Experiment 11
Oxidation-Reduction
Chemistry
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Purpose
To observe and predict the EMF of a voltaic cell and a
concentration cell. Also, to balance redox equations in
acidic and alkaline solutions.
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Background
Voltaic Cells
• An electrochemical cell that uses spontaneous chemical reactions to
produce a voltage.
• The voltage is produced by potential difference between two
substances.
• The transfer of electrons from one ion or molecule to another occurs:
• Oxidation – when one substance loses electrons.
• Reduction – when one substance gains electrons.
• Remember – LEO the lion goes GER (Loss of Electrons is
Oxidation, Gain of Electrons is Reduction)
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
• The interesting part is converting this equation to a workable cell.
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Background
Voltaic Cells
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Background
Voltaic Cells
Current is carried by:
• Electrons (through wire)
• Ions (solution, salt bridge)
Both pathways are required
to form a completed circuit.
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Background
Voltaic Cells
V
Notice that each beaker
(half-cell) contains the
complete half-reaction.
• Anode (Oxidation)
Zn(s)  Zn2+(aq) + 2 e-
• Cathode (Reduction)
Cu2+(aq) + 2 e- Cu(s)
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• Remember the phrase Red
Cat (Reduction occurs at
the Cathode) to help you
remember this.
Background
Voltaic Cells – Cell EMF
Cell electromotive force (Ecell) – the voltage a voltaic cell
generates.
The voltage generated by a voltaic cell depends on a
number of factors, a few of which are:
• the half-cells used
• the concentrations of the reagents
• the temperature of the cell.
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Background
Voltaic Cells – Standard Cell Potential
Cell electromotive force (Ecell) – the voltage a voltaic cell
generates.
If the cell is operated at standard state (298K, 1M solution
concentrations and 1 atmosphere pressure) the voltage
o
E
generated is referred to as a standard cell potential ( cell)
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Background
Voltaic Cells – Standard Cell Potential
Cell electromotive force (Ecell) – the voltage a voltaic cell
generates.
If the cell is operated at standard state (298K, 1M solution
concentrations and 1 atmosphere pressure) the voltage
o
E
generated is referred to as a standard cell potential ( cell)
Zn(s) +
Cu2+(aq)
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
Zu2+(aq)
+ Cu(s)
E
o
cell
  1.10V
Background
Voltaic Cells – Standard Reduction Potential
• Reduction potentials are summarized in a table.
• These potentials can be used to determine the cell potential.
o
o
o
Ecell
 Ered
(cathode)  Ered
(anode)
• If you do the calculation for Eocell and get a negative
number, then you chose the wrong anode and wrong
cathode and need to switch both of them.
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Background
Voltaic Cells – Concentration Cell
• Solution concentration has an impact on cell emf.
• The Nernst equation can be used to predict the emf.
RT
EE 
ln Q
nF
o
• Simplifying the equation yields (at T = 298.15K):
0.0592V
EE 
log Q
n
o
*Ions move towards diluted solution
Products/Reactants
Diluted/Concentrated
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Background
Balancing Oxidation-Reduction Reactions
1. Break up the overall reaction into two half reactions omitting H+, OH-, and H2O.
2. For each half reaction, balance all of the elements except O and H.
3. Balance for oxygen. For each half reaction, balance the O's by adding H2O to the
side of the reaction that is deficient in oxygen.
4. Balance for hydrogen. For each half reaction, add H+ to the side of the reaction that
is deficient in hydrogen. If you are balancing a reaction in acid solution, skip to step 5
BASE ONLY
a) For each half reaction, add OH to both sides of the equation. Add enough OHto neutralize the H+ ions added in step 4.
b) For each half reaction, take the H+ and OH and combine them to form water.
Cancel out water molecules that appear on both sides of the reaction.
5. For each half reaction, balance the charge by adding electrons. The electrons are
added to the more positive side, and the difference in charge gives the number of
electrons to add.
6. Multiply one (or both) half reactions by a whole number so that both half reactions
have the same number of electrons.
7. Add the half reactions. The electrons will cancel, as will some of the H+, OH, and
H2O.
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Background
Balancing Oxidation-Reduction Reactions (Acid)
1. Break up the overall reaction into two half reactions omitting H+, OH-, and H2O.
H2S(aq) + NO3-(aq)  NO(g) + S(s)
H2S(aq)  S(s)
Oxidation half reaction
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NO3-(aq)  NO(g)
Reduction half reaction
Background
Balancing Oxidation-Reduction Reactions (Acid)
2. For each half reaction, balance all of the elements except O and H.
H2S(aq) + NO3-(aq)  NO(g) + S(s)
H2S(aq)  S(s)
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NO3-(aq)  NO(g)
Background
Balancing Oxidation-Reduction Reactions (Acid)
3. Balance for oxygen. For each half reaction, balance the O's by adding H2O to the
side of the reaction that is deficient in oxygen.
H2S(aq) + NO3-(aq)  NO(g) + S(s)
H2S(aq)  S(s)
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NO3-(aq)  NO(g)
Background
Balancing Oxidation-Reduction Reactions (Acid)
4. Balance for hydrogen. For each half reaction, add H+ to the side of the reaction that
is deficient in hydrogen.
H2S(aq) + NO3-(aq)  NO(g) + S(s)
H2S(aq)  S(s)
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NO3-(aq)  NO(g)
Background
Balancing Oxidation-Reduction Reactions (Acid)
5. For each half reaction, balance the charge by adding electrons. The electrons are
added to the more positive side, and the difference in charge gives the number of
electrons to add.
H2S(aq) + NO3-(aq)  NO(g) + S(s)
H2S(aq)  S(s)
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NO3-(aq)  NO(g)
Background
Balancing Oxidation-Reduction Reactions (Acid)
6. Multiply one (or both) half reactions by a whole number so that both half reactions
have the same number of electrons.
H2S(aq) + NO3-(aq)  NO(g) + S(s)
H2S(aq)  S(s)
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NO3-(aq)  NO(g)
Background
Balancing Oxidation-Reduction Reactions (Acid)
6. Multiply one (or both) half reactions by a whole number so that both half reactions
have the same number of electrons.
H2S(aq) + NO3-(aq)  NO(g) + S(s)
H2S(aq)  S(s)
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NO3-(aq)  NO(g)
Background
Balancing Oxidation-Reduction Reactions (Acid)
7. Add the half reactions. The electrons will cancel, as will some of the H+, OH, and
H2O.
3H2S(aq)  3S(s) + 6H+(aq) + 6e6e- + 8H+(aq) + 2NO3-(aq)  2NO(g) + 4H2O(l)
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Background
Balancing Oxidation-Reduction Reactions (Acid)
7. Add the half reactions. The electrons will cancel, as will some of the H+, OH, and
H2O.
3H2S(aq)  3S(s) + 6H+(aq) + 6e6e- + 8H+(aq) + 2NO3-(aq)  2NO(g) + 4H2O(l)
2H+(aq) + 2NO3-(aq) + 3H2S(aq)  2NO(g) + 3S(s) + 4H2O(l)
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Background
Balancing Oxidation-Reduction Reactions (Alkaline)
1. Break up the overall reaction into two half reactions omitting H+, OH-, and H2O.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq)  CrO42-(aq)
Oxidation half-reaction
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MnO2(s)  Mn2+(aq)
Reduction half-reaction
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
2. For each half reaction, balance all of the elements except O and H.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq)  CrO42-(aq)
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MnO2(s)  Mn2+(aq)
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
3. Balance for oxygen. For each half reaction, balance the O's by adding H2O to the
side of the reaction that is deficient in oxygen.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq)  CrO42-(aq)
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MnO2(s)  Mn2+(aq)
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
4. Balance for hydrogen. For each half reaction, add H+ to the side of the reaction that
is deficient in hydrogen.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq)  CrO42-(aq)
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MnO2(s)  Mn2+(aq)
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
4. Balance for hydrogen. For each half reaction, add H+ to the side of the reaction that
is deficient in hydrogen.
a) For each half reaction, add OH to both sides of the equation. Add enough OHto neutralize the H+ ions added in step 4.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq)  CrO42-(aq)
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MnO2(s)  Mn2+(aq)
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
4. Balance for hydrogen. For each half reaction, add H+ to the side of the reaction that
is deficient in hydrogen.
b) For each half reaction, take the H+ and OH and combine them to form water.
Cancel out water molecules that appear on both sides of the reaction.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq)  CrO42-(aq)
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MnO2(s)  Mn2+(aq)
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
4. Balance for hydrogen. For each half reaction, add H+ to the side of the reaction that
is deficient in hydrogen.
b) For each half reaction, take the H+ and OH and combine them to form water.
Cancel out water molecules that appear on both sides of the reaction.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq)  CrO42-(aq)
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MnO2(s)  Mn2+(aq)
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
5. For each half reaction, balance the charge by adding electrons. The electrons are
added to the more positive side, and the difference in charge gives the number of
electrons to add.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq) + 8OH-(aq) CrO42-(aq) + 4H2O(l)
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MnO2(s) + 2H2O(l) 
Mn2+(aq) + 4OH-(aq)
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
6. Multiply one (or both) half reactions by a whole number so that both half reactions
have the same number of electrons.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq) + 8OH-(aq) CrO42-(aq) + 4H2O(l)
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MnO2(s) + 2H2O(l) 
Mn2+(aq) + 4OH-(aq)
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
6. Multiply one (or both) half reactions by a whole number so that both half reactions
have the same number of electrons.
Cr3+(aq) + MnO2(s)  Mn2+(aq) + CrO42-(aq)
Cr3+(aq) + 8OH-(aq) CrO42-(aq) + 4H2O(l)
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MnO2(s) + 2H2O(l) 
Mn2+(aq) + 4OH-(aq)
Background
Balancing Oxidation-Reduction Reactions (Alkaline)
7. Add the half reactions. The electrons will cancel, as will some of the H+, OH, and
H2O.
2Cr3+(aq) + 16OH-(aq)  2CrO42-(aq) + 8H2O(l) + 6e6e- + 3MnO2(s) + 6H2O(l) 3Mn2+(aq) + 12OH-(aq)
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Background
Balancing Oxidation-Reduction Reactions (Alkaline)
7. Add the half reactions. The electrons will cancel, as will some of the H+, OH, and
H2O.
2Cr3+(aq) + 16OH-(aq)  2CrO42-(aq) + 8H2O(l) + 6e6e- + 3MnO2(s) + 6H2O(l) 3Mn2+(aq) + 12OH-(aq)
6e- + 2Cr3+(aq) + 3MnO2(s) + 16OH-(aq) + 6H2O(l) 
2CrO42-(aq) + 3Mn2+(aq) + 12OH-(aq) + 8H2O(l) + 6e-
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Background
Balancing Oxidation-Reduction Reactions (Alkaline)
7. Add the half reactions. The electrons will cancel, as will some of the H+, OH, and
H2O.
2Cr3+(aq) + 16OH-(aq)  2CrO42-(aq) + 8H2O(l) + 6e6e- + 3MnO2(s) + 6H2O(l) 3Mn2+(aq) + 12OH-(aq)
2Cr3+(aq) + 3MnO2(s) + 4OH-(aq)  2CrO42-(aq) + 3Mn2+(aq) + 2 H2O(l)
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Procedure – Experiment 11
- For this experiment, work in pairs.
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Procedure – Experiment 11
Voltaic Cells
Computer Set-Up
1. Hook the rectangular end of the gray cord into the back of the laptop
computer and the round end into Science Workshop black box interface.
2. Plug in and turn on the Pasco Science Workshop black box interface. You
should see a green LED on the front face of the computer.
3. Plug the DIN plug of the voltage sensor into Channel A on the Scientific
Workshop black box interface.
4.Turn on the computer. When the computer has booted-up, double click on
the “Data Studio” icon. The program should start.
5. On the opening screen, select “Create Experiment”.
6. Click and drag the “Voltage” sensor to Channel A of the interface box as
shown in the experimental setup window.
7. Click and drag the "digits" icon labeled “Voltage Ch A (V)”.
8. To have the computer measure the voltage, press “Start”.
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Procedure – Experiment 11
Voltaic Cells
Cell Set-Up (Cu / Zn)
1. Get Zn, Cu, and Mg electrodes. Polish each electrode with sandpaper.
2. Pour approximately 50 mL of 0.1 M CuSO4 into a 150 mL beaker. Pour 0.1 M
Zn(NO3)2 into the porous cup until the cup is about 2/3 full. Put the porous
cup into the CuSO4 solution.
3. Place a Zn electrode into the Zn(NO3)2 solution and a Cu electrode into the
CuSO4 solution.
4. Hook one of the alligator clips to the Zn electrode, and the other alligator clip
to the Cu electrode.
5. Record the voltage on your report sheet. If the voltage is negative, reverse
the alligator clips.
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Procedure – Experiment 11
Voltaic Cells
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Procedure – Experiment 11
Voltaic Cells
Cell Set-Up (Cu / Mg)
6. Remove the alligator clips and empty the contents of the porous cup into the
waste container. Rinse the inside of the cup with distilled water
7. Place 0.1 M Mg(NO3)2 into the porous cup until the cup is about 2/3 full. Put
the Mg electrode into the Mg(NO3)2 solution. Place the porous cup into the
CuSO4 solution.
8. Hook one of the alligator clips to the Cu electrode, and the other alligator clip
to the Mg electrode.
9. Record the voltage on your report sheet. If the voltage is negative, reverse
the alligator clips.
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Procedure – Experiment 11
Voltaic Cells
Cell Set-Up (Zn / Mg)
10. Remove the alligator clips. DO NOT EMPTY ANY CONTAINERS!!
11. Get another 150 mL beaker and put 50 mL of 0.1 M Zn(NO3)2 in the beaker.
Put the porous cup containing the Mg(NO3)2 solution into the beaker
containing the Zn(NO3)2.
12. Place the Zn electrode into the Zn(NO3)2 solution and make sure the Mg
electrode is in the Mg(NO3)2 solution.
13. Hook one of the alligator clips to the Zn electrode, and the other alligator clip
to the Mg electrode.
14. Record the voltage on your report sheet. If the voltage is negative, reverse
the alligator clips.
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Procedure – Experiment 11
Concentration Cells
Cell Set-Up (Concentration Cell)
15. Remove the alligator clips and put both the Mg(NO3)2 solution and the
Zn(NO3)2 solution into the waste container. SAVE THE 0.1 M CuSO4
SOLUTION. YOU WILL USE IT NEXT.
16. Wash the porous cup with distilled water and fill it approximately 2/3 full with
0.001 M CuSO4 solution.
17. Hook each of the alligator clips to the Cu electrodes in each solution.
18. Record the voltage on your report sheet. If the voltage is negative, reverse
the electrodes.
19. To the porous cup containing the 0.001 M CuSO4 solution, add 10 drops 6 M
NH3 and mix well. Record the voltage on your report sheet.
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Safety
• The 6 M NH3 gives off irritating fumes.
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Waste Disposal
• All solutions used in this experiment must be placed in
the container marked "Recovered Metals and Metal Ions".
• Electrode materials should be returned to the “Used
Electrodes” container.
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