Oxidation - Reduction Reactions
Redox Terminology—REVIEW CHAPTER 4!!!
redox reaction -- electron transfer process
e.g.,
2 Na + Cl2 --> 2 NaCl
Overall process involves two Half Reactions:
oxidation -- loss of electron(s)
reduction -- gain of electron(s)
e.g.
Na --> Na+ + e–
(oxidation)
Cl2 + 2 e– --> 2 Cl–
(reduction)
Related terms:
oxidizing agent = the substance that gets reduced (Cl2)
reducing agent = the substance that gets oxidized (Na)
Oxidation and reduction always occur together so that there is no
net loss or gain of electrons overall.
OILRIG
Redox Reactions
• The transfer of electrons between species, meaning that
one species is oxidized and one is reduced. The two
processes will always occur together.
• When has a redox reaction occurred?
– If there is a change in the oxidation state of any element in the
reaction, a redox reaction has happened.
– Remember that if something is oxidized, something else must
be reduced! (and vice-versa)
More examples:
2 Ca(s) + O2(g) --> 2 CaO(s)
2 Al(s) + 3 Cl2(g) --> 2 AlCl3(s)
HALF-REACTION METHOD for balancing Redox Equations
1. Write unbalanced ionic equations for the two half-reactions. (Look
at oxidation numbers in the “skeleton equation.”)
2. Balance atoms other than H and O.
3. Add appropriate number of electrons.
4. Balance O with H2O.
5. Balance H with H+.
6. If in acidic solution, then skip to step 7. If in basic solution, then add
equal number of OH– to both sides to cancel all of the H+.
7. Multiply balanced half-reactions by appropriate coefficients so that
the number of electrons are equal.
8. Rewrite the balanced half reactions.
9. Add the half-reactions together.
10. Cancel species that appear on both sides to get the balanced Net
Ionic Equation.
11. If necessary, add spectator ions to get the balanced molecular
equation.
Check the Final Balance (atoms and charges)!
Example Problem
Balance the following redox equation.
Ce4+(aq) + Sn2+(aq) --> Ce3+(aq) + Sn4+(aq)
The reaction can be separated into a reaction involving the
substance being reduced;
Ce4+(aq) + e– --> Ce3+(aq)
And the substance being oxidized;
Sn2+(aq) --> Sn4+(aq) + 2e–
We can see that the equations don’t balance; you must
multiply the top equation by two, then add them to get
2 Ce4+(aq) + Sn2+(aq) --> 2 Ce3+(aq) + Sn4+(aq)
Sample Problem
The following redox reaction occurs in basic solution. Write
complete, balanced equations for the individual halfreactions and the overall net ionic equation.
MnO4–(aq) + N2H4(aq) --> MnO2(s) + NO(g)
Oxidation Half Reaction:
Reduction Half Reaction:
Net Ionic Equation:
Sample Problem
The following redox reaction occurs in basic solution. Write
complete, balanced equations for the individual halfreactions and the overall net ionic equation.
MnO4–(aq) + N2H4(aq) --> MnO2(s) + NO(g)
Oxidation Half Reaction:
8 OH–(aq) + N2H4(aq) --> 2 NO(g) + 6 H2O(l) + 8 e–
Reduction Half Reaction:
3 e– + 2 H2O(l) + MnO4–(aq) --> MnO2(s) + 4 OH–(aq)
Net Ionic Equation:
3 N2H4(aq) + 8 MnO4–(aq) --> 6 NO(g) + 8 MnO2(s) + 2 H2O(l) +
8 OH–(aq)
Sample Problems
•
•
Balance the following redox equations in acidic solution.
Al(s) + I2(s) --> AlI3(s)
•
KClO3(aq) + HNO2(aq) --> KCl(aq) + HNO3(aq)
•
Cl–(aq) + MnO2(s) --> Mn2+(aq) + Cl2(g)
•
Balance the above redox equations in basic solution.
Sample Problems
•
•
Balance the following redox equations in acidic solution.
Al(s) + I2(s) --> AlI3(s)
2 Al(s) + 3 I2(s) --> 2 AlI3(s)
•
KClO3(aq) + HNO2(aq) --> KCl(aq) + HNO3(aq)
KClO3(aq) + 3 HNO2(aq) --> KCl(aq) + 3 HNO3(aq)
•
Cl–(aq) + MnO2(s) --> Mn2+(aq) + Cl2(g)
4H+(aq) + 2 Cl–(aq) + MnO2(s) --> Cl2(g) + Mn2+(aq) + 2 H2O(l)
•
Balance the above redox equations in basic solution.
– First doesn’t change.
– Second: use NO2– and NO3– rather than the acids.
– Third: Add 4 OH- to each side,
2 H2O(l) + 2 Cl–(aq) + MnO2(s) --> Cl2(g) + Mn2+(aq) + 4 OH–(aq)
Electrochemistry Terminology
Electrochemical Cell --
a device that converts electrical
energy into chemical energy or
vice versa
• Two Types:
Electrolytic cell
Converts electrical energy into chemical energy
Electricity is used to drive a non-spontaneous reaction
Galvanic (or voltaic) cell
Converts chemical energy into electricity (a battery!)
A spontaneous reaction produces electricity
• Conduction:
Metals:
metallic (electronic) conduction
Solutions:
(or molten salts)
-- free movement of electrons
electrolytic (ionic) conduction
-- free movement of ions
Galvanic Cells
Galvanic Cells (batteries) -- produce electrical energy
e.g. a spontaneous reaction:
Cu(s) + Ag+(aq) --> Cu2+(aq) + Ag(s)
(Ag metal will be deposited on a Cu wire dipped into aqueous
AgNO3 solution)
In a galvanic cell, the half reactions are occurring in separate
compartments (half-cells)
Electrochemical Conventions
Charges on the electrodes
Galvanic cell:
Electrolytic cell:
(+)
(–)
(+)
(–)
cathode ~ reduction
anode ~ oxidation
anode ~ oxidation
cathode ~ reduction
Cell Notation -- summary of cell description
e.g.
Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s)
anode
cathode
Cell Potential
Cell potential ~ Eºcell (an electromotive force, emf)
Units of Eºcell are volts: 1 volt = 1 joule/coulomb
Eºcell is a measure of the relative spontaneity of a cell reaction
Positive (+) Eºcell --> spontaneous reaction
Eºcell depends on:
– Nature of reactants
– Temperature -- superscript º means 25 ºC
– Concentrations -- superscript º means all conc are at 1.00
M and gases are at 1.00 atm
but, Eºcell is independent of amounts of reactants
Standard Potentials
Standard Reduction Potential:
the potential of a half-cell relative to a standard reference
(Pt)
Sample Problem; Write the cell notation for this cell.
Standard Cell Potential
Standard Reduction Potentials
Eº (volts)
F2(g) + 2 e– --> 2 F–(aq)
+ 2.87
Ag+(aq) + e– --> Ag(s)
+ 0.80
Cu2+(aq) + 2 e– --> Cu(s)
+ 0.34
2 H+(aq) + 2 e– --> H2(g)
0.00
Zn2+(aq) + 2 e– --> Zn(s)
– 0.76
Li+(aq) + e– --> Li(s)
– 3.05
• Ease of Reduction ~ increases with Eº
e.g. F2 is easiest to reduce, Li+ is the hardest
• Standard Cell Potential ~ Eºcell can be determined from
standard reduction potentials:
Eºcell = Eºred – Eºoxid
= [reduction potential of substance reduced]
– [reduction potential of substance oxidized]
Standard Potential Example Problem
Is the following a galvanic or an electrolytic cell? Write the
balanced cell reaction and calculate Eºcell.
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Cathode:
Anode:
Cell rxn:
Cu2+ + 2 e– --> Cu
Eº = 0.34 V
Zn --> Zn2+ + 2 e–
Eº = – 0.76 V
Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)
Eºcell = Eºred – Eºox = 0.34 V – (– 0.76 V)
= + 1.10 V ∴ galvanic.
Another Example Problem
Given the standard reduction potentials:
O2(g) + 4 H+(aq) + 4 e– --> 2 H2O(l) Eº = 1.23 V
Cl2(g) + 2 e– --> 2 Cl–(aq)
Eº = 1.36 V
Write a balanced equation and calculate Eºcell for a galvanic
cell based on these half reactions. Sketch the galvanic cell.
“galvanic” implies a positive value for Eºcell
so, Cl2/Cl– should be the reduction half reaction,
since 1.36 – 1.23 = + 0.13 V
∴ reverse the O2 half reaction and make it the oxidation
Multiply the Cl2 reaction by 2, to make e–’s cancel, hence:
2 Cl2(g) + 2 H2O(l) --> 4 Cl–(aq) + O2(g) + 4 H+(aq)
Note: When a half-reaction is multiplied by a coefficient, the Eº
IS NOT MULTIPLIED by the coefficient.
Diagram of cell
e–
e–
voltmeter
Pt
electrode
Pt electrode
salt bridge
cathode
(reduction)
O2 gas
Cl2 gas
K+
K+
Cl–
Cl–
K+
K+
NO3–
Cl–
K+
K+
NO3–
supporting electrolyte
NO3–
anode
(oxidation)
Sample Problem
Under standard conditions, is the following a galvanic or an
electrolytic cell? Support your conclusion with appropriate
calculations and balanced chemical reactions.
Ag(s), AgBr(s) | Br–(aq) || Au3+(aq) | Au(s)
--From table,
Au3+(aq) + 3e –  Au(s) Eº = 1.50 V
AgBr(s) + e –  Ag(s) + Br –(aq) Eº = 0.071 V
Sample Problem
Under standard conditions, is the following a galvanic or an
electrolytic cell? Support your conclusion with appropriate
calculations and balanced chemical reactions.
Ag(s), AgBr(s) | Br–(aq) || Au3+(aq) | Au(s)
--From table,
Au3+(aq) + 3e –  Au(s) Eº = 1.50 V
AgBr(s) + e –  Ag(s) + Br –(aq) Eº = 0.071 V
3 Ag(s) + 3 Br–(aq) + Au3+(aq) --> 3 AgBr(s) + Au(s), Eºcell = + 1.43
V, so galvanic.
Reactions of Metals with Acids
Generally:
metal + acid --> salt + H2(g)
e.g. Zn(s) + H2SO4(aq) --> ZnSO4(aq) + H2(g)
Metal is oxidized to yield the metal cation
H+ from the acid is reduced to yield H2
Some acids (especially HNO3) are “oxidizing acids” -- the NO3–
ion is reduced to form NO2 or NO depending on
concentration of the acid.
e.g. reactions of Cu metal with HNO3
Cu(s) + concentrated HNO3 --> NO2(g)
Cu(s) + dilute HNO3 --> NO(g)
21
Cell Potential and Thermodynamics
Free Energy Change (DG)
DG = – Wmax (maximum work)
• For an electrochemical cell:
W = nFEcell
where n = # moles e–
F = 96,500 coulombs/mole e–
joule
Units: volt = coulomb
coulombs
joules = (moles e–)
mole e–
joule
coulomb
∴ free energy is related to Ecell as follows:
DG = – nFEcell
under standard conditions: use DGº and Eºcell
Equilibrium Constant
since DGº = – RTlnKc = – nFEºcell can rearrange to:
RT
Eºcell = nF ln Kc
(Use 8.314 J mol-1 K-1 for R!)
at 25 ºC, use values for R, T, and F, then convert to base-10 log:
Eºcell =
Overall,
0.0592
log Kc
n
Sig Fig Review: Log, Ln
Rules are the same for log and ln!
When you report the log or ln of a number, the significant digits
are the ones after the decimal point, e.g.
ln (24.3) = 3.190
[3 sig fig!]
log (0.068) = –1.17
[2 sig fig!]
Antilog is the opposite; when you convert back to natural
numbers, you will “appear” to lose significant digits, e.g.
Antilog (23.32) = 1023.32 = 2.1 x 1023
[2 sig fig!]
Anti(natural log) of 23.32 = e23.32 = 1.3 x 1010
[2 sig fig!]
Sample Problem
Given the following standard reduction potentials, calculate
the solubility product constant (Ksp) for lead sulfate, PbSO4.
PbSO4(s) + 2 e– --> Pb(s) + SO42–(aq) Eº = – 0.360 V
Pb2+(aq) + 2 e– --> Pb(s)
Eº = – 0.130 V
Sample Problem
Given the following standard reduction potentials, calculate
the solubility product constant (Ksp) for lead sulfate, PbSO4.
PbSO4(s) + 2 e– --> Pb(s) + SO42–(aq) Eº = – 0.360 V
Pb2+(aq) + 2 e– --> Pb(s)
Eº = – 0.130 V
Kc = 2 x 10–8
Effect of Concentration on Cell Potential
For a general reaction:
aA + bB  cC + dD
DG = DGº + RTln Q
where Q = “concentration quotient” or “reaction quotient”
= [C]c[D]d / [A]a[B]b (use given concentrations)
Since DG = – nFE :
– nFE = – nFEº + RTln Q
Rearrange to get the “Nernst Equation”
E = Eº – [RT/nF]ln Q
(review from
Ch. 14-16)
The Nernst Equation
E = Eº – [RT/nF]ln Q
The Nernst Equation shows the relationship between
the standard cell potential (Eº) and the cell potential
(E) under actual, non-standard conditions.
this can be simplified at 25 ºC to:
E = Eº – (0.0592/n)log Q
Major use of the Nernst Equation:
• Determine concentrations from standard reduction
potentials
• Use actual concentrations (i.e. Q) to calculate Ecell
Sample Problem
A silver wire coated with AgCl is sensitive to the chloride ion
concentration because of the following half-cell reaction.
AgCl(s) + e– --> Ag(s) + Cl–(aq)
Eº = 0.2223 V
Calculate the molar concentration of Cl– when the potential
of this half-cell is measured to be 0.4900 volts relative to a
standard hydrogen electrode.
Sample Problem
A silver wire coated with AgCl is sensitive to the chloride ion
concentration because of the following half-cell reaction.
AgCl(s) + e– --> Ag(s) + Cl–(aq)
Eº = 0.2223 V
Calculate the molar concentration of Cl– when the potential
of this half-cell is measured to be 0.4900 volts relative to a
standard hydrogen electrode.
3 x 10–5 M
Quantitative Aspects of Electrochem ~ the Faraday
e.g. consider a cell with the cathode reaction:
Cr3+(aq) + 3 e– --> Cr(s)
Stoichiometry: 1 mole Cr3+ ~ 3 moles e– ~ 1 mole Cr
– If a current equivalent to 3 moles of electrons is passed
through the solution of Cr3+, then 1 mole of Cr metal will be
produced
By definition: 1 Faraday (F) = 1 mole electrons
but, how much electrical current is this?
1 F = 9.65 x 104 coulombs
and
1 coulomb = 1 amp sec
or, 1 amp =
1 coulomb
1 sec
Batteries
• Dry cells (paste)
– Metal anode is oxidized, solid MnO2 in
paste is reduced
– Supporting electrolyte is either
• acid (can corrode more easily, cannot
recharge), or
• alkaline (base, lasts longer)
• Lead storage (car batteries!)
– Lead is oxidized, PbO2 is reduced
– Acid electrolyte (H2SO4)
– Rechargeable, heavy
Light-Weight Rechargeable Batteries
• NiCAD
– Cadmium is oxidized, NiO2 is reduced
– Supporting electrolyte is KOH
• NiMH
– Hydrogen is oxidized (stored as metal
hydride), NiO2 is reduced
– Supporting electrolyte is KOH
• Lithium ion
– “Concentration” cell
– Migration of lithium ions from graphite
anode to metal oxide cathode
– Supporting electrolyte is KOH
Fuel Cells
• like batteries in which reactants are constantly
being added
– so it never runs down!
• Pt electrodes, KOH electrolyte
Simpler diagram
Electrolysis
-- chemical reactions that occur during electrolytic conduction
e.g. molten NaCl
reduction:
Na+(l) + e– --> Na(s)
Na+ ions migrate toward
the negative electrode
and are reduced
oxidation:
2 Cl–(l) --> Cl2(g) + 2e–
Cl– ions migrate toward
the positive electrode
and are oxidized
Electrolysis Cell
• Electrodes:
cathode -- where reduction occurs
anode -- where oxidation occurs
• Net Cell Reaction:
add anode & cathode half-reactions so that # of
electrons cancel
2 Cl– --> Cl2 + 2 e– (anode ~ oxidation)
2 [ Na+ + e– --> Na ] (cathode ~ reduction)
2 Cl–(l) + 2 Na+(l) --> Cl2(g) + 2 Na(s) (Cell Reaction)
Overvoltage, or overpotential — in some cases additional voltage must
be applied in order to get an electrolytic reaction to occur (kinetic
reasons)
Electrolysis of Water
In aqueous solution electrolysis of water may occur:
• oxidation (H2O --> O2) -- at anode
2 H2O --> O2 + 4 H+ + 4e– Eº = 1.23 V
• reduction (H2O --> H2) -- at cathode
in neutral or basic solution
2 H2O + 2 e– --> H2 + 2 OH– Eº = -0.83 V
• reduction (H+ --> H2) -- at cathode
in acidic solution
2 H+ + 2 e– --> H2 Eº = 0.00 V
Higher Eº  easier to reduce!
Lower Eº  easier to oxidize!
Some Industrial Applications of Electrolysis
Preparation of Aluminum from molten Al2O3
(can’t use Al3+ in solution since H2O is easier to reduce)
anode: 3 [O2– --> 1/2 O2 + 2 e– ]
cathode: 2 [ Al3+ + 3 e– --> Al ] (Eº = –1.66 V)
Cell: 2 Al3+(l) + 3 O2–(l) --> 3/2 O2(g) + 2 Al(s)
Electroplating
reduction of metal ions (from solution) to
pure metals (Ag, Cr, Cu, etc.)
e.g.
Cr3+(aq) + 3 e– --> Cr(s) (chrome plating)
Example Problem
How long (in hours) would it take to produce 25 g of Cr from a
solution of Cr3+ by a current of 2.75 amp?
Cr3+(aq) + 3 e– --> Cr(s)
1 mole Cr ~ 3 moles e– ~ 3 F ~ 3 x 9.65 x 104 coulombs
1 mole Cr
25 g Cr x
1.44 F x
52 g Cr
x
96,500 coulombs
F
139,000 amp sec
2.75 amp
x
3F
1 mole Cr
= 1.44 F
= 139,000 coulombs = 139,000 amp sec
1 hr
3600 sec
= 14 hr
Review Problems
1. A large electrolytic cell that produces metallic aluminum
from Al2O3 ore is capable of making 250 kg of aluminum in
24 hours. Determine the current (in amps) that is required
for this process. Include appropriate chemical reactions.
2. An aqueous solution of NaBr was electrolyzed with a
current of 2.50 amps for 15.0 minutes. What volume (in mL)
of 0.500 M HCl would be required to neutralize the resulting
solution? (Hint: H2 is produced at the cathode and Br2 at
the anode.)
Review Problems
1. A large electrolytic cell that produces metallic aluminum
from Al2O3 ore is capable of making 250 kg of aluminum in
24 hours. Determine the current (in amps) that is required
for this process. Include appropriate chemical reactions.
3.1 x 104 amps (Al3+(aq) + 3 e– --> Al(s) is ½-reaction, overall
reaction is 2 Al2O3(l)  3 O2(g) + 4 Al(s))
2. An aqueous solution of NaBr was electrolyzed with a
current of 2.50 amps for 15.0 minutes. What volume (in mL)
of 0.500 M HCl would be required to neutralize the resulting
solution? (Hint: H2 is produced at the cathode and Br2 at
the anode.)
46.6 mL HCl soln
Review Problems, cont.
3. Consider the following electrochemical cell in which the
volume of solution in each half-cell is 100 mL.
Zn(s) | Zn2+ (1.00 M) || Ag+ (1.00 M) | Ag(s)
(a) Write balanced chemical equations for the anode,
cathode, and overall cell reactions.
(b) Determine Eºcell, DGº, and the equilibrium constant (Kc)
for the cell reaction.
(c) If current is drawn from this cell at a constant rate of
0.10 amp, what will the cell potential be after 15.0 hours?
Answers
3. (a) Anode: Zn(s) --> Zn2+(aq) + 2 e–
Cathode: Ag+(aq) + e– --> Ag(s)
Overall: Zn(s) + 2 Ag+(aq) --> Zn2+(aq) + 2 Ag(s)
(b) Eºcell = + 1.56 V
DG = - 301 kJ/mol
Kc = 6 x 1052
(c) 1.54 V