, just as the problem statement says is observed here for a
dilute aqueous NaCl solution. Thus, we might assume that the electrolysis half reactions are the
same in this case as in the investigation:
cathode (reduction): 4H2O(l) + 4e– (from electrode)  2H2(g) + 4OH–(aq)
anode (oxidation): 2H2O(l)  O2(g) + 4H+(aq) + 4e– (to electrode)
The gas produced at the anode is oxygen, which does not burn, so this is also consistent with the
observations reported in the problem statement.
(b) In the concentrated aqueous NaCl solution, the cathode reaction produces a basic solution, so
the half reaction is still probably the same as in the electrolysis of the dilute solution. At the
anode, however, the gas produced causes a choking sensation when inhaled. The only elements
present in the solution are H, O, Na, and Cl. The anode reaction is an oxidation and the most
likely species to be oxidized (aside from water) is the chloride ion, Cl–(aq), in the solution. The
product of the oxidation is chlorine gas, Cl2(g), which produces a choking sensation when
inhaled. Solutions of chlorine gas also act as bleach, so the bleaching of the indicator dye color
at the anode (reported in the problem statement) is consistent with this identification of the
product. The reactions are:
cathode (reduction): 2H2O(l) + 2e– (from electrode)  H2(g) + 2OH–(aq)
anode (oxidation): 2Cl–(aq)  Cl2(g) + 2e– (to electrode)
The ease of oxidation of water and chloride ion are similar with chloride being a bit easier to
oxidize. When there is only a small amount of chloride in the solution, the large number of water
molecules relative to chloride ions overwhelms the chloride and water is oxidized. When the
concentration of chloride ion is large, it can compete successfully with water and is oxidized. At
both concentrations, the gas at the anode is probably a mixture that contains both oxygen and
chlorine, but in quite different proportions in the two cases.
Problem 10.2.
(a) When an aqueous solution of KI is electrolyzed, bubbles of gas are formed at the cathode and
the solution around the anode becomes orange. The orange color at the anode appears to be I3–
(aq) (which the problem statement tells us is formed in a solution containing iodine, I2(aq), and
iodide, I–(aq)), so iodine, I2(aq), must have been produced at this electrode and reacted with I–
(aq). Thus, the anodic reaction is probably the oxidation of I–(aq) to give I2(aq):
2I–(aq)  I2(aq) + 2e–
The cathodic reaction must be the reduction of water to yield hydrogen gas and hydroxide in
solution:
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
This is the reaction that occurs at the cathode in aqueous solution whenever there is no species
present that is more easily reduced (such as some metal ions like Ag+, Cu2+, etc.) than water.
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Chapter 10
(b) A small amount of added sodium thiosulfate does not affect either electrode reaction directly
and, as observed, gas (hydrogen) is still produced at the cathode as soon as electrolysis is begun.
When iodine, I2(aq), is formed at the anode, however, it reacts rapidly with the thiosulfate and is
not, at first, observed. After some time of electrolysis, enough I2(aq) has been produced to react
with all the thiosulfate and the subsequent I2(aq) produced reacts with I–(aq) to form the I3–(aq)
color that is observed.
Problem 10.3.
(a) The half reaction at the cathode in the electrolysis of molten sodium chloride is
Na+(melt) + e–  Na(l)
Twenty-three grams of sodium metal is one mole of sodium, so one mole of electrons are
required for the electrolysis to produce of 23 g of Na(l). Calculate the time required to pass one
mole of electrons through the electrolysis cell at a current of 12 A:
it
(12 A)  (t)
1 mol electrons =
=
F
9.6485  10 4 C  mol –1
t = 8.0  103 s = 2.2 hr
(b) The half reaction at the cathode (reduction) is written in part (a). The oxidation half reaction
at the anode to produce chlorine gas is:
2Cl–(melt)  Cl2(g) + 2e–
The net cell reaction is:
2Na+(melt) + 2Cl–(melt)  2Na(l) + Cl2(g)
One mole of chlorine gas is produced for every two moles of sodium metal produced. Thus, onehalf mole of chlorine gas, 35.5 g [= (0.5 mol)·(71.0 g·mol–1)], is produced during this
electrolysis.
Problem 10.4.
(a)/(b) One of the reactions in the Dow process (production of pure magnesium metal by
electrolysis of molten magnesium chloride) is the formation of Mg metal from Mg2+ ions:
Mg2+ (melt) + 2e–  Mg(l)
Since this is a reduction half-reaction, it takes place at the cathode. The other reaction is
formation of Cl2 gas from Cl– ions:
2 Cl–(melt)  2e– + Cl2(g)
Since this is an oxidation half-reaction, it takes place at the anode.
(c) The net reaction in the Dow procell is:
2 Cl–(aq) + Mg2+(aq)  Mg(l) + Cl2(g)
(d) The mass of magnesium metal produced by a 95,000-A current passed through a molten
magnesium chloride cell for 8.0 hours is obtained by calculating the number of moles of
electrons that pass through the cell and then the number moles and mass of magnesium metal
from the stoichiometry of electrolysis in part (a):
mol e– =
70
(95000 A)(8.0 hr)(60 min  hr –1 )(60 s  min –1 )
= 2.8  104 mol e–
4
–1
9.65  10 C  mol
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Reduction-Oxidation: Electrochemistry
 1 mol Mg   24 g Mg 
2.8  104 mol e– = (2.8  104 mol e–) 
= 3.4  102 kg Mg
 2 mol e–   1 mol Mg 
(e) Solid salts do not conduct electricity, because the ions are fixed in place in the crystal lattice.
Thus, the solid magnesium chloride must be melted, which requires a high temperature, in order
to make its electrolysis possible.
Problem 10.5.
(a) The description of this gold electroplating system gives the reduction half reaction that occurs
at the surface to be plated, so the object to be plated must be the cathode in the electroplating
cell. The sheet of gold must be the anode, so the anode reaction is oxidation of gold, which
enters the solution as the cyanide complex, Au(CN)4–(aq). The electrode reactions are:
cathode (reduction): Au(CN)4–(aq) + 3e–  Au(s) + 4CN–(aq)
anode (oxidation): Au(s) + 4CN–(aq)  Au(CN)4–(aq) + 3e–
Note that the reactions are just the reverse of one another. As the electroplating is carried out, the
concentration of the gold cyanide complex remains constant as the gold cation is reduced at the
cathode and produced at the anode.
(b) Calculate the moles of gold deposited by a 2.5 A current passing through the electroplating
cell for 7.5 minutes and convert to grams:
mol Au
= 1/
3(mol
–
e)
 i t 
= 1/3 


F
= 1/
 (2.5 A)(7.5 min)(60 s  min –1 ) 
–3
3
 = 3.9  10 mol
9.65  10 4 C  mol–1

mass Au = ( 3.9  10–3 mol)(197.0 g·mol–1) = 0.77 g
(c) The ideal electroplating process would deposit 0.77 g [part (b)]. If only 0.65 g was deposited,
the efficiency is (0.65 g) (0.77 g)  ·100% = 84%.


Problem 10.6.
(a) The number of moles of electrons passed through the silver and cobalt cells in series is equal
to the number of moles of Ag deposited, since one electron is required for each Ag atom.
 1 mol Ag 
mol e– = mol Ag = (0.2789 g Ag) 
= 2.586  10–3 mol
 107.87 g Ag 
(b) The ratio of the number of moles of electrons that pass through the cobalt cell [from part (a)]
to the number of moles of cobalt deposited gives us the number of electrons required to reduce
one atom of cobalt and hence the oxidation number of cobalt in the solution:
 1 mol Co 
mol Co = (0.0502 g Co) 
= 8.52  10–4 mol
 58.93 g Co 
mol e –
2.586  10 –3 mol
=
= 3.04
mol Co
8.52  10 –4 mol
Thus, the oxidation number of the cobalt in the complex is +3.
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Chapter 10
Problem 10.7.
To determine how much time will it take to electroplate 0.0353 g of chromium from an aqueous
potassium dichromate, K2Cr2O7, solution, with a current of 0.125 A passing through the cell, we
need to know how many electrons are required to reduce the Cr in Cr2O72– ions to Cr metal. That
is, we need to know the oxidation number of chromium in dichromate. The ion has a 2– charge
and, recall from Chapter 6, that each of the oxygens can be assumed to have an oxidation number
of –2, so the seven oxygens contribute an overall –14 oxidation number. The two chromiums
must have a combined oxidation number of +12, in order to give the correct ionic charge.
Therefore, each Cr has an oxidation number of +6. Each mole of Cr deposited requires six moles
of electrons. The number of moles of Cr deposited is:
 1 mol Cr 
0.0353 g Cr = (0.0353 g Cr) 
= 6.79  10–4 mol Cr
 52.00 g Cr 
To get the time required, we express the required number of moles of electrons in terms of
current, time, and the Faraday constant and solve for the time:
(0.125 A)t
 i t  

mol e– = 6(6.79  10–4 mol) = 
=

 F   9.65  10 4 C  mol–1 
t = 3.14  103 sec = 52.4 min
Problem 10.8.
(a) To find the current used by this electric company customer, we need to find the number of
moles of electrons that entered the meter, divide by the time, and account for the percentage this
current is of the total. The data we have are the mass of zinc deposited, 57 g, the time, 30 days,
and the percentage of the current that passed through the meter, 8%. Since Zn2+(aq) is reduced to
Zn(s), two moles of electrons are required for each mole of zinc deposited:
 1 mol Zn 
57 g Zn = (57 g Zn) 
= 0.87 mol Zn
 65.37 g Zn 
mol e– = 2(0.87 mol) = 1.74 mol e– (into the meter)
This is the number of moles of electrons that are used by the meter, but is only 8% of the total
moles of electrons that enter the house in 30 days; the total is:
 100% 
total mol e– = (1.74 mol e–) 
= 21.8 mol e– ≈ 22 mol e–
 8% 
To find the average current (in ampere) used during the month, convert the total moles of
electrons to coulombs (using the Faraday constant) and divide by the number of seconds in 30
days:
current =
22 mol9.65  10 4
C  mol–1

30 da 24 hr  da –1 60 min  hr –1 60 s  min –1 
= 0.82 A
(b) In order to operate, the cathode and anode in Edison’s meter had to retain these identities
during the entire time of the measurement, months and years. Since the current he supplied was
direct current, this criterion was met. Now, however, with alternating current, the electrodes in
an Edison meter would be changing identities sixty times per second. For half of each second,
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Reduction-Oxidation: Electrochemistry
one of the electrodes would be the cathode where Zn2+ is reduced to Zn and for the other half
second would be the anode where Zn is oxidized to Zn2+. The net effect would be no plating of
the zinc, so the meter would be useless (as well as pretty inconvenient).
Problem 10.9.
We look at each of these unbalanced reactions to see which species on the left is oxidized (the
anode reaction) and which reduced (the cathode reaction). All the species are metals and their
cations, so it is easy to tell which are the oxidized forms (cations). The net cell reaction is then
obtained by adding the two half reactions after each is multiplied by the appropriate
stoichiometric factor to cancel the electrons. Reactions are shown as reversible, since we have no
information about the direction of the reactions. The conventional cell notation is written with
the anode on the left.
(a)
anode reaction:
cathode reaction:
net cell reaction:
cell notation:
Mn(s)  Mn2+(aq) + 2e–
Ti2+(aq) + 2e–  Ti(s)
Mn(s) + Ti2+(aq) æ Mn2+(aq) + Ti(s)
Mn(s) | Mn2+(aq) || Ti2+(aq) | Ti(s)
(b)
anode reaction:
cathode reaction:
net cell reaction:
cell notation:
U(s)  U3+(aq) + 3e–
V2+(aq) + 2e–  V(s)
2U(s) + 3V2+(aq)  2U3+(aq) + 3V(s)
U(s) | U3+(aq) || V2+(aq) | V(s)
(c)
anode reaction:
cathode reaction:
net cell reaction:
cell notation:
Zn(s)  Zn2+(aq) + 2e–
Ni2+(aq) + 2e–  Ni(s)
Zn(s) + Ni2+(aq)  Zn2+(aq) + Ni(s)
Zn(s) | Zn2+(aq) || Ni2+(aq) | Ni(s)
(d)
anode reaction:
cathode reaction:
net cell reaction:
cell notation:
Mg(s)  Mg2+(aq) + 2e–
Cr3+(aq) + 3e–  Cr(s)
3Mg(s) + 2Cr3+(aq)  3Mg2+(aq) + 2Cr(s)
Mg(s) | Mg2+(aq) || Cr3+(aq) | Cr(s)
Problem 10.10.
The left-hand half of the line notation represents the anode reaction, so the metal shown there is
oxidized to the cation in the anode reaction. The cation in the right-hand half of the line notation
is reduced to the metal in the cathode reaction. The net cell reaction is then obtained by adding
the two half reactions after each is multiplied by the appropriate stoichiometric factor to cancel
the electrons and also canceling species, if any, that appear on both sides of the reaction
equation.
(a)
anode reaction:
cathode reaction:
net cell reaction:
Cu(s)  Cu2+(aq) + 2e–
Cu+(aq) + e–  Cu(s)
2Cu+(aq)  Cu(s) + Cu2+(aq)
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(b)
anode reaction:
cathode reaction:
net cell reaction:
Co(s)  Co2+(aq) + 2e–
Ag+(aq) + e–  Ag(s)
Co(s) + 2Ag+(aq)  Co2+(aq) + 2Ag(s)
(c)
anode reaction:
cathode reaction:
net cell reaction:
Al(s)  Al3+(aq) + 3e–
Fe2+(aq) + 2e–  Fe(s)
2Al(s) + 3Fe2+(aq)  2Al3+(aq) + 3Fe(s)
(d)
anode reaction:
cathode reaction:
net cell reaction:
Pb(s)  Pb2+(aq) + 2e–
Cu2+(aq) + 2e–  Cu(s)
Pb(s) + Cu2+(aq)  Pb2+(aq) + Cu(s)
Chapter 10
Problem 10.11.
In this manganese(II)-chromium(II) cell, the manganese electrode is negative, so electrons must
be leaving the cell at this electrode. Since manganese metal is being oxidized (leaving its
electrons behind to enter the external part of the circuit), the Mn(s)–Mn2+(aq) half cell must be
the anode of the cell:
anode reaction:
Mn(s)  Mn2+(aq) + 2e–
cathode reaction: Cr2+(aq) + 2e–  Cr(s)
net cell reaction:
Mn(s) + Cr2+(aq)  Mn2+(aq) + Cr(s)
cell notation:
Mn(s) | Mn2+(aq) || Cr2+(aq) | Cr(s)
Since the positive charge increases in the anode half cell (as Mn2+(aq) cations are produced) and
decreases in the cathode half cell (as cations are used up), the sulfate anions must migrate from
the cathode to the anode half cell, in order to maintain charge neutrality in the solutions.
Electrons are produced at the anode and used up at the cathode, so the flow of electrons in the
wire connecting the electrodes is from the manganese metal electrode to the chromium metal
electrode.
Problem 10.12.
(a) The observations described in this problem statement suggest that some product is darkening
the surface of the Mg metal and that the concentration of Cu2+ is decreasing in the solution.
(Solutions containing Cu2+ are blue and the color is fading.) A redox reaction that would account
for these observations is:
Cu2+(aq) + Mg(s)  Cu(s) + Mg2+(aq)
The darkening of the Mg surface is caused by the solid Cu reduced by the Mg metal at its
surface. Mg2+ in solution is colorless, so we have no proof that it is being formed, but this
reaction is analogous to the reaction between Cu and Ag+ (Investigate This 6.62 and 10.14).
(b) This is a sketch of an electrochemical cell that would take advantage of the reaction in part
(a) to provide a flow of electrons in an external circuit plus its cathodic and anodic half reactions,
net cell reaction, and line notation are:
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Chapter 10
anode reaction:
cathode reaction:
net cell reaction:
cell notation
Reduction-Oxidation: Electrochemistry
Mg(s)  Mg2+(aq) + 2e–
Cu2+(aq) + 2e–  Cu(s)
Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
Mg(s) | Mg2+(aq) || Cu2+(aq) | Cu(s)
Problem 10.13.
(a) The observations described in this problem statement suggest that chromium metal can react
with iron(II) cations to reduce the cations to iron metal (the dark deposit on the metal surface).
Concomitantly, chromium metal atoms must be oxidized to chromium(III) cations. A cell that
can take advantage of this spontaneous reaction is a chromium metal-chromium(III) cation anode
and iron(II) cation-iron metal cathode connected by a salt bridge:
(b) The half reactions and net cell reaction for this cell are:
anode reaction:
Cr(s)  Cr3+(aq) + 3e–
cathode reaction: Fe2+(aq) + 2e–  Fe(s)
net cell reaction:
2Cr(s) + 3Fe2+(aq)  2Cr3+(aq) + 3Fe(s)
(c) The line notation for this cell is:
Cr(s) | Cr3+(aq) || Fe2+(aq) | Fe(s)
The rationale for the anode and cathode reactions is given in part (a) and the conventional cell
notation follows naturally with the anode written on the left.
Problem 10.14.
(a) In order to maintain the neutrality of the solutions around the electrodes in an electrochemical
cell, ions have to move in the solutions, as in Figure 10.4. If electrons are drawn too quickly
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Chapter 10
from the cell, the ions may not be able to move fast enough to prevent charge separation or
polarization of the cell. Charge separation like this always opposes the flow of electrons (since
the separation is set up by the reactions that cause the flow). If the polarization becomes too
great, the flow of electrons falls to an unusable level and the battery appears to have “run down,”
that is, come to equilibrium. Allowing the battery to rest for some time gives the ions inside a
chance to move and eliminate the charge separation. The battery can then be used again, because
the opposing charge separation is no longer present.
(b) The slow movement of ions inside the battery is exactly the condition that will favor
polarization, if the reactions are required to go quickly to produce a lot of electrons.
Problem 10.15.
(a) Since the zinc metal electrode in these simple galvanic cells loses mass, the zinc atoms must
be being oxidized to zinc cations, so the zinc electrode is the anode of the cell. A reduction
reaction must be going on at the copper electrode (cathode). Water is the common ingredient in
all the cells discussed in the problem, so it is likely to be water that is reduced. We know that the
reduction of water produces hydrogen gas, so this fits with the observed production of hydrogen
gas at the copper cathode. The half reactions are:
anode (zinc) reaction:
Zn(s)  Zn2+(aq) + 2e–
cathode (copper) reaction: 2H2O(l) + 2e–  H2(g) + 2OH–(aq)
(b) The net cell reaction is:
Zn(s) + 2H2O(l)  Zn2+(aq) + H2(g) + 2OH–(aq)
(c) The line notation for the cell is:
Zn(s) |Zn2+(aq), ionic solution in H2O(l), OH–(aq) |H2(g) |Cu(s)
The choice of anode and cathode for this cell is described in part (a). Note that the cell has no
salt bridge since the two electrodes are in the same solution. An aqueous ionic solution is
specified, since the water is required as a reactant and the ions as charge carriers to complete the
circuit within the cell.
Problem 10.16.
(a) Consider an electrochemical cell in which a zinc and carbon electrode dip into the same
solution and chlorine gas bubbles into the cell solution and reacts at the surface of a graphite
(carbon) electrode. The cell is based on this spontaneous reaction:
Zn(s) + Cl2(g)  Zn2+(aq) + 2Cl–(aq)
The half reactions are:
oxidation (anode): Zn(s)  Zn2+(aq) + 2e–
reduction (cathode): Cl2(g) + 2e–  2Cl–(aq)
(b) The zinc electrode is the anode in this cell, since, as shown in part (a), the oxidation reaction
occurs at that electrode. The carbon electrode is the cathode, since the electrons are furnished to
the chlorine molecules to reduce them to chloride ions as they bubble against its surface.
(c) Initially, the cell solution can be essentially any aqueous ionic solution (to provide charge
carriers). As the cell reaction proceeds, Zn2+(aq) and Cl–(aq) ions are produced in a one-to-two
ratio and the solution becomes a solution of zinc chloride dissolved in the aqueous ionic solution,
which is shown in this sketch of the cell and its line notation:
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Reduction-Oxidation: Electrochemistry
Zn(s) | Zn2+(aq), Cl–(aq), aqueous ionic solution | Cl2(g) | C(s)
(d) Electrons move from the anode to the cathode in the external part of the circuit, so they move
from the zinc to the carbon electrode.
(e) The number of moles of electrons required to produce a 0.12 A current for exactly 7 days is:
it
(0.12 A)(7 day)(24 hr  day –1 )(60 min  hr –1 )(60 s  min –1 )
mol e =
=
F
9.65  10 4 C  mol–1
mol e– = 0.75 mol
One mole of chlorine molecules requires two moles of electrons for its reduction, so 0.38 mol
[= 0.5·(0.75 mol)] of chlorine molecules is required to produce the current in this case.
–
Problem 10.17.
The measured voltage of an electrochemical cell made by connecting a Fe2+|Fe half cell with a
Pb2+|Pb half cell is positive when the anode input from the voltmeter is connected to the lead
metal electrode. The Pb electrode is the anode, so oxidation of Pb(s) to Pb2+(aq) must be
occurring at this electrode. The cell reaction consistent with this observation is:
Pb(s) + Fe2+(aq)  Pb2+(aq) + Fe(s)
You could test the direction of this reaction by immersing a clean piece of Pb metal in a solution
of Fe2+. Discoloration of the Pb surface is an indication that some reaction is occurring to affect
the Pb. You could test the solution after some time to see whether it contains Pb2+ ions. You
could also place a very tiny sliver of Pb in Fe2+ solution to see if it disappears. In all cases, you
need to do control experiments with water and with alkali metal solutions of the anions, to be
sure that it is the Fe2+ that is responsible for your observations.
Problem 10.18.
Both cell reactions shown involve the silver-silver ion half reaction, so they can be combined to
cancel the silver species, leaving the desired cell reaction:
Ni(s) + 2Ag+(aq)  Ni2+(aq) + 2Ag(s)
Eo = 1.03 V
+
2+
–[Cu(s) + 2Ag (aq)  Cu (aq) + 2Ag(s)
Eo = 0.46 V]
Ni(s) + Cu2+(aq)  Ni2+(aq) + Cu(s)
ACS Chemistry FROG
Eo = 0.57 V
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Chapter 10
Problem 10.19.
(a) When a clean strip of lead is placed in a 0.2 M aqueous solution of silver nitrate, the surface
of the metal in contact with the solution soon turns dark and then small silvery needles begin to
grow on the surface. This behavior is similar to what you observe when a strip of copper metal is
immersed in an aqueous solution of silver nitrate (Investigate This 6.62 and 10.14.). What will
happen if a strip of copper is placed in a solution of lead nitrate? What will happen if a strip of
lead is placed in a solution of copper nitrate? We know that both copper metal and lead metal
reduce silver cations to silver metal while themselves being oxidized to copper(II) and lead(II)
cations, respectively. Unless both metals have the same potential to bring about this reduction,
one of them will be able to reduce cations of the other. That is, either lead metal can reduce
copper(II) cations or copper metal can reduce lead(II) cations, but we do not know which. Thus,
we cannot predict what will happen in the two experiments proposed.
(b) The cell data given show that the potentials for lead and copper to reduce silver cations are
not the same. The potential of the cell with lead as the anode is more positive, so lead has the
greater ability to reduce silver cations. We can subtract the copper-silver cell from the lead-silver
cell to get the potential and cell notation for the lead-copper cell:
Pb(s) | Pb2+(aq, 0.2 M) || Ag+(aq, 0.2 M) | Ag(s)
E = 0.90 V
2+
+
– [Cu(s) | Cu (aq, 0.2 M) || Ag (aq, 0.2 M) | Ag(s)
E = 0.44 V]
Pb(s) | Pb2+(aq, 0.2 M) || Cu2+(aq, 0.2 M) | Cu(s)
E = 0.46 V
(c) With the knowledge from part (b), we see that the oxidation of lead metal by copper(II)
cations (or equivalently, the reduction of copper(II) cations by lead metal) is a favored process
(positive cell potential), so now we can answer part (a). The reduction of copper(II) cations by
lead metal is the process that will occur. The surface of a piece of lead dipped into a solution of
copper(II) cations should darken as finely divided copper metal deposits on the surface. The
initially blue solution of Cu2+(aq) will get lighter as the concentration of the cation decreases.
Problem 10.20.
(a) The negative measured potential is an indication that the voltmeter is connected incorrectly to
the cell. That is, the lead electrode is the cathode and the nickel electrode is the anode. Thus, we
can write:
anode reaction:
Ni(s)  Ni2+(aq) + 2e–
cathode reaction: Pb2+(aq) + 2e–  Pb(s)
net cell reaction:
Ni(s) + Pb2+(aq)  Ni2+(aq) + Pb(s)
cell notation:
Ni(s) | Ni2+(aq) || Pb2+(aq) | Pb(s)
(b) The favored reaction in this system is the reduction of lead(II) ion by nickel metal. Therefore,
placing a strip of lead metal in a nickel(II) cation solution will result in no reaction. The lead
strip will get wet, but no other changes will be observed.
Problem 10.21.
From Appendix C, we find the standard reduction potentials for copper and magnesium ions and
combine the half reactions and standard reduction potentials to get the cell reaction and standard
cell potential for the cell sketched in Problem 10.12:
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Cu2+ + 2e–  Cu
–{Mg2+ + 2e–  Mg}
Eo = 0.337 V
Eo = –(–2.38 V)
Cu2+ + Mg  Cu + Mg2+
Eo = 2.72 V
Problem 10.22.
(a) In Appendix B, we find that the standard reduction potentials are 0.799 V for the Ag+|Ag half
cell and –0.41 V for the Fe2+|Fe half cell. Thus, the more positive silver reaction will proceed as
written as a reduction and the iron reaction will proceed in reverse as an oxidation. The iron
metal in the Fe2+|Fe half cell will be the anode and the silver metal in the Ag+|Ag half cell will be
the cathode of this cell. The assumption we make to get these predictions is that the sign of the
potential for the actual cell (with concentrations that might not be standard concentrations) will
be the same as the sign of the standard cell potential. Since the standard cell potential is so large
[see part (b)], it is unlikely that any reasonable concentrations in the half cells will change its
sign. The line notation for this cell is:
Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s)
(b) To get the net cell reaction, combine the half-cell reactions appropriately to cancel out
electrons. To get the standard cell potential from the standard reduction potentials, use the
equation from Figure 10.9.
2·[Ag+(aq) + e–  Ag(s)]
– [Fe2+(aq) + 2e–  Fe(s)]
Fe(s) + 2Ag+(aq)  Fe2+(aq) + 2Ag(s)
E° = Eº(Ag+, Ag) – Eo(Fe2+, Fe) = 0.799 V – (–0.41 V) = 1.20 V
Problem 10.23.
The table of standard reduction potentials in Appendix B is labeled showing that the strongest
reducing agents (the species on the right-hand side of the reduction half reaction) are those at the
end of the list, that is, those whose formation has the most negative standard reduction potential.
This makes sense, since the driving force to go in reverse (giving up electrons and reducing other
species) is the largest. The half reactions and standard reduction potentials of interest in this
problem are:
Al3+(aq) + 3e–  Al(s)
E° = –1.66 V
2+
–
Ca (aq) + 2e  Ca(s)
E° = –2.76 V
2+
–
Mg (aq) + 2e  Mg(s) E° = –2.38 V
Na+(aq) + e–  Na(s)
E° = –2.71 V
2+
–
Zn (aq) + 2e  Zn(s)
E° = –0.763 V
Thus the order of increasing strength as reducing agents (and increasing |E°|) is:
Zn < Al < Mg < Na < Ca (with Na and Ca being very nearly the same)
Problem 10.24.
The table of standard reduction potentials in Appendix B is labeled showing that the strongest
oxidizing agents (the species on the left-hand side of the reduction half reaction) are those at the
beginning of the list, that is, those whose reaction has the most positive standard reduction
ACS Chemistry FROG
79
Reduction-Oxidation: Electrochemistry
Chapter 10
potential. This makes sense, since the driving force to go as written (taking up electrons and
oxidizing other species) is the largest. The half reactions and standard reduction potentials of
interest in this problem are:
Br2(l) + 2e–  2Br–(aq)
E° = 1.06 V
–
–
Cl2(g) + 2e  2Cl (aq)
E° = 1.359 V
–
–
F2(g) + 2e  2F (aq)
E° = 2.87 V
I2(s) + 2e–  2I–(aq)
E° = 0.5355 V
+
–
O2(g) + 4H (aq) + 4e  2H2O(l) E° = 1.230 V
Thus the order of increasing strength as oxidizing agents (and increasing E°) is:
I2 < Br2 < O2 < Cl2 < F2 < Cl2
Problem 10.25.
(a) Chromium metal, Cr(s), reduces lead(II) ion, Pb2+(aq) to lead metal, Pb(s). The reduction half
reaction of lead(II) cation has a more positive standard reduction potential than that for the
reduction half reaction of chromium(III), so the process is spontaneous with this net cell reaction
and standard cell potential:
2Cr(s) + 3Pb2+(aq)  2Cr3+(aq) + 3Pb(s)
E° = E°(Pb2+, Pb) – E°(Cr3+, Cr) = –0.13 V – (–0.74 V) = 0.61 V
(b) In basic solution, mercury metal, Hg(l), reduces cadmiun hydroxide, Cd(OH)2(s), to
cadmium metal, Cd(s), and forms mercury(II) oxide, HgO(s). The reduction half reaction of
mercury(II) oxide to mercury metal has a more positive standard reduction potential than that for
the reduction half reaction of cadmium hydroxide to cadmium metal. The process in the
statement is not spontaneous. In basic solution, HgO(s) oxidizes Cd(s) to Cd(OH)2(s) and forms
Hg(l) with this net cell reaction and standard cell potential:
Cd(s) + HgO(s) + H2O(l)  Cd(OH)2(s) + Hg(l)
E° = E°(HgO, Hg) – Eo(Cd(OH)2, Cd) = 0.098 V – (–0.81 V) = 0.91 V
(c) Nickel(II) ion, Ni2+(aq), is reduced to nickel metal, Ni(s), by hydrogen gas, H2(g) in acidic
solution. The reduction half reaction of hydronium ion to hydrogen gas has a more positive
standard reduction potential than that for the reduction half reaction of nickel cation to nickel
metal. The process in the statement is not spontaneous. In acidic solution, H3O+(aq) oxidizes
Ni(s) to Ni2+(aq) and forms H2(g) with this net cell reaction and standard cell potential:
Ni(s) + 2H3O+(aq)  Ni2+(aq) + H2(g) + 2H2O(l)
E° = E°(H3O+, H2) – E°(Ni2+, Ni) = 0.000 V – (–0.23 V) = 0.23 V
(d) Sodium metal, Na(s), reduces water, H2O(l), at pH 7 to form hydrogen gas H2(g). The
reduction half reaction of water (hydronium ion) at pH 7 has a more positive standard reduction
potential than that for the reduction half reaction of sodium cation, so the process is spontaneous
with this net cell reaction and standard cell potential:
2Na(s) + 2H3O+(aq)  2Na+(aq) + H2(g) + 2H2O(l)
E°´ = E°´(H3O+, H2) – E°(Ni2+, Ni) = –0.414 V – (–2.71 V) = 2.30 V
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Problem 10.26.
(a) We interpret what is going on according to the changes in color observed upon mixing the
solutions and write the redox reaction, if any, that occurs in each case.
(i) yellow green + colorless  orange solution: Br2 is formed in the solution
Cl2(aq) + 2Br–(aq)  2Cl–(aq) + Br2(aq)
(ii) yellow green + colorless  red solution: I2 is formed in the solution
Cl2(aq) + 2I–(aq)  2Cl–(aq) + I2(aq)
(iii) orange + colorless  orange solution: no apparent reaction
Br2(aq) + 2Cl–(aq)  2Br–(aq) + Cl2(aq) DOES NOT OCCUR
(iv) orange + colorless  red solution: I2 is formed in the solution
Br2(aq) + 2I–(aq)  2Br–(aq) + I2(aq)
The standard reduction potentials for these three halogens are:
Cl2(aq) + 2e–  2Cl–(aq) E° = 1.359 V
Br2(aq) + 2e–  2Br–(aq) E° = 1.087 V
I2(aq) + 2e–  2I–(aq)
E° = 0.536 V
The strongest oxidizing agent (reactant with the most positive reduction potential) is Cl2, which
can oxidize both bromide and iodide to the halogen molecule (zero oxidation number), as shown
in reactions (i) and (ii). Bromine cannot oxidize chloride; the bromine reduction potential is
lower than that for the chlorine half reaction. Thus, mixture (iii) gave no apparent reaction.
Bromine can oxidize iodide and this is shown in case (iv).
(b) Since the reduction potential for iodine is lower than that for chlorine, iodine cannot oxidize
chloride. This would be the reverse of reaction (ii) and we know from the experimental results
that reaction (ii) is the favored direction. If you mixed a red I2/I–(aq) solution with a colorless Cl–
(aq) solution the resulting mixture would retain the red color, as no reaction would occur.
(c) Since the reduction potential for chlorine is greater than that for bromine, chlorine can
oxidize bromide. Thus, if you bubble chlorine gas, Cl2(g), into an aqueous solution of sodium
bromide, NaBr(aq) [which contains Br–(aq)], the initially clear, colorless solution will become
orange as Br2(aq) is produced and forms Br2/Br–(aq). This is the reverse of reaction (iii) in part
(a), which we found experimentally does not procede as proposed. Now we see why and predict
that the reverse reaction, 2Br–(aq) + Cl2(aq)  Br2(aq) + 2Cl–(aq), will be observed (and it is).
Problem 10.27.
The data in Appendix B show that oxygen gas is a stronger oxidant than elemental bromine, so it
should be possible, in acidic solution, to oxidize bromide ion to bromine by bubbling oxygen
through a brine solution containing bromide ion:
4Br–(aq) + O2(g) + 4H+(aq)  2Br2(l) + 2H2O(l)
E° = E°(O2, H2O) – E°(Br2, Br–) = 1.229 V – (1.087 V) = 0.142 V
This reaction is apparently relatively slow (possibly because the interaction of the gaseous
oxygen with the ions in solution is inefficient), so this is not a commercially useful way to
extract the bromine.
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Problem 10.28.
The assumptions we have to make to get an estimate of the cell potential for the reaction between
molten aluminum chloride, AlCl3, and sodium metal are rather bold and difficult to justify, but
will probably rationalize the direction of the reaction. Assume that AlCl3 is ionized to Al3+ and
Cl– in its liquid, and that the aluminum ion acts as if it is in aqueous solution. Assume that we
can ignore the formation of the solid sodium chloride and make our analysis as though the
product is sodium ion in aqueous solution. Obviously, these assumptions are all suspect, but the
only data we have in Appendix B are for aqueous solutions. We have to use them to see what
they predict, knowing that the actual system is quite different. In these reactions, the state of the
species has been omitted, because of the great uncertainty about what it is in this system:
Al3+ + 3e–  Al
Eo = –1.66 V
+
–
–3{Na + e  Na}
Eo = –(–2.71 V)
Al3+ + 3Na  Al + 3Na+ Eo = 1.05 V
The cell potential is favorable for this reaction. Formation of solid products that probably
separate from the molten reaction mixture will also favor the formation of the products. This
reaction is not one that can be run readily on a large scale, so the amount of pure aluminum metal
available before the end of the nineteenth century was small. (Nowadays, aluminum beverage
cans are so inexpensive and ubiquitous that you might even be tempted to discard them rather
than recycle. Resist the temptation; it is still the case that a great deal of energy is required to
reduce aluminum ions to aluminum metal, as its reduction potential shows, and recycling saves
this energy—as well as the metal.)
Problem 10.29.
(a) The reduction half reaction of chlorine gas to chloride anion has a more positive standard
reduction potential, 1.359 V, than that for the reduction half reaction of zinc cation to zinc metal,
–0.763 V. Therefore, in a cell made by placing a sheet of zinc metal, Zn(s), in an aqueous
solution of zinc sulfate, ZnSO4(aq), that is connected by a salt bridge to an aqueous solution of
sodium chloride, NaCl(aq), containing a coil of platinum wire, Pt(s), over which chlorine gas,
Cl2(g), is bubbled, the platinum electrode in contact with the chlorine gas and chloride ion
solution will be the cathode (where chlorine gas is reduced) and the zinc electrode will be the
anode (where zinc metal is oxidized to zinc cation).
(b) The explanation for these half reactions is in part (a):
anode reaction:
Zn(s)  Zn2+(aq) + 2e–
cathode reaction: Cl2(g) + 2e–  2Cl–(aq)
net cell reaction:
Zn(s) + Cl2(g)  Zn2+(aq) + 2Cl–(aq)
(c) The line notation for the cell and its standard potential are:
Zn(s) | Zn2+(aq) || Cl–(aq) | Cl2(g) | Pt(s)
E° = E°(Cl2, Cl–) – E°(Zn2+, Zn) = 1.359 V – (–0.763 V) = 2.122 V
Problem 10.30.
(a) The relevant reduction half reactions and their combination for the net reaction for the
titration of Fe2+(aq) with MnO4–(aq) are:
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MnO4–(aq) + 8H+(aq) + 5e–  Mn2+(aq) + 4H2O(l)
–5·[Fe3+(aq) + e–  Fe2+(aq) ]
5Fe2+(aq) + MnO4–(aq) + 8H+(aq)  5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
(b) The standard cell potential for the reactions in part (a) is:
E° = E°(MnO4–, Mn2+) – E°(Fe3+, Fe2+) = 1.507 V – (0.771 V) = 0.736 V
(c) Since the permanganate reduction from Mn(VII) to Mn(II) is a five electron change and the
oxidation of iron(II) to iron(III) is a one electron change, it takes one-fifth, 0.200, mol of
permanganate to oxidize 1.000 mol of iron. Or, put the other way, one mole of permanganate
reacts with five moles of iron(II). If 23.46 mL of 0.0200 M KMnO4(aq) is required to titrate a
sample containing iron(II) ion, the number of moles of iron in the sample is:
mol Mn(VII) = (0.02346 L)(0.0200 mol·L–1) = 4.69  10–4 mol
 5 mol Fe(II) 
mol Fe(II) = [4.69  10–4 mol Mn(VII)] 
= 2.35  10–3 mol Fe(II)
 1 mol Mn(VII) 
Problem 10.31.
(a) The relevant reduction half reactions and their combination to give the net reaction producing
chlorine gas by dropping hydrochloric acid solution, HCl(aq), onto crystals of potassium
permanganate, KMnO4(s) are:
2·[MnO4–(aq) + 8H+(aq) + 5e–  Mn2+(aq) + 4H2O(l)]
–5·[Cl2(g) + 2e–  2Cl–(aq)]
10Cl–(aq) + 2MnO4–(aq) + 16H+(aq)  5Cl2(g) + 2Mn2+(aq) + 8H2O(l)
(b) The standard cell potential for this reaction is:
E° = E°(MnO4–, Mn2+) – E°(Cl2, Cl–) = 1.507 V – (1.359 V) = 0.148 V
Problem 10.32.
(a) Will reaction occur between the pair of reactants Fe3+(aq) and Br–(aq)? The standard
reduction potential for the iron(III) cation to iron(II) cation half reaction, 0.771 V, is more
negative than that for the elemental bromine to bromide anion half reaction, 1.087 V, so no
reaction will occur between Fe3+(aq) and Br–(aq).
(b) Will reaction occur between the pair of reactants Fe3+(aq) and I–(aq)? The standard reduction
potential for the iron(III) cation to iron(II) cation half reaction, 0.771 V, is more positive than
that for the elemental iodine to iodide anion half reaction, 0.5355 V, so reaction can occur
between Fe3+(aq) and I–(aq). The net reaction and its standard cell potential are:
2Fe3+(aq) + 2I–(aq)  2Fe2+(aq) + I2(s)
E° = E°(Fe3–, Fe2+) – E°(I2, I–) = 0.771 V – (0.5355 V) = 0.235 V
(c) Will reaction occur between the pair of reactants Fe2+(aq) and Br2(aq)? The standard
reduction potential for the elemental bromine to bromide anion half reaction, 1.087 V, is more
positive than that for the iron(III) cation to iron(II) cation half reaction 0.771 V, so reaction can
occur between Fe2+(aq) and Br2(aq). The net reaction and its standard cell potential are:
Br2(aq) + 2Fe2+(aq)  2Br–(aq) + 2Fe3+(aq)
E° = E°(Br2, Br–) – E°(Fe3+, Fe2+) = 1.087 V – (0.771 V) = 0.316 V
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(d) Will reaction occur between the pair of reactants Fe2+(aq) and Ag+(aq)? The standard
reduction potential for the silver cation to silver metal half reaction, 0.799 V, is more positive
than that for the iron(III) cation to iron(II) cation half reaction, 0.771 V, so reaction can occur
between Fe2+(aq) and Ag+(aq). The net reaction and its standard cell potential are:
Ag+(aq) + Fe2+(aq)  Ag(s) + Fe3+(aq)
E° = E°(Ag+, Ag) – E°(Fe3+, Fe2+) = 0.799 V – (0.771 V) = 0.028 V
Problem 10.33.
(a) Will the reaction, Cu2+(aq) + H2O2(aq)  Cu(s) + O2(g), occur? The standard reduction
potential for the copper(II) cation to copper metal half reaction, 0.337 V, is more negative than
that for the elemental oxygen to hydrogen peroxide half reaction, 0.69 V, so no reaction will
occur between Cu2+(aq) and H2O2(aq).
(b) Will the reaction, Cu(s) + NO3–(aq) + H+(aq)  Cu2+(aq) + NO(g), occur? The standard
reduction potential for the nitrate cation to nitric oxide half reaction, 0.955 V, is more positive
than that for the copper(II) cation to copper metal half reaction, 0.337 V, so reaction can occur
between Cu(s) and NO3–(aq) in acidic solution. The balanced net reaction and its standard cell
potential are:
3Cu(s) + 2NO3–(aq) + 8H+(aq)  3Cu2+(aq) + 2NO(g) + 4H2O(aq)
E° = E°(NO3–, NO) – E°(Cu2+, Cu) = 0.955 V – (0.337 V) = 0.618 V
(c) Will the reaction, MnO2(s) + I–(aq)  Mn2+(aq) + I2(aq), occur? The standard reduction
potential for the manganese dioxide to manganese(II) cation half reaction, 1.230 V, is more
positive than that for the elemental iodine to iodide ion half reaction, 0.5355 V, so reaction can
occur between MnO2(s) and I–(aq) in acidic solution. The balanced net reaction and its standard
cell potential are:
MnO2(s) + 2I–(aq) + 4H+(aq)  Mn2+(aq) + I2(aq) + 2H2O(aq)
E° = E°(MnO2, Mn2+) – E°(I2, I–) = 1.230 V – (0.5355 V) = 0.694 V
(d) Will the reaction, Cr2O72–(aq) + CH3CH2OH(aq) + H+(aq)  Cr3+(aq) + CH3CHO(aq),
occur? The standard reduction potential for the dichromate anion to chromium(III) cation half
reaction, 1.36 V, is more positive than that shown for the ethanal (acetaldehyde) to ethanol half
reaction, –0.20 V, so reaction can probably occur between Cr2O72–(aq) and CH3CH2OH(aq). The
ambiguity in this case arises because the potential for the dichromate reduction is given for
acidic solution, with pH = 0, and the ethanal potential is for the reaction at pH = 7. Le Chatelier’s
principle predicts that the ethanal reduction potential will become more positive (closer to that of
the dichromate reduction) as the pH decreases, since the reaction involves hydronium ions as
reactants. The difference between the two potentials is so large that this effect is not likely to
change our prediction of the direction of the overall reaction. This reaction is, in fact, the basis
for the Breathalyzer® test for alcohol in a person’s breath, so it does work and goes in the
direction we write here. The balanced net reaction and its standard cell potential are:
3CH3CH2OH(aq) + Cr2O72–(s) + 8H+(aq)  3CH3CHO(aq) + 2Cr3+(aq) + 7H2O(aq)
E° = E°(Cr2O72–, Cr3+) – E°(ethanal, ethanol) = 1.36 V – [> (–0.20 V)] = < 1.56 V
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Problem 10.34.
(a) For the cell reaction, Zn(s) + Pb2+(aq)  Zn2+(aq) + Pb(s), use the measured cell potential,
0.660 V, to find the free energy change (with n =2, since the reaction is a two-electron transfer)
per mole of reaction:
Greaction = –nFE = – 2·(9.6485  104 C·mol–1)(0.660 V) = –127.4 kJ·mol–1
The reaction is spontaneous. The free energy change is negative and the cell potential is positive,
either of which can be taken as an indication of spontaneity.
(b) The negative of the free energy change gives the maximum amount of work that can be done
by the cell on its surroundings. In part (a) we calculated the free energy change for a mole of
reaction (equivalent in this case to a mole of zinc metal reacting). Thus, for the reaction of 0.125
mol Zn(s):
maximum work = – (0.125 mol)(–127.4 kJ·mol–1) = 15.9 kJ
(c) For every mole of Zn(s) that reacts a mole of Pb(s) is formed, so 0.125 mol of lead metal
forms when the work in part (b) is produced:
mass Pb(s) = (0.125 mol)(207.2 g·mol–1) = 25.9 g
Problem 10.35.
(a) In a copper-zinc cell, each Zn atom that goes into solution as Zn2+ leaves behind two
electrons for the external circuit. Therefore, the number of moles of electrons produced is twice
the number of moles of Zn that react. The molar mass of Zn is 65.37 g·mol–1 and the mass of Zn
that has reacted in the cell described here is 270.7 g (= 486.5 – 215.8). The moles of Zn reacted
and e– produced are:
 1 mol Zn 
mol Zn reacted = (270.7 g) 
= 4.141 mol Zn
 65.37 g Zn 
mol e– produced = 2(mol Zn) = 8.282 mol e–
(b) The Faraday constant, F, is the amount of charge on one mole of electrons, so the total
charge transferred during the lifetime of this cell is:
charge transferred = (8.282 mol)(96485 C·mol–1) = 7.991  105 C
To get the amount of work that could be produced by this amount of charge, we need to know
the cell potential, E, for this cell. Figure 10.6 shows a copper-zinc cell with a cell potential of
1.10 V (with 0.10 M solutions for both metal ions). From the data in Appendix B we find that the
standard cell potential is:
E° = E°(Cu2+, Cu) – E°(Zn2+, Zn) = 0.337 V – (–0.763 V) = 1.100 V
It looks like 1.10 V is a reasonable approximation for the cell potential to get an estimate of the
work that the cell can produce. (If you jump ahead to Problem 10.74, you find that one version of
a cell like this, often called a Daniell cell, has an intial potential of 1.15 V that decreases slowly
over its lifetime.)
work = (charge transferred)(cell potential) = (7.991  105 C)(1.10 V) ≈ 8.8  102 kJ
(Note that this is enough energy to lift a loaded moving van to the top of a one-story building.)
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Problem 10.36.
(a) In the cell described in this problem, the cell potential is positive when the anode lead from a
digital voltmeter is connected to the platinum wire, so the platinum wire is the anode of the cell
and the reaction there must be the oxidation of hydrogen gas (bubbled over the wire) to
hydronium ions. The cathode must be the silver wire and the cathode reaction must be the
reduction of the silver cation in solid silver chloride to silver metal with the release of chloride
anion into the solution. The line notation for this cell is:
Pt | H2(g, 1 bar) | H+(aq, 1 M), Cl–(aq, 1 M) | AgCl(s) | Ag(s)
(b) The rationale for the choices of reaction are given in part (a) and based on the makeup of the
cell and the identity of the anode. The reduction half reactions and cell reaction are:
anode reaction:
2H+(aq) + 2e–  H2(g)
cathode reaction: AgCl(s) + e–  Ag(s) + Cl–(aq)
net cell reaction:
H2(g) + 2AgCl(s)  2H+(aq) + 2Ag(s) + 2Cl–(aq)
(c) All of the species in the cell are present at their standard concentrations (or pressure), so the
measured potential, 0.22 V, is the standard potential, E, for this cell and the standard free energy
change for one mole of cell reaction is:
Greaction = –nFE = – 2·(9.6485  104 C·mol–1)(0.22 V) = –42 kJ·mol–1
The free energy change is for one mole of hydrogen gas and two moles of silver chloride solid
reacting as shown in the net cell reaction, which involves two moles of electrons.
(d) The anode of this cell is the standard hydrogen electrode, SHE, which, by definition, has a
reduction potential of zero volts. Other reduction potentials are determined by measuring the cell
potential of a cell made by combining the appropriate half cell with the SHE (taken as the
anode). This is the way this cell is set up, so the standard reduction potential for the AgCl|Ag
electrode is 0.22 V, which is the value you find (to two significant figures) in Appendix B.
Problem 10.37.
(a) In the reaction, O2(g) + 2H2S(aq)  2H2O(l) + 2S(s), the oxygen atoms in O2 are being
reduced from an oxidation number of zero to an oxidation number of –2 in water. The opposite is
the case for sulfur atoms, which are being oxidized from an oxidation number of –2 to 0. The
reduction half reactions for O2 and for S are:
O2(g) + 4H+(aq) + 4e–  2H2O(l)
S(s) + 2H+(aq) + 2e–  H2S(aq)
The hydronium ions cancel out when the two half reactions are combined (first minus double the
second) to give the overall reaction.
(b) The standard cell potential for the reaction can be calculated from the standard free energy
given for the reaction:
–418.8  10 J mol  = 1.085 V
G o
E° = –
=–
nF
4 96485 C mol –1 
3
–1
Problem 10.38.
(a) The redox reaction between oxalic acid and permanganate ion is:
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5HOOCCOOH(s) + 2MnO4–(aq) + 6H+(aq)  10CO2(g) + 2Mn2+(aq) + 8H2O(l)
The standard free energy change for this redox reaction is:
G°rxn = (10 mol)G°f(CO2) + (2 mol)G°f(Mn2+) + (8 mol)G°f(H2O)
– (5 mol)G°f(HOOCCOOH) – (2 mol)G°f(MnO4–) – (6 mol)G°f(H+)
G°rxn = 10(–394.36 kJ) + 2(–223 kJ) + 8(–237.13 kJ)
– 5(–697.9 kJ) – 2(–425 kJ) – 6(0 kJ)
G°rxn = –1.95  103 kJ
The reaction, with this very high negative standard free energy change, is definitely spontaneous
under standard conditions.
(b) The standard cell potential for this reaction is obtained by rearranging –G°rxn = nFE° to
solve for E°. The value of n is 10 in this case because two permanganate anions [containing
manganese(VII)] are reduced to two manganese(II) cations.
o
1.95 106 J
–Grxn
E° =
=
= 2.02 V
nF
10  (9.6485 104 C  mol –1)
(c) From part (b), we know the standard cell potential for the reaction that involves the oxidation
of oxalic acid by permanganate and we can find the standard reduction potential for the
permanganate anion reduction, 1.507 V, in Appendix B. These values can be combined to find
the standard reduction potential for the reduction of carbon dioxide gas to oxalic acid:
2CO2(g) + 2H+(aq) + 2e–  HOOCCOOH(s):
E° = 2.02 V = E°(MnO4–, Mn2+) – E°(CO2, oxalic acid) = 1.507 V – E°(CO2, oxalic acid)
E°(CO2, oxalic acid) = –0.51 V
Problem 10.39.
For each of these cells, we write the cathodic and anodic half cell reactions (as reductions), the
net cell reaction, the standard cell potential (from the reduction potentials in Appendix B), and
the standard free energy change (from G°rxn = –nFE°).
(a) Cr(s) | Cr2+(aq) || Cr3+(aq) | Cr(s)
cathodic:
2·[Cr3+(aq) + 3e–  Cr(s)]
anodic:
–3·[Cr2+(aq) + 2e–  Cr(s)]
net reaction: Cr(s) + 2Cr3+(aq)  3Cr2+(aq)
E° = E°(Cr3+, Cr) – E°(Cr2+, Cr) = –0.74 V – (–0.89 V) = 0.15 V
G°rxn = –nFE° = – 6·(9.6485  104 C·mol–1)(0.15 V) = –87 kJ·mol–1
(b) Pt(s) | Fe3+(aq),Fe2+(aq) || Cr2O72–(aq), Cr3+(aq) | Pt(s)
cathodic:
Cr2O72–(aq) + 14H+(aq) + 6e–  2Cr3+(aq) + 7H2O(l)
anodic:
–6·[Fe3+(aq) + e–  Fe2+(aq)]
net reaction: 6Fe2+(aq) + Cr2O72–(aq) + 14H+(aq)  6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
E° = E°(Cr2O72–, Cr3+) – E°(Fe3+, Fe2+) = 1.36 V – (0.771 V) = 0.59 V
G°rxn = –nFE° = – 6·(9.6485  104 C·mol–1)(0.59 V) = –3.4  102 kJ·mol–1
(c) Zn(s) | Zn2+(aq) || Fe2+(aq) | Fe(s)
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cathodic:
Fe2+(aq) + 2e–  Fe(s)
anodic:
–[Zn2+(aq) + 2e–  Zn(s)]
net reaction: Zn(s) + Fe2+(aq)  Zn2+(aq) + Fe(s)
E° = E°(Fe2+, Fe) – E°(Zn2+, Zn) = –0.41 V – (–0.763 V) = 0.35 V
G°rxn = –nFE° = – 2·(9.6485  104 C·mol–1)(0.35 V) = –68 kJ·mol–1
(d) Hg(l) | Hg2Cl2(s) | Cl–(aq) || Hg22+(aq) | Hg(l)
cathodic:
Hg22+(aq) + 2e–  2Hg(l)
anodic:
–[Hg2Cl2(s) + 2e–  2Hg(l) + 2Cl–(aq)]
net reaction: Hg22+(aq) + 2Cl–(aq)  Hg2Cl2(s)
E° = E°(Hg22+, Hg) – E°(Hg2Cl2, Hg) = 0.796 V – (0.268 V) = 0.528 V
G°rxn = –nFE° = – 2·(9.6485  104 C·mol–1)(0.528 V) = –101.9 kJ·mol–1
Note that the net reaction in this case, part (d), is not a redox reaction. It is a precipitation
reaction and the substantial negative standard free energy change suggests that mercury(I)
chloride, Hg2Cl2(s), is quite insoluble. This insolubility is what drives the reaction in the cell.
Problem 10.40.
(a) The standard reduction potential for the copper(II) cation to copper metal half reaction, 0.337
V, is more positive than that for the iron(II) cation to iron metal half reaction, –0.41 V, so
copper(II) cation will be reduced to copper metal and iron metal will be oxidized to iron(II)
cation in the cell described in this problem. The loss of atoms from the iron metal will result in
its losing mass. Addition of the copper metal to the copper electrode will increase its mass.
(b) Iron(II) cations are produced in the iron half cell, so the metal ion concentration will increase
in this half cell.
(c) Use the standard reduction potentials (which are given at 298 K) to get the initial (standard)
cell potential:
E° = E°(Cu2+, Cu) – E°(Fe2+, Fe) = 0.337 V – (–0.41 V) = 0.75 V
Consider what happens to each reduction half reaction, as the reaction proceeds:
cathode:
Cu2+(aq) + 2e–  Cu(s)
anode:
Fe2+(aq) + 2e–  Fe(s)
In the copper half cell, the concentration of copper(II) declines, so the reduction potential
becomes less positive. In the iron half cell, the concentration of iron(II) increases, so the
reduction potential becomes more positive. (Use Le Chatelier’s principle to make these
predictions.) The result is that the reduction potentials for the half reactions approach one
another as the more positive becomes less positive and the less positive becomes more positive.
Thus, the cell potential, the difference between these two reduction potentials becomes less and
less as the cell reaction goes on. Ultimately, the reduction potentials will come to the same value
and the cell potential will become zero. Equilibrium will have been reached. Dead cells are at
equilibrium.
Problem 10.41.
(a) The cell notation, Ni(s) | Ni2+(aq, 0.1 M) || Fe3+(aq, 0.1 M) , Fe2+(aq, 0.1 M) | Pt(s), implies
that Ni, the left-hand electrode, is the anode. The measured positive potential means that this is
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correct, since the assumption we have to make is that the anode input to the voltmeter was
connected to the Ni metal electrode. Thus, oxidation of Ni occurs at the Ni electrode and
reduction of Fe3+ at the Pt cathode:
2Fe3+ + Ni  2Fe2+ + Ni2+
(b) Since Fe2+ is a product of the reaction, increasing its concentration has the effect of shifting
the reaction back toward the reactants (to reduce the effect of the disturbacne, the addition of a
reaction product—Le Chatelier’s principle). This shift will decrease the cell potential, since the
shift is in a direction to lower the driving force for the cell reaction.
(c) Ni2+ is also a product of the reaction, so increasing its concentration will have the same
directional effect as for the addition of Fe2+ in part (b), that is, decreasing the cell potential.
(d) If a strip of nickel metal is immersed in an equimolar aqueous solution of Fe3+ and Fe2+ ions,
we would you expect, on the basis of the cell reaction we have written, some of the Ni metal to
react and go into solution as Ni2+ ions while twice as many moles of Fe3+ are reduced to Fe2+.
Thus, you would expect to find Ni2+ in the solution with the amount of Fe3+ reduced by two
moles for every mole of Ni2+ present and the amount of Fe2+ increased by two moles for every
mole of Ni2+.
Problem 10.42.
(a) For this cell, Cd(s) | Cd2+(aq, ?? M) || Ni2+(aq, 1.0 M) | Ni(s), the measured cell potential, E,
is 0.225 V at 298 K. The cadmium metal electrode is the anode in this cell, so cadmium metal is
being oxidized to cadmium cation. The cathodic reaction must be the reduction of nickel(II)
cation to nickel metal. The half reactions (as reductions) and net cell reaction are:
cathodic:
Ni2+(aq) + 2e–  Ni(s)
anodic:
Cd2+(aq) + 2e–  Cd(s)
net reaction: Cd(s) + Ni2+(aq)  Cd2+(aq) + Ni(s)
(b) The standard cell potential is the difference between the cathodic and anodic standard
reduction potentials:
E° = E°(Ni2+, Ni) – E°(Cd2+, Cd) = –0.23 V – (–0.40 V) = 0.17 V
(c) If the cadmium cation concentration were 1 M, the cell potential would be the standard cell
potential (because the nickel(II) cation concentration is 1 M as well). The cell potential is more
positive than the standard potential, so the net reaction has a greater driving force than if all
concentrations were 1 M. Since cadmium cation is a product of the cell reaction, Le Chatelier’s
principle predicts that its concentration is lower than 1 M, as this would make the driving force
for reaction greater, as observed. Use the measured potential for the cell, 0.225 V, the standard
cell potential from part (b), and the known concentration of nickel(II) cation, 1.0 M, in the
Nernst equation to get a numerical value for the unknown cadmium cation concentration:
0.05916 V (Cd 2+ ) 
0.05916 V
E = 0.225 V = E° –
logQ = 0.17 V –
log 

 (Ni 2+ ) 
n
2
–
(Cd 2+ ) 
(Cd 2+ ) 
2  (0.225 V)
= log 
=
log


 = log(Cd2+) = –1.86
 (Ni 2+ ) 
 1 
0.05916 V
(Cd2+) = 0.014
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The concentration of cadmium ion is 0.014 M. This numerical result is consistent with the
directional reasoning based on Le Chatelier’s principle.
(d) We know that the equilibrium constant for the reaction is related to the standard cell
potential:
2 (9.6485 104 C mol –1 )(0.17 V)
nFE o
lnK =
=
= 13.2
 K = 5.4  105
–1
–1
RT
(8.314 J K  mol )(298 K)
Problem 10.43.
For the cell of interest, Pt(s) | H2(g) | H+(aq), SO42–(aq) | PbSO4(s) | Pb(s), the columns in this
table are labeled with the effect of each change on the cell potential.
change in the cell
increase
increase in pH of the solution
decrease
X
dissolving Na2SO4 in the solution
X
increase in size of the Pb electrode
X
decrease in H2 gas pressure
addition of water to the solution
X
X
increase in the amount of PbSO4
dissolving a bit of NaOH in the solution
no effect
X
X
To analyze the results of the changes, use the net cell reaction:
H2(g) + PbSO4(s)  2H+(aq) + Pb(s) + SO42–(aq)
The first and last changes both decrease the concentration of hydronium ion, H+(aq), which is a
product of the reaction. Le Chatelier’s principle predicts that the system will react to this
disturbance by forming more hydronium ion, which drives the reaction toward products and
increases the cell potential. Similarly, addition of water reduces the concentrations of both
hydronium ion and sulfate anion. Since these are products, the reaction will be driven toward
products, thus increasing the cell potential. Changing the amount of a pure solid in an
equilibrium system has no effect on the reaction quotient because the dimensionless
concentration ratio is unchanged. Thus increasing either the lead or the lead sulfate has no effect
on the cell potential. Adding sulfate anion increases the concentration of a product and the
system will respond by decreasing its concentration, that is, the reaction proceeds in reverse,
which decreases the cell potential. Finally, a decrease in the hydrogen gas pressure is a decrease
in the “concentration” of a reactant and the system responds by increasing the amount of
hydrogen, that is, the reaction proceeds in reverse, which decreases the cell potential.
Problem 10.44.
For the cell of interest, Cu(s) | Cu2+(aq) || Cr2O72–(aq), Cr3+(aq), H+(aq) | Pt(s), the columns in
this table are labeled with the effect of each change on the cell potential.
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change in the cell
increase
decrease in pH in the Cr solution
decrease
no effect
X
decrease in size of the Cu electrode
X
addition of water to the Cu solution
X
addition of water to the Cr solution
X
dissolving Cr(NO3)3 in the Cr solution
X
dissolving K2Cr2O7 in the Cr solution
X
To analyze the results of the changes, use the net cell reaction:
3Cu(s) + Cr2O72–(aq) + 14H+(aq)  3Cu2+(aq) + 2Cr3+(aq) + 7H2O(l)
Decreasing the pH in the chromium half cell increases the hydronium ion concentration.
Hydronium ion is a reactant, so the system responds by reacting to decrease its concentration.
The reaction is driven toward products, thus increasing the cell potential. Note that this can be a
very large effect since the concentration of hydronium ion in the reaction quotient is raised to the
14th power. The size or amount of any solid reactant or product has no effect on the reaction
quotient, because its dimensionless concentration ratio is unaffected. Thus decreasing the size of
the copper electrode (as long as it is not decreased to zero) has no effect on the cell potential.
Addition of water to the copper half cell decreases the concentration of copper(II) cation, a
reaction product. The system responds by making more copper(II) cation, thus driving the
reaction toward products and increasing the cell potential. Addition of water to the chromium
half cell is a more difficult change to analyze, since it decreases the concentrations of both
reactants, Cr2O72–(aq) and H+(aq), and a product, Cr3+(aq). To figure out what will happen, we
consider how the reaction quotient for this reaction will be affected and how its change will
affect the cell potential, via the Nernst equation:
E = E° –
0.05916 V
n
o
logQ = E –
0.05916 V
6
 (Cu2 + )3(Cr3+ )2 
log 

(Cr2 O72 – )(H+ )1 4 
The overwhelming effect of the hydronium ion concentration raised to the 14th power is the
major factor here. A small decrease in the concentration of hydronium ion will cause a large
increase in the numerical value of Q, because it cannot be wholly compensated by the
concomitant decrease in the concentration of the chromium(III) cation, which is only raised to
the 2nd power in the numerator. Thus, the logQ term will increase and this will make the negative
term on the right hand side of the equation more negative, thus decreasing E. Addition of more
chromium(III) cation (the fifth change in the table) to the chromium half cell increases the
concentration of a reaction product. The system responds by reacting in reverse to decrease this
concentration, thus driving the reaction toward reactants and decreasing the cell potential. The
last change, addition of more dichromate anion to the chromium half cell, increases the
concentration of one of the reactants. The system responds by reacting to decrease its
concentration, so the reaction is driven toward products, thus increasing the cell potential.
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Problem 10.45.
(a) The standard potential for the quinhydrone-silver cell is obtained by subtracting the standard
reduction potential for the anode reaction (quinhydrone) from the standard potential for the
cathode reaction (silver):
E° = E°(Ag+, Ag) – E°(Qu, H2Qu) = 0.799 V – (0.699 V) = 0.100 V
(b) Write the Nernst equation (at 298 K) for the net cell reaction with the known values for E, Eo,
and concentrations and solve for the unknown hydronium ion concentration:
Pt(s) | H2Qu(aq, c), Qu(aq, c), H3O+(pH) || Ag+(aq, 0.10 M) | Ag(s)
H2Qu(aq) + 2Ag+(aq)  Qu(aq) + 2H+(aq) + 2Ag(s)
0.05916 V  (Qu)(H + )2 
0.256 V = 0.100 V –
log 
+ 2
2
 (H 2Qu)(Ag ) 
0.156 V = –
0.05916 V
2
 (c)(H + )2 
0.05916 V
log 
=–
[2·log(H+) – 2·log(0.10)]
2 
2
 (c)(0.10) 
 0.156 V 
–log(H+) = pH = 
 – log(0.10) = 3.64
 0.05916 V 
Problem 10.46.
(a) The line notation for the cell with net reaction, Zn(s) + Hg22+(aq)  Zn2+(aq) + 2Hg(l), is:
Zn | Zn2+(aq) || Hg22+(aq) | Hg(l)
(b) Substitute the known values, [Hg22+(aq)] = 0.010 M, [Zn2+(aq)] = 0.50 M, and E = 1.51 V,
into the Nernst equation (at 298 K) and solve to find E° for the cell reaction:
1.51 V = E° –
0.05916 V
2
0.05916 V
(Zn 2+ ) 
log 

(Hg 22 + ) 
 0.50 
= 1.51 V + 0.05 V = 1.56 V
0.010 
2
(You can check this result using the standard reduction potentials in Appendix B.) To get the
equilibrium constant, substitute E° in:
nE o
2 (1.56 V)
lnK =
=
= 52.7;
 K = 8.0  1022
0.05916 V
0.05916 V
E° = 1.51 V +
log
(c) The equilibrium constant for the reaction is quite large, so the equilibrium lies very far to the
side of the products. Thus, almost no zinc cations will be reduced by the addition of mercury to a
solution containing 1.0 M zinc cations. Assuming that there is a negligible change in the
concentration of zinc cations, we can use K to calculate the tiny concentration of mercury(I)
dication, Hg22+(aq), that will be formed.
(Zn 2+ )   1.0 
2+
–23
K = 8.0  1022 = 
 = 
 ; (Hg2 (aq)) = 1.3  10
(Hg 22 + )  (Hg 22 + ) 
[Hg22+(aq)] = 1.3  10–23 M
This tiny concentration confirms that very little reaction occurs, so [Zn2+(aq)] = 1.0 M is an
excellent approximation.
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Problem 10.47.
(a) For the cell, Sn(s) | Sn2+(aq, 0.10 M) || Sn4+(aq, 0.010 M) , Sn2+(aq, 1.0 M) | Pt(s), the anodic
and cathodic half reactions (as reductions) and the net cell reaction are:
anode:
Sn2+an(aq) + 2e– Sn(s)
cathode:
Sn4+(aq) + 2e–  Sn2+cath(aq)
cell reaction: Sn(s) + Sn4+(aq)  Sn2+an(aq) + Sn2+cath(aq)
Here, for the moment, we distinguish between tin(II) ion on the anode and cathode sides of the
cell, since they do not mix with one another. The reaction quotient and its numeric value are:
Sn
Q=
2+
an
(aq)Sn 2+
cath (aq)
Sn
4+
(aq)
=
0.101.0 
= 10.
0.010
(b) Use the Nernst equation with our known values of E from the problem statement and Q from
part (a) to get E° and G°rxn:
0.059 V
0.059 V
E = 0.25 V = E° –
logQ = E° –
log(10);
E° = 0.28 V
n
2
G°rxn = –nFE° = –2(96485 C·mol–1)(0.28 V) = 5.4  104 J·mol–1 = 54 kJ·mol–1
(c) The equilibrium constant expression is written for the reaction as though all species are
together in the same solution:
Sn
K=
Sn
2+
(aq)
4+
2
(aq)
The equilibrium constant is obtained either from E° or G°reaction; from E°:
2 96485 C  mol–1 0.28 V
nFE o
lnK =
=
= 21.8
RT
8.314 J  mol–1  K –1 298K 
K = e21.8 = 2.9  109
(d) The cell potential (and free energy change) for the reaction of tin metal with tin(IV) are
favorable, so we expect that tin metal added to the tin(IV) solution will react to reduce most of
the tin(IV) to tin(II). If all the tin(IV) reacts, an equivalent number of moles of tin metal must be
oxidized to tin(II). There are 2.5  10–4 mol [= (0.050 L)(0.005 M)] of tin(IV) which can react
with 0.03 g of tin [= (2.5  10–4 mol)(119 g·mol–1)]. Thus there is tin metal remaining in the
solution, so we can use our equilibrium constant, which is written for a reaction solution that
contains tin(IV) and tin(II) in equilibrium with tin metal. The equilibrium constant is so large
that we can probably assume that “all” the tin in solution is tin(II) and use the equilibrium
constant expression to calculate the tiny concentration of tin(IV) remaining. The concentration of
tin(II) is 0.010 M, because it includes the original tin present as tin(IV) plus the equivalent
amount formed by oxidation of the tin metal. The concentration of tin(IV), [Sn4+(aq)], is
numerically equivalent to (Sn4+(aq)):
(Sn
4+
Sn
(aq)) =
2+
(aq)
K
0.010
2
=
2.9 10
9
= 3.4  10–14
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Our assumption that essentially all of the tin(IV) has reacted is justified. The solution contains
almost all tin(II) with only a minuscule amount of tin(IV) remaining.
Problem 10.48.
[NOTE: This problem should be moved to the next section (assigned after Section 10.7 has been
introduced), because part (c) is most easily done by applying the Nernst equation to the half cell
reaction, which is done in Section 10.7.]
(a) When a saturated calomel electrode (S.C.E.), E = 0.241 V, is connected to a Co2+|Co half cell
in which the concentration of Co2+ is 0.050 M, the measured cell potential at 298 K is 0.561 V
and the cobalt half cell is the anode. The cell potential is the difference between the cathode
reduction potential and the anode reduction potential. Since the cobalt half cell is the anode we
can write:
E = 0.561 V = E(S.C.E.) – E(Co2+, Co) = 0.241 V – E(Co2+, Co)
E(Co2+, Co) = –0.320 V
(b) The net cell reaction for this cell is:
Co(s) + 2Hg(l) + 2Cl–(saturated aqueous KCl solution)  Co2+(aq) +Hg2Cl2(s)
The free energy change for the cell reaction, under the conditions stated, is:
Grxn = –nFE = – 2·(9.6485  104 C·mol–1)(0.561 V) = –108.3 kJ·mol–1
(c) Applying the Nernst equation (at 298 K) to the cobalt half cell reduction reaction and
reduction potential for the cell here gives the standard reduction potential and thence the standard
free energy change:
0.05916 V  1 
E(Co2+, Co) = –0.320 V = E°(Co2+, Co) –
log  2 + 
(Co )
2
0.05916 V
 1 
= –0.282 V
0.050 
2
G°rxn = –nFE° = – 2·(9.6485  104 C·mol–1)(–0.282 V) = 54.4 kJ·mol–1
E°(Co2+, Co) = –0.320 V +
log
Problem 10.49.
(a) The half reactions (as reductions) in mercury cells can be represented as:
ZnO(s) + H2O(l) + 2e–  Zn(s) + 2OH–(aq)
HgO(s) + H2O(l) + 2e–  Hg(l) + 2HO–(aq)
The problem states that Zn is the anode and, since oxidation occurs at the anode, the first
reaction must be going in reverse as an oxidation in the cell. The cell reaction is the first half
reaction subtracted from the second which gives (with redundant species cancelled out):
Zn(s) + HgO(s)  ZnO(s) + Hg(l)
(b) As the product concentrations get higher, Le Chatelier’s principle suggests that the system
will respond in a way to decrease the effect of the change, that is, by reacting to use up the
products. The reaction would go in reverse (or not go forward as forcefully as before the change)
and this would decrease the driving force for the reaction, which shows up as a decrease in the
cell potential. (This argument has to be made with care and used cautiously because Le
Chatelier’s principle really applies to equilibrium systems and cells producing a potential are not
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at equilibrium.) For the mercury cell, you see that the cell reaction involves only solids and a
pure elemental liquid. The concentrations (or activities) of these species do not change as they
are produced or used up. Only when one or the other of the reactants, Zn and HgO, is used up (or
almost used up) does the cell potential decline.
(c) The reaction quotient for the cell reaction in part (a) is:
(Hg(l))(ZnO (s) ) 11
Q=
=
=1
(HgO (s) )(Zn (s)) 11
The dimensionless concentration ratio for pure solids and liquids is unity, so this substitution
gives a constant reaction quotient of unity. The logarithm of unity is zero, so the cell potential
does not vary until the reactants are almost used up in the cell.
Problem 10.50.
(a) The net cell reaction, Al(s) + Fe3+(aq)  Al3+(aq) + Fe(s), is given in the problem statement.
The corresponding Q is:
Q=
(Al3+ (aq))(Fe(s))
(Al3+ (aq))
=
(Al(s))(Fe 3+ (aq)) (Fe 3+ (aq))
The dimensionless concentration ratio for pure solids and liquids is unity, so (Al(s)) and (Fe(s))
are both unity, which leaves us with only the aqueous ionic concentrations in Q.
(b) Use the Nernst equation (at 298 K) and the cell potential and concentration data to get E°
and, hence, G°reaction:
0.05916 V (Al 3 +(aq)) 
1.59 V = E° –
log  3 +

(Fe (aq)) 
3
0.05916 V
 0.250 
log 
= 1.59 V + 0.03 V = 1.62 V
0.0050 
3
G°rxn = –nFE° = – 3·(9.6485  104 C·mol–1)(1.62 V) = –469 kJ·mol–1
E° = 1.59 V +
(c) The equilibrium constant expression is identical to the expression for Q (at equilibrium) in
part (a). The equilibrium constant at 298 K can be obtained from either E° or G°rxn:
nE o
3 (1.62 V)
lnK =
=
= 82.2;
 K = 4.8  1035
0.05916 V
0.05916 V
(d) Although the thermite reaction occurs between solid aluminum metal and solid iron(III)
oxide, 2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(l), the redox chemistry is the same as that analyzed
in parts (a)–(c). We can estimate the standard free energy change by assuming it is the same as
we calculated in part (b), taking into account that the thermite reaction involves two moles of
aluminum metal and our calculation was for only one mole in the cell reaction. Therefore, the
free energy of the thermite reaction would be approximately –940 kJ for every two moles of
molten iron metal produced or –470 kJ for one mole of iron. Free energy is related to the thermal
energy of the reaction by G = H – TS. The reaction of interest involves three moles of solids
as reactants and produces a mole of solid and two moles of liquid products. Thus, the entropy
change is probably positive but not very large, so the TS term is relatively small. To a rough
approximation, it’s likely that G  H. This means that there would be somewhat more than
five times as much energy released by the reaction as are needed to melt the iron (about
ACS Chemistry FROG
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Reduction-Oxidation: Electrochemistry
Chapter 10
81 kJ·mol–1). It certainly makes sense that molten iron is produced and the reaction is sometimes
used to make small amounts of molten iron to weld iron (steel) pieces together.
Problem 10.51.
(a) To find the cell potential (at 298 K) for Mg(s) | Mg2+(aq, 0.60 M) || Cu2+(aq, 0.60 M) | Cu(s),
use the standard reduction potentials from Appendix B to find E° and then the Nernst equation to
get E for the conditions here:
E° = E°(Cu2+, Cu) – Eo(Mg2+, Mg) = 0.377 V – (–2.38 V) = 2.76 V
E = 2.76 V –
0.05916 V
 (Mg 2+ (aq)) 
0.60 
log 
= 2.76 V – (0.02958 V)·log 
2+


0.60 
 (Cu (aq)) 
2
E = 2.76 V
(b) At equilibrium, the cell potential (and the free energy change for the change from reactants to
products at their equilibrium concentrations) is zero. This is the criterion for equilibrium in an
electrochemical cell. The state of equilibrium will be reached, only if enough magnesium metal
is present for the reaction to go as far as necessary to reach equilibrium with some left over.
Although the “concentration” of the solids does not appear in the reaction quotient, the reaction
depends upon their presence.
(c) The equilibrium criterion is E = 0, so substitute into the Nernst equation and solve for
(Mg2+(aq))/(Cu2+(aq)) = [Mg2+(aq)]/[Cu2+(aq)]:
0 = 2.76 V –
0.05916 V
2
 (Mg 2+ (aq)) 
log 
 (Cu 2+ (aq))  eq
 (Mg 2+ (aq)) 
log 
= 93.3
 (Cu 2+ (aq))  eq
 (Mg 2+ (aq)) 

= 2  1093
2+

 (Cu (aq))  eq
At equilibrium “all” the copper(II) ions have been reduced to copper metal.
Problem 10.52.
(a) The line notation and net reaction for a zinc-chlorine cell are:
Zn(s) | Zn2+(aq, sat’d ZnCl2), Cl–(aq, sat’d ZnCl2) | Cl2(g, P) | Pt(s),
Zn(s) + Cl2(g, P)  Zn2+(aq, sat’d ZnCl2) + 2Cl–(aq, sat’d ZnCl2)
Use the standard reduction potentials from Appendix B to determine the standard cell potential
(at 298 K):
E° = E°(Cl2, Cl–) – E°(Zn2+, Zn) = 1.359 V – (–0.763 V) = 2.122 V
(b) The reaction quotient for this cell reaction is:
(Zn 2+ (aq, sat'd ZnCl2 ))(Cl – (aq, sat'd ZnCl2 ))2
Q=
(Cl2 (g, P))
(c) Use the Nernst equation with E° from part (a) and to get the cell potential under the solution
conditions specified in the problem statement. The stoichiometry of zinc chloride dissolution
gives solution concentrations of [Zn2+(aq, sat’d ZnCl2)] = 3 M and [Cl–(aq, sat’d ZnCl2)] = 6 M,
and the chlorine gas pressure is 1.0 bar.
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E = 2.122 V –
0.05916 V
2
(3)(6)2
log
=2V
(1)
The uncertainty in the concentrations and nonideality of these concentrated solutions suggest that
one significant figure in this result is about the best we can justify.
(d) The work available per mole of cell reaction (equivalent to one mole of Zn(s) reacting) is:
welec [J·mol–1] = nFE = 2·(9.6485  104 C·mol–1)(2 V) = 4  102 kJ·mol–1
To get 75 kJ of work, we need 0.2 mol [= (75 kJ)/(4  102 kJ·mol–1)] of Zn(s) to react. The
reaction requires one mole of chlorine gas for each mole of zinc metal, so 0.2 mol of Cl2(g) is
also required.
(e) The concentrations of the ions are unchanged after 75 kJ of work is produced by this cell. The
reaction mixture is saturated with ZnCl2(s). The Zn2+(aq) and Cl–(aq) produced (in the
stoichiometric 1:2 ratio) by the reaction simply react to form more ZnCl2(s), since the solution is
saturated and already contains the maximum possible concentrations of these ions.
Problem 10.53.
(a) In the cell represented as Zn(s) | ZnO(s) | KOH(aq, 40%) | Ag2O(s) | Ag(s), the cathodic
reaction is reduction of silver(I) in solid silver(I) oxide to silver metal:
Ag2O(s) + H2O(aq) + 2e–  2Ag(s) + 2OH–(aq)
(b) The anodic half reaction (written as an oxidation) is the oxidation of zinc metal to Zn(II) in
solid zinc(II) oxide:
Zn(s) + 2OH–(aq)  ZnO(s) + H2O(aq) + 2e–
(c) Combining the half reactions in parts (a) and (b) gives the net cell reaction:
Zn(s) + Ag2O(s)  ZnO(s) + 2Ag(s)
(d) The KOH electrolyte solution is not represented in the net cell reaction in part (c) but is
required to provide the mobile charge carriers, OH–(aq) anions, that complete the electrical
circuit in the cell. This is like the aluminum-air cell in Worked Example 10.29.
Problem 10.54.
The reduction potential for methylene blue reduction, MB+(aq) + H+(aq) + 2e–  MBH(aq),
should vary as the pH of the half reaction is varied. This table gives reduction potential data for
methylene blue at different pH’s.
pH
0
E, V 0.53
1
2
3
4
5
6
7
0.47
0.38
0.29
0.21
013
0.07
0.01
8
9
10
–0.03 –0.07 –0.11
Hydronium ion is a reactant in the methylene blue, MB, reduction reaction (10.93). As the pH at
which the half reaction occurs is increased (hydronium ion concentration decreased), a reactant
concentration is decreased. We would predict that the system would respond to reduce the effect
of the change by not going forward as forcefully and therefore that the reduction potential would
decrease with increasing pH. The table shows that this is what occurs.
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Problem 10.55.
[NOTE: To be consistent with our nomenclature throughout the rest of the text, the formula for
hypochlorous acid should be written HOCl, to make clear that the H is bonded to the O; this is an
oxyacid.]
(a) Hypochlorous acid, HOCl, is a strong oxidizing agent:
2HOCl(aq) + 2H+(aq) + 2e–  Cl2(g) + 2H2O(l) E° = 1.630 V
Two other components of a bleach solution are oxidizable by hypochlorous acid, chloride ion
and water:
Cl2(g) + 2e–  2Cl–(aq)
E° = 1.359 V
+
–
O2(g) + 4H (aq) + 4e  6H2O(l)
E° = 1.229
Oxidation of water would produce oxygen gas, which is not “hazardous.” Oxidation of chloride
ion would produce chlorine gas, which is toxic and, hence, “hazardous.” Reduction of
hypochlorous acid also produces chlorine. It seems likely that the reaction to “produce a
hazardous gas” referred to on the warning label is the one with chloride:
HOCl(aq) + H+(aq) + Cl–(aq)  Cl2(g) + H2O(l) E° = 0.271 V
(Although both oxidation reactions are favored, the oxidation of chloride is fast and the
oxidation of water is slow, which is why you can make solutions of hypochlorite.) The standard
free energy change for one mole of this reaction is:
G° = –nFE° = –(1)(96485 C·mol–1)(0.271 V) = –26.1 kJ
The value of n is one, because only one electron is transferred in the net reaction (from Cl– with
an oxidation number of –1 to the Cl with an oxidation number +1 in HClO, to give them both an
oxidation number of zero in the Cl2 product).
(b) The equilibrium constant expression and equilibrium constant for the net redox reaction are:
(Cl2 (g))(H 2O(l))
(Cl2 (g))
K=
=
+
–
(HOCl(aq))(H (aq))(Cl (aq))
(HOCl(aq))(H + (aq))(Cl– (aq))
nE o
1 (0.271 V)
lnK =
=
= 4.58;
0.05916 V
0.05916 V
 K = 98
(c) Apply the Nernst equation (at 298 K) to the net redox reaction with (H+(aq)) = 10–14 (pH 14)
and all other species still at their standard concentrations:
(Cl2 (g))
0.05916 V log
E(pH 14) = E° –
(HOCl(aq))(H + (aq))(Cl– (aq))
n
E(pH 14) = 0.271 V –
0.05916 V log
1
(1)
= –0.557 V
(1)(10 –14 )(1)
This potential, E(pH 14) = –0.557 V, is the standard cell potential for the reaction in basic
solution, where [OH–(aq)] = 1.00 is the standard condition. In basic solution (at pH 14) the
reaction is quite unfavorable, as written. Under these conditions, chlorine dissolved in the basic
solution reacts with itself to form hypochlorite anion, ClO–(aq), and chloride ion (which is pretty
much how bleach is made and why the solution contains the chloride ion necessary for the
reaction in part (a) – under acidic conditions).
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(d) In basic solution, the concentration of hydronium ion is low. Hydronium ion is a reactant in
the net reaction, so lowering its concentration, lowers the driving force of the reaction, as we saw
in part (c). We can use the Nernst equation again to get the cell potential when the hydronium
ion concentration is 10–8 M (pH 8) as in a bottle of bleach, and all other species are still at
approximately unit activity.
(Cl2 (g))
0.05916 V log
E(pH 8) = E° –
(HOCl(aq))(H + (aq))(Cl– (aq))
n
E(pH 8) = 0.271 V –
0.05916 V log
1
(1)
= –0.202 V
(1)(10 –8 )(1)
Thus, making the solution basic makes the reaction unfavorable. Furthermore, in basic solution,
the HOCl reacts with hydroxide ion to lose its proton, so its concentration also gets quite small,
which decreases the cell potential even more. We could use the Ka for HClO to find out how
much lower, but there is no need to do that for this problem.
Problem 10.56.
(a) For this cell, Pb(s) | Pb2+(aq, 1.0 M) || H+(aq, 1.0 M) | H2(g, 1 bar) | Pt(s), the standard cell
potential is:
E° = E°(H+, H2) – E°(Pb2+, Pb) = 0 V – (–0.13 V) = 0.13 V
(b) Use Ksp and the solubility product expression for PbSO4(s) to find the concentration of
lead(II) ion in equilibrium with [SO4(aq)] = 0.50 M:
Ksp = 6.3  10–7 = (Pb2+(aq))(SO4(aq)) = (Pb2+(aq))(0.50)
(Pb2+(aq)) = 1.3  10–6
(c) Substitute the value for (Pb2+(aq)) from part (b) in the Nernst equation for the lead ion-lead
metal half cell reduction potential to find the reduction potential after enough sulfate ion has
been added to the lead ion solution in the cell in part (a) to make [SO4(aq)] = 0.50 M:
(Pb(s))
0.05916 V
E(Pb2+, Pb) = E°(Pb2+, Pb) –
log
2
(Pb 2+ (aq))
1
0.05916 V
E(Pb2+, Pb) = –0.13 V –
log
= –0.30 V
1.3  10 –6
2
(d) Since the cell in part (a) is set up with the SHE as the cathode, the cell potential is the
negative of the reduction potential for the lead ion-lead metal half cell, that is, E = 0.30 V in this
case. Note that, since oxidation occurs in the lead ion-lead metal half cell, the potential of the cell
increases, larger driving force, as the concentration of the product, lead ion, decreases. This is in
accord with Le Chatelier’s principle.
Problem 10.57.
Use the Nernst equation to determine the standard cell potential from the measured cell potential
and the conditions in the Clark cell:
Zn(s) | Zn2+(aq, 0.100 M), SO42–(aq) | Hg2SO4(s) | Hg(l)
Zn(s) + Hg2SO4(s)  Zn2+(aq) + 2Hg(l) + SO42–(aq)
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1.435 V = E° –
0.05916 V
E° = (1.435 V) +
2
 (Zn 2+ (aq))(Hg(l))2 (SO2–

4 (aq))
log 

(Zn(s))(Hg 2SO4 (s))

0.05916 V
2
Chapter 10
 (0.100)(1)2 (0.100) 
log 
 = 1.376 V
(1)(1)

The standard cell potential is the difference between the desired standard reduction potential,
E°(Hg2SO4, Hg), and the standard zinc reduction potential, E°(Zn2+, Zn):
E° = 1.376 V = E°(Hg2SO4, Hg) – Eo(Zn2+, Zn)
E°(Hg2SO4, Hg) = (1.376 V) + (–0.763 V) = 0.613 V
Problem 10.58.
(a) The NAD+ reaction with ethanol and its standard cell potential (pH 7) are:
NAD+(aq) + CH3CH2OH(aq)  NADH(aq) + CH3CHO(aq) + H+(aq) E°´ = –0.12 V
We can get the standard free energy change at pH 7, G°´, from the standard cell potential at pH
7:
G°´ = –nFE°´ = –(2)(96485 C·mol–1)(–0.12 V) = 23 kJ·mol–1
The positive free energy and negative cell potential both show that the reaction, in the direction
written, is not favored under standard conditions at pH 7.
(b) The reaction quotient for this reaction (at pH 7 which is accounted for already) is:
Q´ =
NADH CH 3CHO
NAD CH CH OH 
+
3
2
The equilibrium constant for the reaction is less than one; K´eq < 1, because G°´ > 0. The
addition of a reagent that reacts with the aldehyde, CH3CHO, lowers its concentration, so Q´,
after addition of this reagent, is smaller than K´eq in this solution with added reagent. Both lnK´eq
and lnQ´ are negative (< 0), but lnQ´ is smaller (more negative). The equation for the free energy
is:
G´ = G°´ + RTlnQ´
The negative logarithmic term will make G´ smaller than G°´ and could even make G´
negative. Note that it is not necessary that G´ become negative to make the analysis for ethanol
work. As the aldehyde reacts with the added reagent, more reaction of ethanol with NAD+ is
required to try to restore equilibrium conditions, even if they are not very favorable for product
formation. The newly produced aldehyde can react with more reagent and more reaction will
have to occur to replace it. As this process continues, the NADH product is building up. If an
excess of the reagent that reacts with aldehyde is present, the reaction above will be forced to
continue producing aldehyde until the ethanol is essentially used up and an equivalent amount of
NADH produced to measure spectrophotometrically.
(c) H3O+ is a product of the reaction, so raising the pH [decreasing the (H3O+(aq)] will have the
affect of decreasing G´ and possibly even making it negative. The problem is that the enzyme,
alcohol dehydrogenase, might not be active at a higher pH; it could be denatured.
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Problem 10.59.
(a) The electrode reactions and net cell reaction for zinc and platinum electrodes implanted in
our oxygenated and ionic body fluids are:
anodic:
Zn(s)  Zn2+(aq) + 2e–
cathodic
O2(aq) + 2H+(aq) + 2e–  H2O(aq)
net reaction: Zn(s) + O2(aq) + 2H+(aq)  Zn2+(aq) + H2O(aq)
(b) The standard cell potential for this net reaction under biological conditions (pH 7) is:
E°´ = E°´(O2, H2O) – E°(Zn2+, Zn) = 0.816 V – (–0.763 V) = 1.579 V
(c) The amount of charge (moles of electrons) available from 4.5 g of zinc metal is equal to twice
the number of moles of zinc metal, since each atom oxidized produces two electrons:


4.5 g
mol e– = 2 
= 0.138 mol
-1 
 65.37 g  mol 
Convert moles of electrons to coulombs, and calculate the amount of time current would flow at
35  10–6 A to equal this many coulombs:
number coulombs = ( 0.138 mol)(9.6485  104 C·mol–1) = 1.33  104 C
1.33  104 C = (35  10–6 A)·t
 t = 3.8  108 s = 4.4  103 day = 12 yr
Twelve years is a reasonable length of time for an implant to last before maintenance (or
replacement) is required.
Problem 10.60.
(a) We can use standard reduction potentials and standard free energies of formation to find the
standard free energy changes for each of these three reactions and their sum:
(i) NO3–(aq) + 4H+(aq) + 3e–  NO(g) + 2H2O(l)
(ii) NO(g) + 1/2O2(g)  NO2(g)
(iii) H2O(l)  1/2O2(g) + 2H+(aq) + 2e–
(iv) NO3–(aq) + 2H+(aq) + e–  NO2(g) + H2O(l)
G°i = –nFE°i = – 3·(9.6485  104 C·mol–1)(0.955 V) = –276 kJ·mol–1
G°ii = G°f(NO2) – G°f(NO) – 1/2G°f(O2)
–1
G°ii
) – (86.55 kJ·mol–1) – 1/2(0 kJ·mol–1) = –35 kJ·mol–1
G°iii = –nFE°iii = – 2·(9.6485  104 C·mol–1)(–1.229 V) = 237 kJ·mol–1
Note that the potential used for half reaction (iii) is the negative of the standard reduction
potential, because the reaction is written in reverse as an oxidation. The sum of these standard
free energy changes is the standard free energy change for half reaction (iv):
G°iv = (–276 kJ·mol–1) + (–35 kJ·mol–1) + (237 kJ·mol–1) = –74 kJ·mol–1
(b) Get the standard cell potential for half reaction (iv) from its standard free energy change in
part (a):
Go
(–74  103 J  mol – 1)
E°iv = – iv = –
= 0.77 V
nF
1 (9.6485  104 C  mol – 1)
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Problem 10.61.
(a) The reaction of interest must be oxidation of chloride anion, by the manganese(IV) in
manganese(IV) dioxide, MnO2(s), to produce the desired chlorine gas. In this process, the
manganese(IV) is reduced, probably to manganese(II), Mn2+(aq), which is the lowest oxidation
state of manganese ion in the solution. The net redox reaction is:
2Cl–(aq) + MnO2(s) + 4H+(aq)  Cl2(g) + Mn2+(aq) + 2H2O(aq)
(b) The standard cell potential for this reaction is:
E° = Eo(MnO2, Mn2+) – E°(Cl2, Cl–) = 1.230 V – (1.359 V) = –0.129 V
The reaction has a negative standard cell potential, which indicates that it is not spontaneous
under standard conditions.
(c) Use the Nernst equation to find the reduction potentials if [H+(aq)] =[Cl–(aq)] = 6 M, and all
other concentrations are still standard concentrations:
–
–
E(Cl2, Cl ) = E°(Cl2, Cl ) –
E(Cl2, Cl–) = 1.359 V –
0.05916 V
2
0.05916 V
2
log
E(MnO2, Mn2+) = E°(MnO2, Mn2+) –
E(MnO2, Mn2+) = 1.230 V –
(Cl– (aq))2
log
(Cl2 (g))
62
= 1.313 V
1
0.05916 V
0.05916 V
2
log
log
(Mn 2+ (aq))(H 2O(aq))
(MnO2 (s))(H + (aq))4
11
= 1.322 V
1 64
2
The cell potential for the reaction in 6 M HCl is:
E = E(MnO2, Mn2+) – E(Cl2, Cl–) = 1.322 V – (1.313 V) = 0.009 V
The reaction is spontaneous (but without a large driving force) when 6 M HCl is used as the
reactant. Higher concentrations of the acid would provide a more positive cell potential and more
driving force. Once again, we need to be aware that these calculations based on ideal behavior
for solutions at high concentrations are liable to be inaccurate and provide only directional
information. In particular, the activity of water in this solution is probably less than one, which
will further increase E(MnO2, Mn2+) and the cell potential.
Problem 10.62.
(a) The reaction of vanillin with Benedict’s reagent transforms the aldehyde carbon into a
carboxylic acid carbon:
H 3CO
HO
H 3CO
C
O + 2Cu(cit)– + 5OH–
O
HO
H
C
+ Cu2O + 2cit3– + 3H 2O
O
In this basic solution the carboxylic acid group transfers its proton to hydroxide ion, so it is
shown as a carboxylate anion in this reaction equation. (If the solution is basic enough, at least
some of the protons from the phenolic –OH group will also be transferred to hydroxide, but this
is not shown here.)
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(b) Although vanillin does give a positive Benedict’s test and there is almost surely vanillin in
the bottle of vanilla extract, you can’t be positive about this, just from the Benedict’s test. The
reason is that the mixture contains corn syrup, last ingredient on the list. Corn syrup is derived
from corn starch and contains glucose and other reducing sugars, as well as longer chain
molecules that make the liquid “syrupy.” The positive Benedict’s test can be caused by these
sugars as well as by the vanillin.
Problem 10.63.
(a) The reactions involved in the analysis of bleach are:
(1) HClO(aq) + 3I–(aq) + H+(aq)  Cl–(aq) + I3–(aq) + H2O(l)
(2) I3–(aq) + 2S2O32–(aq)  3I–(aq) + S4O62–(aq)
The first reaction is the reduction of chlorine in hypochlorous acid from an oxidation number of
+1 to chloride ion with and oxidation number of –1. Appendix B does not give a reduction
potential for this reaction, so we have to calculate it by combining two half reactions that give
the overall half reaction of interest:
(3) 2HClO(aq) + 2H+(aq) + 2e–  Cl2(g) + 2H2O(l)
E°3 = 1.630 V
–
–
(4) Cl2(g) + 2e  2Cl (aq)
E°4 = 1.359 V
(5) 2HClO(aq) + 2H+(aq) + 4e–  2Cl–(aq) + 4H2O(l) E°5 = ?
The approach we use to get the unknown reduction potential for half reaction (5) is to calculate it
using the free energies of the reactions:
–n5FE°5 = G°5 = G°3 + G°4 = –n3FE°3 – n4FE°4
Half reaction (5), written as the sum of two other half reactions, involves four electrons, so its
potential is:
E°5 =
2(1.630 V)  2(1.359 V)
–G5o
n FE o  n4 FE4o
n E o  n4 E4o
= 3 3
= 3 3
=
4
n5 F
n5 F
n5
E°5 = 1.495 V
NOTE: This last equation is general. You can determine the reduction potential of any half
reaction that is the sum of two or more other half reactions by using a weighted combination of
their reduction potentials. The n in the denominator is the sum of the n values for the half
reactions that are combined.
The reaction sequence in the problem involves three half reactions [with reaction (5) written in
its conventional simplest form with the smallest possible set of integer stoichiometric
coefficients]:
(5) HClO(aq) + H3O+(aq) + 2e–  Cl–(aq) + 2H2O(l)
E°5 = 1.495 V
–
–
–
(6) I3 (aq) + 2e  3I (aq)
E°6 = 0.5355 V
2–
–
2–
(7) S4O6 (aq) + 2e  2S2O3 (aq)
E°7 = 0.09 V
Since the iodine (triiodide) reduction potential is intermediate between the other two, iodide, I–
(aq), can be oxidized by hypochlorous acid and, in turn, the iodine (triiodide) product can
oxidize thiosulfate. Thus, reaction (6) is the coupling reaction in this sequence. The cell
potentials for reactions (1) and (2) are:
E°1 = E°5 – E°6 = (1.495 V) – (0.5355 V) = 0.959 V
E°2 = E°6 – E°7 = (0.5355 V) – (0.09 V) = 0.45 V
ACS Chemistry FROG
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Chapter 10
Just as we reasoned, both are favored.
(b) The iodine (triiodide)-iodide half reaction couples the other two. The necessary property for
coupling is that the reduction potential for the coupling reaction be intermediate between the
other two. This is the basis for the reasoning in part (a). It also has to be the case that the
reactions are quantitative and go essentially to “completion.” The equilibria have to lie far
toward the products. The cell potentials, in this case, give equilibrium constants of about 1032
and 1015.
(c) The net reaction is the sum of reactions (1) and (2) and, as you see, the reactant and product
in the coupling reaction do not appear in the net reaction:
(1) HClO(aq) + 3I–(aq) + H+(aq)  Cl–(aq) + I3–(aq) + H2O(l)
(2) I3–(aq) + 2S2O32–(aq)  3I–(aq) + S4O62–(aq)
(8) HClO(aq) + 2S2O32–(aq) + H+(aq)  Cl–(aq) + S4O62–(aq) + H2O(l)
(d) Net reaction (8) shows that the equivalent of one mole of hypochlorous acid reacts with two
moles of thiosulfate ion. We can use this equivalence to work backward from the number of
moles of thiosulfate ion reacted in the analysis to the number of moles of hypochlorous acid in
the sample:
mol S2O32– reacted = (32.56  10–3 L)(0.200 M) = 6.51  10–3 mol S2O32–
1 mol HClO 
–3
mol HClO reacted = (6.51  10–3 mol S2O32–) 
mol
2–  = 3.26  10
2
mol
S
O
2 3 

The concentration of hypochlorite in the bleach is the same as the concentration of hypochlorous
acid found in the analysis reaction:
conc’n of ClO– = conc’n of HClO =
3.26  10 –3 mol
= 0.652 M
5.00  10 –3 L
Problem 10.64.
(a) Two reactions that are important in considering the silver mirror or Tollens’ test for
aldehydes are:
(1) 2Ag+(aq) + RCHO(aq) + 3OH–(aq)  2Ag(s) + RCOO–(aq) + 2H2O(l)
(2) 2Ag+(aq) + 2OH–(aq)  Ag2O(s) + H2O(l) K  1016
Equation (2) represents the precipitation reaction for Ag2O(s). The solubility product for the
dissolution of Ag2O(s), the reverse of reaction equation (2), is:
Ksp = (Ag+(aq))2(OH–(aq))2 = 10–16
Recall that, when a reaction is reversed, the equilibrium constant for the reversed reaction is the
inverse of that for the forward reaction. In the Tollen’s solution with pH = 12 (pOH = 2), the
(OH–(aq)) = 10–2. Substituting for (OH–(aq)) in the solubility product equation and solving gives
(Ag+(aq)) = 10–6. An Ag+(aq) concentration of 10–6 M is the highest concentration of silver ion
that can be present in the solution without precipitating as the oxide.
(b) Use the Nernst equation to calculate E(Ag+,Ag) from E(Ag+,Ag) = 0.799 V and, from part
(a), (Ag+(aq)) = 10–6:
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

1
 0.05916 V 
E(Ag+,Ag) = E(Ag+,Ag) – 
log 

+


 (Ag (aq)) 
1
 0.05916 V 
 1 
E(Ag+,Ag) = (0.799 V) – 
log  –6  = (0.799 V) – (0.355 V) = 0.444 V



 10 
1
(The uncertainties in the solubility product, the hydroxide ion concentration, and, hence, the
silver ion concentration certainly do not justify three significant figures in this cell potential, but
the conclusion in part (c) is unaffected, even if this value is only good to one significant figure.)
(c) To get the minimum value for the cell potential for reaction (1), when RCHO is glucose, we
combine E(Ag+,Ag) with Egluco nate,gluco se . Recall equation (10.85):
RCOO–(aq) + 2H2O(l) + 2e–  RCHO(aq) + 3OH–(aq)
o
Egluco
nate,gluco se = – 0.44 V
The reduction potential here is for pH = 7. As we argued in the text, Egluco nate,gluco se will be more
negative at higher pH where the concentration of OH–(aq) is higher and drives the reaction even
further toward the left, that is, makes the oxidation of glucose even more favorable. Thus, the
minimum value for the oxidation of glucose by silver ion in Tollens reagent is:
E = E(Ag+,Ag) – Egluco nate,gluco se = (0.444 V) – (–0.44V) = 0.88 V
For reaction (1), two moles of electrons are transferred from glucose to silver for every mole of
glucose that reacts, so the minimum free energy change, – nFE, for the reaction is:
G = –2·(9.6485  104 C·mol–1)·(0.88 V) = –1.7  102 kJ·mol–1
Problem 10.65.
The standard reduction potentials, E°´, for the cytochromes in the electron transport pathway in
mitochondria are given alphabetically in this list.
cytochrome
a
a3
b
c
c1
E°´, V
0.29
0.55
0.077
0.254
0.22
The order of the cytochromes in the electron transport pathway, from the one that oxidizes
ubiquinol to the one that is oxidized by oxygen, is the order from weakest to strongest oxidizing
agent, since the weakest one must oxidize ubiquinol and then be reoxidized by the next and so on
until the last one is oxidized by oxygen. The order of increasing oxidizing power (of the oxidized
cytochrome) is: b < c1 < c < a < a3.
Problem 10.66.
(a) Appendix B gives the half reactions and reduction potentials (at pH 7, physiological
conditions) for the two half reactions that make up reaction (i), the reduction of 3phosphoglycerate, 3PG (a carboxylate, RCOO–), to glyceraldehyde-3-phosphate, G3P (an
aldehyde, RCHO), by NADPH:
ACS Chemistry FROG
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Reduction-Oxidation: Electrochemistry
Chapter 10
RCOO– + 3H+ + 2e–  RCHO + H2O
–{NADP+ + H+ + 2e–  NADPH}
E°´ = –0.55 V
–E°´ = 0.32 V
RCOO– + 2H+ + NADPH  RCHO + NADP+ + H2O
E°´ = –0.23 V
The negative cell potential means that the reaction is not favored, as written. The standard free
energy change, G°´ = –(2)(96485 C·mol–1)(–0.23 V) = 44 kJ·mol–1 (of reaction), is positive.
(b) From Appendix B, the two half reactions that make up reaction (iii), reduction of
1,3-diphosphoglycerate, 1,3-DPG (RCOO-PO32–) by NADPH, are:
RCOO-PO32– + 2H3O+ + 2e–  RCHO + HOPO32– + 2H2O
E°´ = –0.29 V
+
+
–
–{NADP + H3O + 2e  NADPH + H2O}
–E°´ = 0.32 V
RCOO-PO32– + NADPH + H3O+  RCHO + NADP+ + HOPO32– + H2OE°´ = 0.03 V
The standard free energy change, G°´ = –(2)(96485 C·mol–1)(0.03 V) = –6 kJ·mol–1 (of
reaction), is small, but negative.
(c) The net reaction for the pathway in part (b) of the problem is the sum of the two reactions:
(2) RCOO– + ATP4–  RCOO-PO32– + ADP3–
(3) RCOO-PO32– + NADPH + H+  RCHO + NADP+ + HOPO32–
RCOO– + ATP4– + NADPH + H+  RCHO + ADP3– + NADP+ + HOPO32–
The standard free energy change for the net reaction is the sum of the standard free energy
changes for reactions (2)—given as 19 kJ·mol–1 in the problem statement—and (3):
G°´ = 19 kJ·mol–1 + (–6 kJ·mol–1)  13 kJ·mol–1
The free energy is somewhat positive, so the net reaction is not favored in the direction written.
(d) We get the [G3P]/[3PG] ratio at equilibrium for each case by using G°´ to get the
equilibrium constant and substituting the cellular conditions of the reaction, [ATP]/[ADP] = 10;
[NADPH]/[NADP+] = 10; and [HOPO32–] = 10–2 M, into the equilibrium constant expression.
For reaction (i):
 G3P  NADP +  
Go 
lnKi = –
= –18 = ln 

RT
 3PG NADPH 
eq
 G3P  NADP +  
 G3P 
1.5  10 = 
 =

 3PG NADPH  eq  3PG  eq
–8
 G3P 
 3PG 
eq
  NADP +  


 NADPH  eq
 NADPH 
= 1.5  10–8 
= 1.5  10–8(10) = 1.5  10–7
+ 
  NADP   eq
For the combination of reactions in part (b) of the problem:
lnK(b) = –
106
 G3P  ADP 3–   NADP +   HOPO2–
Go 
3 

= –5.2 = ln 

RT
3PG  ATP 4–  NADPH 


ACS Chemistry FROG
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Chapter 10
Reduction-Oxidation: Electrochemistry
 G3P  ADP 3–   NADP +   HOPO2–
3 

5.5  10 = 

3PG  ATP 4–  NADPH 

 eq
–3
 G3P 
 3PG 
 G3P 
 3PG 
 NADPH    ATP 4–   

1
= 5.5  10 




+
3–
2– 
  NADP   eq   ADP   eq   HOPO 3   eq
–3
eq
= 5.5  10–3(10)(10)(100) = 55
eq
The indirect pathway in part (b) can provide almost 400 million times more of the desired
product, G3P, glyceraldehyde-3-phosphate, than the direct pathway. This is because the overall
free energy change for the indirect pathway, although unfavorable, is less than one-third as large
as that for the direct pathway. Phosphorylation of the acid requires a large input of free energy,
but produces a product whose reduction is favored. A substantial part of the free energy for the
phosphorylation is provided by the reaction with ATP to transfer a phosphate group to the acid.
(e) The reactions in part (b) are coupled in the sense that we can imagine the hydrolysis free
energy for ATP is used to drive the net reactions. Although reaction (2) occurs directly, when
catalyzed by the appropriate enzyme, we can think of it as being the sum of two reactions:
RCOO– + HOPO32– + H3O+  RCOO-PO32– + 2H2O
ATP4– + 2H2O  ADP3– + HOPO32– + H3O+
Under physiological conditions, G°´ for the first reaction is about 49 kJ·mol–1 and for the
second about –30 kJ·mol–1. Therefore, the direct reaction has G°´  19 kJ·mol–1. This is another
example of the way in which the hydrolysis free energy of ATP is “coupled” to other reactions to
provide a more favorable free energy change for an overall reaction, including the pathway here.
Problem 10.67.
(a) The cell described in the problem is set up for measuring half cell reduction potentials
relative to the SHE, so the cell potential is the reduction potential for the silver half cell with the
silver ion concentration controlled by the solubility of AgSCN(s) in a solution that is 0.10 M in
SCN–(aq). Use the Nernst equation to find the silver ion concentration from the measured cell
potential 0.45 V with the SHE as the anode:
(Ag(s))
(1)
0.05916 V
0.05916 V
0.45 V = E(Ag+,Ag) –
log
= 0.799 V –
log
+
(Ag (aq))
(Ag + (aq))
1
1
log
(0.799 V) – (0.45 V)
(1)
=
= 5.9;
+
0.05916 V
(Ag (aq))
(Ag+(aq)) = 1.3  10–6
[Ag+(aq)] = 1.3  10–6 M
(b) The solubility product expression and solubility product for silver thiocyanate are:
Ksp = (Ag+(aq))(SCN–(aq)) = (1.3  10–6)(0.10 M) = 1.3  10–7
Problem 10.68.
(a) The half reaction in a half cell that contains Ag(s), AgCl(s), and Cl–(aq) is the sum of the
silver ion-silver half reaction and the solubility reaction for solid silver chloride:
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Reduction-Oxidation: Electrochemistry
Chapter 10
Ag+(aq) + e–  Ag(s)
AgCl(s)  Ag+(aq) + Cl–(aq)
AgCl(s) + e–  Ag(s) + Cl–(aq)
(b) Use the standard reduction potential for the silver ion-silver half reaction to get G°(Ag+,
Ag) for the first half reaction in part (a) and the solubility product for silver chloride to get
G°(AgCl Ksp) for the solubility reaction:
G°(Ag+, Ag) = –nFE° = –(1)(9.6485  104 C·mol–1)(0.799 V) = –77.1 kJ·mol–1
G°(AgCl Ksp) = –RTlnKsp = –(8.314 J·K–1·mol–1)·(298 K)·ln(1.8  10–10) = 55.6
–1
kJ·mol
The standard free energy change for the silver chloride-silver half reaction is the sum of these
two values:
G°(AgCl, Ag) =G°(Ag+, Ag) + G°(AgCl Ksp)
G°(AgCl, Ag) = (–77.1 kJ·mol–1) + (55.6 kJ·mol–1) = –21.5 kJ·mol–1
(c) Use G°(AgCl, Ag) to calculate
°(AgCl, Ag):
G (AgCl, Ag)
(–21.5 103 J  mol –1 )
=–
= 0.223 V
nF
(1)(9.6485 10 4 C  mol –1)
This value is identical, within the precision of the data, to the one in Appendix B. The reduction
potential for this half-cell reaction is the information that is usually used to calculate the
solubility product for silver chloride (and other solubility products) by the reverse of the
reasoning for this problem.
o
°(AgCl, Ag) = –
Problem 10.69.
(a) The measured cell potential, E, for this cell is 0.608 V at 298 K:
Ag(s) | AgBr(s) | Br–(aq, 0.050 M) || Ag+(aq, 0.200 M) | Ag(s)
The measured cell potential is the difference between the cathode and anode half-cell potentials:
E = 0.608 V = E(Ag+, Ag) – E(AgBr, Ag)
Each half-cell potential can be expressed in terms of the Nernst equation and the resulting
expression solved for E°(AgBr, Ag) (which solves part (b) of the problem):
 o
0.05916 V log  (Ag(s))  
+
0.608 V = E (Ag ,Ag) –
 (Ag+ (aq))  
1


 o
0.05916 V log  (Ag(s))(Br – (aq))  
– E (AgBr, Ag) –


1
(AgBr(s))  


0.05916 V  1  
log 
0.608 V = (0.799 V) –
 0.200  
1

 o
0.05916 V  1 (0.050)  
log 
– E (AgBr, Ag) –
 

1
1


E°(AgBr, Ag) = 0.073 V
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Convert this standard half-cell reduction potential to the standard free energy change for the
reaction and subtract the standard free energy change for the silver ion-silver half-cell reaction to
get the standard free energy of the solubility reaction (the reverse of the procedure in Problem
10.68) and hence the solubility product:
G°(AgBr, Ag) = –nFE° = –(1)(9.6485  104 C·mol–1)(0.073 V) = –7.0 kJ·mol–1
G°(Ag+, Ag) = –nFE° = –(1)(9.6485  104 C·mol–1)(0.799 V) = –77.1 kJ·mol–1
Therefore,
G°(AgBr Ksp) = G°(AgBr, Ag) – G°(Ag+, Ag) = (–7.0 kJ·mol–1) – (–77.1 kJ·mol–1)
G°(AgBr Ksp) = 70.1 kJ·mol–1 = –RTlnKsp = –(8.314 J·K–1·mol–1)·(298 K)·lnKsp
Ksp = 5.1  10–13
(b) See part (a): E°(AgBr, Ag) = 0.073 V
Problem 10.70.
(a) The anodic half cell is the one containing the silver ion complex with thiosulfate dianion,
Ag(S2O3)23–(aq, from the dissolved Na2S2O3·5H2O, so the cell described in the problem is:
Ag(s) | Ag+(aq), Ag(S2O3)23–(aq), S2O32–(aq) || Ag+(aq, 0.050 M) | Ag(s)
(b) To get the equilibrium constant for complex formation, we need the equilibrium values for
(Ag+(aq)), (S2O32–(aq)), and (Ag(S2O3)23–(aq)) in the complex-forming solution. We can use the
cell potential to find (Ag+(aq)) in the anodic half cell:
E = 0.618 V = E –
0.05916 V
1
log
0.05916 V
0.618 V = (0.000 V) –
1
(Ag + (aq, anode))
(Ag + (aq, cathode))
log
(Ag + (aq, anode))
0.050
(Ag (aq)) = 1.8  10
The number of moles of Na2S2O3·5H2O in the anodic half cell and the concentration of S2O32–
(aq) that would be present in the absence of reaction is:
+
–12
mol Na2S2O3·5H2O =
0.500 g 
248 g  mol  = 2.02  10
–1
–3
mol
2.02  10 –3 mol
= 0.202 M
0.0100 L
Some of this thiosulfate dianion is used by the reaction to form the complex;
Ag+(aq) + 2S2O32–(aq)  Ag(S2O3)23–(aq)
Essentially all of the silver ion initially present has reacted to form the complex (almost none is
left uncomplexed), so [Ag(S2O3)23–(aq)] = 0.050 M. The ratio of thiosulfate dianion to silver ion
reacting is 2:1, so 0.100 M thiosulfate has reacted, leaving [S2O32–(aq)] = 0.102 M in the
complex-forming solution. The equilibrium constant for complex formation is:
[S2O32–(aq)] =
K=
0.050
(Ag(S2O 3 )2–
2 (aq))
=
= 2.7  1012
+
2–
2
–12
2
(Ag (aq))(S2O 3 (aq))
(1.8  10 )(0.102)
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Chapter 10
(c) The half reaction of interest, Ag(S2O3)23–(aq) + e–  Ag(s) + 2S2O32–(aq), is the difference
between the half reaction for silver ion reduction to silver metal and the complexation reaction
whose equilibrium constant we calculated in part (b).
Ag+(aq) + e–  Ag(s)
–[Ag+(aq) + 2S2O32–(aq)  Ag(S2O3)23–(aq)]
Ag(S2O3)23–(aq) + e–  Ag(s) + 2S2O32–(aq)
Convert the standard reduction potential for the silver ion-silver half reaction and the equilibrium
constant in part (b) to the standard free energy changes for these reactions and subtract them to
get the standard free energy change for the half reaction of interest:
G°(Ag+, Ag) = –nFE° = –(1)(9.6485  104 C·mol–1)(0.799 V) = –77.1 kJ·mol–1
G°(Ag(S2O3)23– formation) = –RTlnK = –(8.314 J·K–1·mol–1)·(298 K)·ln(2.7  1012)
G°(Ag(S2O3)23– formation) = –70.9 kJ·mol–1
G°(Ag(S2O3)23–, Ag) = G°(Ag+, Ag) – G°(Ag(S2O3)23– formation)
G°(Ag(S2O3)23–, Ag) = –77.1 kJ·mol–1 – (–70.9 kJ·mol–1) = –6.2 kJ·mol–1
Convert this standard free energy to the standard reduction potential for the reaction:
°(Ag(S2O3)23–, Ag) = –
G o (Ag(S2O2 )2–
(–6.2  10 3 J  mol–1 )
2 , Ag)
=–
= 0.064 V
nF
(1)(9.6485  10 4 C  mol–1 )
Problem 10.71.
(a) Write and solve the Nernst equation for the cell potential as a function of pH for this cell and
cell reaction:
Pt(s) | H2(g, 1 bar) | H+(aq, pH) || Cl–(aq, 1.00 M) |AgCl(s) | Ag(s)
H2(g) + 2AgCl(s)  2H+(aq) + Cl–(aq) + 2Ag(s)
The cell is set up with a hydrogen electrode as the anode, so the standard potential for the cell is
the standard reduction potential for the silver chloride-silver metal half cell, E° = 0.222 V:
E = (0.222 V) –
0.05916 V
2
(Ag(s))2 (H + (aq))2 (Cl – (aq))2
log
(AgCl(s))2 (H 2 (g))
All of the dimensionless concentration ratios in Q are unity, except (H+(aq)), the variable of
interest:
0.05916 V
+
2
E = (0.222 V) –
log (H (aq)) = (0.799 V) – (0.05916 V)·log(H+(aq))
2
E = (0.222 V) + (0.05916 V)·pH
This is the equation of a straight line with an intercept of 0.222 V at pH 0 and a positive slope of
0.05916 V per pH unit:
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The cell potential increases with pH, which is what you expect from the cell reaction in which
H+(aq) is a product. Increasing pH means decreasing (H+(aq)) and Le Chatelier’s principle
predicts that the reaction proceed in a direction to try to increase (H+(aq)), that is, toward
products, which increases the cell potential.
(b) You can either read the pH from the graph at E = 0.435 V, as shown (poorly) on the graph
above, or (better) use the linear equation to calculate pH:
(0.435 V) – (0.222 V)
pH =
= 3.60
0.05916 V
Problem 10.72.
(a) The half reaction for reduction of methylene blue and Nernst equation corresponding to the
half reaction are:
H
N
N
+ H+ (aq) + 2e–
N
S
N
N
S
N
(aq)
(aq)
+
MB (aq)
MBH(aq),


MBH(aq)
0.059 V



E = E° – 0.059logQ = (0.053 V) –
log
2
2
 MB+(aq)H + (aq) 


(b) The data in Problem 10.54 are E vs. pH for this half cell. In order to plot the data to see
whether they obey the Nernst equation, we have to rewrite the Nernst equation, so we have the
reduction potential as a function of pH:
E = (0.053 V) –


 MBH(aq)  0.059 V
0.059 V
1


log 

–
log
2
+
2
2
 H +(aq) 
 MB (aq) 


E = – (0.059 V)pH + (0.053 V) –
 MBH(aq) 
0.059 V
log 

+
2
 MB (aq) 
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This form of the Nernst equation indicates that E should vary linearly with pH (if the ratio of the
MBH(aq) 
methylene blue forms,
, is constant as pH changes) and the slope of the line should
MB+ (aq) 
be –0.059 V. The data give this plot:
The data do not give a linear dependence over the entire pH range, although the values at high
and low pH seem to fall on straight (but different) lines. Neither line has a slope of – 0.059 V, so
the data do not obey the Nernst equation. It is hard to justify saying that the data at either high or
low pH are consistent with the Nernst equation. The intercept of the plot at pH 0 is E for this
half cell, if the (MBH)/(MB+) ratio is unity, so log[(MBH)/(MB+)] = 0. The value for E in
Appendix B is 0.054 V, which is essentially the same as the intercept here and implies that
(MBH)/(MB+) = 1 under the conditions of the measurements reported in Problem 10.54. On the
other hand, the value for E°´ in Appendix B is 0.01 V, the same as in these data at pH 7. The
difference between the pH 0 and pH 7 values, 0.52 V, is not consistent with the Nernst equation,
which would predict a difference of 0.41 V [= 7·(0.05916 V)].
(c) The oxidized form of methylene blue has a positive charge that is shown on one of the
nitrogens in the structure. However, we know that electrons in the pi-electron system of a
molecule like methylene blue are delocalized over the whole molecule. [You can, for example,
write an equivalent form of the Lewis structure in part (a) with the positive charge on the righthand N atom.] This also spreads the positive charge out over the molecule, so the molecule, as a
whole, repels other positive charges and is a weaker base than the reduced form, which is
electrically neutral.
If solutions are prepared with equal amounts of the oxidized and reduced forms of
methylene blue, we might expect the (MBH)/(MB+) ratio to be unity; log[(MBH)/(MB+)] would
be zero in the E vs. pH equation. However, if MBH reacts as a base to give MBH2+ (or even
MBH32+), the concentration of MBH decreases, the logarithm becomes negative and the last term
in the equation becomes positive. This effect would increase E, make the reduction more
positive, than if the acid-base reaction did not occur. This seems to be what we observe in the
figure at low pH where the potential increases faster than expected from the Nernst equation
(which does not account for the “loss” of MBH). Although MB+ also acts as a base, it is a
weaker base than MBH, so is not as much affected by the changes in pH. Because the MBH is
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Reduction-Oxidation: Electrochemistry
more affected by changes in the pH, the ratio is less than unity at low pH. The problem with this
explanation is that it contradicts the reasoning in part (b), which suggests that the (MBH)/(MB+)
ratio is unity at pH 0. The several discrepancies between the experimental data and predictions
from the Nernst equation for the simple reaction written in part (a) suggest that the reaction is not
that simple and that the values for E° and E°´ may not actually reflect concentration conditions
that are standard in this half cell.
At high pH, both MB+ and MBH should be in their base forms and we might expect the
Nernst equation to be obeyed. However, the potential decreases more slowly than predicted by
the Nernst equation. This is the result we would observe if the (MBH)/(MB+) ratio is greater than
one. It could be the case that the positive MB+ molecule reacts with the increasing amounts of
OH– in the solutions (to form some sort of neutral adduct or complex) and is “lost” from the
solution.
Problem 10.73.
(a) To keep track of which concentration is which, use (H+inter) for the hydronium ion
concentration in the intermembrane space and (H+matrix) for the hydronium ion concentration in
the matrix (with a pH one unit greater than in the intermembrane space). This is a concentration
cell, so the standard cell potential is zero. Write the cell reaction as H+inter  H+matrix. Focus
on the transfer of one unit of charge (one hydronium ion) from the intermembrane side of the
membrane to the matrix, so n = 1 in the Nernst equation:
H+ (matrix) = – (0.05916 V)·log 10 –(pH + 1)
E = – (0.05916 V)·log
10– pH
H+ (inter)
E = – (0.05916 V)·log(10–1) = 0.059 V
The cell potential is positive, so transfer of hydronium ions from the intermembrane space to the
matrix is favorable.
(b) The free energy change for this transfer of hydronium ions is:
G = –nFE = – 1·(9.6485  104 C·mol–1)(0.059 V) = –5.7 kJ·mol–1
(c) The spontaneous direction of hydronium ion transfer should be from the side of the
membrane with the higher concentration to that with the lower. The intermembrane space has the
higher hydronium ion concentration (its pH is lower), so transfer from the intermembrane space
to the matrix, as envisioned in the reaction written in part (a), is spontaneous. The positive cell
potential and negative free energy for this transfer are consistent with this direction.
Problem 10.74.
(a) Since the zinc electrode is the anode (where oxidation occurs) the cell reaction is:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
The work done by an electrodchemical cell is nFE, where, in this case, n = 2, since two moles of
electrons are transferred from Zn to Cu2+ for each mole of Zn oxidized. Thus:
work (1 mol reaction) = 2·(9.6485  104 C·mol–1)·(1.15 V) = 222 kJ·mol–1
(b) Since Zn2+(aq) is produced in the upper layer, its concentration increases there. Cu2+(aq) is
used up in the bottom layer by the electrochemical reaction, but more CuSO4(s) dissolves to take
its place and keep the solution saturated, so the Cu2+(aq) remains unchanged as the cell does
work, but more SO42–(aq) is added to the solution. Either some of the Zn2+(aq) has to leave the
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Reduction-Oxidation: Electrochemistry
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upper layer or SO42–(aq) has to leave the bottom layer (probably both occur) to maintain the
electrical neutrality of the upper and lower solutions. As we have said, the CuSO4(s) dissolves to
keep the concentration of Cu2+(aq) approximately constant.
(c) As the cell discharges, the concentration of Zn2+(aq) builds up in the upper layer. Zn2+(aq) is
a product of the electrochemical reaction, and its build up is a disturbance to the system.
Although the system is not at equilibrium, we can use Le Chatelier’s principle to help understand
how the reaction will be affected by this disturbance. If the system adjusts to reduce the stress it
will adjust in such a way as to reduce the amount of Zn2+(aq) formed. The reaction will have less
driving force to go in the direction written and, thus, a decreasing cell potential, as is observed
when the cell is used.
Problem 10.75.
(a) The cell reaction in the Daniell cell is:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
We can use the Nernst equation to determine the ratio, (Zn2+(aq))/(Cu2+(aq)), that will give the
observed cell potential:
1.15 V = 1.10 V –
0.05916 V
2
 (Zn 2+ (aq)) 
log 
2+

 (Cu (aq)) 
 (Zn 2+ (aq)) 
log 
2+
 = –1.69
 (Cu (aq)) 
 (Zn 2+ (aq)) 
 (Cu 2+ (aq))  = 0.020


The upper, Zn2+(aq), solution is quite a bit more dilute than the lower, Cu2+(aq), solution.
(b) The density of the copper sulfate solution is 1.27 g·mL–1, so one liter of the copper sulfate
solution has a mass of 1270 g. Assume that one liter of solution is one liter of water plus enough
CuSO4 to give the observed density. (This assumption will probably somewhat underestimate the
concentration.) One liter of water has a mass of 1000 g. The extra 270 g in the solution must be
CuSO4. This is (270 g)/(159.6 g·mol–1) = 1.69 mol of CuSO4 in one liter of solution. The copper
sulfate solution is 1.69 M. From part (a), we know the ratio of concentrations, so, for the initial
concentration of zinc cation in the cell, we have:
 (Zn 2+ (aq))   (Zn 2+ (aq)) 
0.020 = 
2+
 = 

 (Cu (aq))   1.69 
(Zn2+(aq)) = 0.034;
[Zn2+(aq)] = 0.034 M
(c) In Problem 10.6, we found that 270.7 g (= 4.140 mol) of zinc had gone into solution as
Zn2+(aq) in a cell that we are now told is a Daniell cell. In 2.5 L of solution, this amount of zinc
 4.140 mol 

would give [Zn2+(aq)] = 1.7 M 
 . To this, we should add the initial
2.5
L



concentration of zinc cation, but it is so low, part (b), that it does not change this final
concentration.
(d) To get the cell potential of the used cell, we once again apply the Nernst equation and
substitute the concentrations of Zn2+(aq) (= 1.7 M) and Cu2+(aq) (= 1.69 M):
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ACS Chemistry FROG
Chapter 10
Reduction-Oxidation: Electrochemistry
0.05916 V
 1.7 
log 
= 1.10 V
2
 1.69 
As you see, the cell potential has not declined too much and Problem 10.6 indicated that the zinc
electrode had not been used up. However, the cell is probably no longer useful because the high
concentration of the upper solution has increased its density, so the solutions can begin to mix
and destroy the separation of the half reactions that makes it possible to get electrical work from
the cell. It’s time to recycle the cell solutions, put in fresh solutions and a fresh zinc electrode,
and use the cell again.
E = 1.10 V –
Problem 10.76.
(a) A major advantage of this set up with a porous glass disc for
ions to migrate through (instead of a salt bridge) is alluded to in the
problem statement. You do not need to construct a salt bridge. The
ions in the half-cell solutions carry the charge through the solutions,
so no extra ions are added. A disadvantage is that ions from the two
half cells mix in this cell as they move to maintain charge
neutrality. If the ions in the two half cells react with one another,
the cell potential could be affected and the cell could become
inoperable.
(b) Metal-metal ion half cells could work well in this set up, if the
anion is the same in both half cells and the solutions do not diffuse too rapidly through the
porous glass disc. (The original design of the Daniell cell shown in Problem 10.74 used this sort
of set up.)
(c) Metal-metal ion half cells with different anions might not work so well, especially if the
anion in one half cell reacts with the metal ion or anion in the other half cell.
Problem 10.77.
The term “fuel cells” is generally applied to electrochemical cells in which a “fuel” gives up
electrons with the ultimate acceptor being oxygen, usually from the air. The overall reaction that
occurs is the same as would occur in the combustion of the fuel in air with no production of
electric potential or electric current. In the aluminum-air cell, the fuel, aluminum metal, gives up
its electrons to oxygen, as shown in reaction equation (10.46). This is similar to the reaction that
occurs when pure aluminum metal is exposed to air and almost immediately forms a thin coat of
aluminum oxide, which protects the metal from further reaction with the air and is the reason that
aluminum is such a useful metal for a variety of purposes, even though the pure metal is very
reactive. The fuel you would have to carry around, in case the cell ran out, is pieces of
aluminum, which will be quite safe for both you and your surroundings.
ACS Chemistry FROG
115