1. Chapter 20 Principles of Reactivity: Electron Transfer Reactions
Chem 106, Chapter 20
Chapter 20 Principles of Reactivity: Electron Transfer Reactions
Copyright 2007, David R. Anderson

2. Oxidation-Reduction Reactions
Balancing oxidation-reduction reactions Balancing Redox Equations by the Oxidation Number Change Method (Other Stuff page)
Voltaic (Galvanic) Cells
Cell construction
Cell potential
Effect of concentration on cell potential
Commercial voltaic cells
Electrolytic Cells
Electrolysis
Quantitative aspects of electrolysis

3. I. Oxidation-Reduction Reactions
Review:
2 Na + Cl2  2 NaCl
(0) (0)
(+1)(–1)
oxidation = increase in oxidation number (loss of electrons)
reduction = decrease in oxidation number (gain of electrons)
e.g., Assign oxidation numbers to the species being oxidized and reduced in the following equation, and label the oxidizing agent and reducing agent.
NaI + 3 HOCl  NaIO3 + 3 HCl

4. I. Oxidation-Reduction Reactions
A. Balancing oxidation-reduction reactions
hard way: ion-electron (half-reaction) method (text)
easy way: oxidation number change method (Web - Other Stuff)
1. Determine the oxidation numbers of the species being oxidized and reduced (and make sure there are the same number of each on each side).
2. Balance the changes in oxidation numbers by multiplying each species by the appropriate coefficient (i.e., balance the electrons gained and lost).
3. Balance charges with: H+ in acidic solution OH– in basic solution
4. Balance H (and O!) with H2O.

5. I. Oxidation-Reduction Reactions
A. Balancing oxidation-reduction reactions
e.g., PH3 + I2  H3PO I– (acidic solution)
e.g., MnO4– + H2SO3  Mn SO42– (acidic)

6. I. Oxidation-Reduction Reactions
A. Balancing oxidation-reduction reactions
e.g., CrO2– + S2O82–  CrO42– SO42– (basic)
e.g., Cl2  Cl– + ClO3– (basic)

7. II. Voltaic (Galvanic) Cells
Produce electricity:
chemical energy  electrical energy
2Na+ + 2Cl–  2Na + Cl2 DG >> 0
requires input of electrical energy (electrolysis)
2Ag+ + Ni  2Ag + Ni2+ DG < 0
produces energy
but with Ag+ and Ni in contact, we can’t generate electricity; electrons just flow from Ni to Ag+
have to use a voltaic (galvanic) cell…

8. II. Voltaic (Galvanic) Cells
electrical device
A. Cell construction e–
anode
(–)
cathode
(+)
salt bridge
KCl in gelatin
allows electrolytic conduction without mixing – +
Cl– K+ Ni
Ni(NO3)2(aq) Ag
AgNO3(aq)
Ni2+
Ag+
Ni  Ni2+ + 2e–
(oxidation)
Ag+ + e–  Ag
(reduction)
net: Ni + 2Ag+  Ni2+ + 2Ag

9. II. Voltaic (Galvanic) Cells
A. Cell construction
Potentiometer (voltmeter)
measures cell potential (voltage or electromotive force)
depends on:
species in redox equation
concentrations
temperature
cell potential, E
= actual cell voltage
standard cell potential, Eº
= voltage at standard state
25ºC
1 atm pressure
1 M concentration Ni Ag
Ni2+
Ag+
salt bridge or porous partition
Shorthand notation:
Ni | Ni2+ || Ag+ | Ag

10. II. Voltaic (Galvanic) Cells
B. Cell potential
1. standard reduction potential, Eº
a. Eº = tendency for a species to be reduced
Ag+ + e–  Ag Eº(Ag+)
Ni2+ + 2e–  Ni Eº(Ni2+)
can’t measure directly
Can only measure difference in a voltaic cell:
2Ag+ + 2e–  2Ag Eº(Ag+)
Ni  Ni2+ + 2e– –Eº(Ni2+)
(not 2 x)
net: Ni + 2Ag+  Ni2+ + 2Ag Eºcell = Eº(Ag+) – Eº(Ni2+)
= 1.05 V
General: Eºcell = Eº(species reduced) – Eº(species oxidized)

11. II. Voltaic (Galvanic) Cells
B. Cell potential
1. standard reduction potential, Eº
b. standard: hydrogen electrode at standard state
1 atm H2
1 M H+
Pt electrode
by definition:
at standard state, the reduction
2H+ + 2e–  H2 has Eº = V (exactly)

12. II. Voltaic (Galvanic) Cells
B. Cell potential
1. standard reduction potential, Eº
c. standard reduction potentials
- measure others against the standard hydrogen electrode:
DVM Ag
1 M
Ag+ H+
1 atm H2
Find:
cathode: 2Ag+ + 2e–  2Ag
anode: H2  2H+ + 2e–
Eºcell = 0.80 V
Since Eºcell = Eº(Ag+) – Eº(H+)
0.80 V = Eº(Ag+) – 0.00 V
 Eº(Ag+) = V
(i.e., more easily reduced than H+)

13. II. Voltaic (Galvanic) Cells
B. Cell potential
1. standard reduction potential, Eº
c. standard reduction potentials
Ni | Ni2+(1 M) || H+(1 M) | H2
Find:
cathode: 2H+ + 2e–  H2
anode: Ni  Ni2+ + 2e–
Eºcell = 0.25 V
Since Eºcell = Eº(H+) – Eº(Ni2+)
0.25 V = 0.00 V – Eº(Ni2+)
 Eº(Ni2+) = –0.25 V
(i.e., less easily reduced than H+ or Ag+)

14. II. Voltaic (Galvanic) Cells
B. Cell potential
1. standard reduction potential, Eº
d. determining cell potentials
Ni | Ni2+(1 M) || Ag+(1 M) | Ag
Ag+ + e–  Ag Eº = V
Ni2+ + 2e–  Ni Eº = –0.25 V
more easily reduced
 2Ag+ + 2e–  2Ag
Ni  Ni2+ + 2e–
Ni + 2Ag+  Ni2+ + 2Ag
Eºcell = Eº(Ag+) – Eº(Ni2+)
= V – (–0.25 V)
= V

15. II. Voltaic (Galvanic) Cells
B. Cell potential
2. spontaneity of redox reactions
easiest to
reduce
(strongest
oxidant)
hardest to
oxidize
(weakest
reductant) Eº
+2.87 V
+0.80 V
0.00 V
–0.25 V
–3.05 V
F2 + 2e–
Ag+ + e–
2H+ + 2e–
Ni2+ + 2e–
Li+ + e–
2F– Ag H2 Ni Li
hardest to
reduce
(weakest
oxidant)
easiest to
oxidize
(strongest
reductant)
A species on the left will react spontaneously with a species on the right that is below it in the table.
Or: The species with the more positive Eº will be reduced, and the species with the more negative Eº will be oxidized (Eºcell always > 0).

16. II. Voltaic (Galvanic) Cells
B. Cell potential
2. spontaneity of redox reactions
Given the two half reactions below, what is the net cell reaction? What is Eº? Draw a galvanic cell using these half cells and label the anode and cathode, their charges, and the direction electrons flow in the circuit.
Fe3+ + 3e–  Fe Eº = –0.04 V
Zn2+ + 2e–  Zn Eº = –0.76 V

17. II. Voltaic (Galvanic) Cells
B. Cell potential
3. cell potential and free energy
DGsys = –wsurr
w = # e–s 
energy e–
= coulombs 
joules
coulomb
(= joules)
= coulombs  volts
coulombs = nF (n = # of moles of e–s in redox reaction)
(F = 96,500 coulombs/mol)
 w = nFE
DG = –nFE
DGº = –nFEº
Eºcell > 0, DGº < 0, spontaneous (voltaic)
Eºcell < 0, DGº > 0, nonspontaneous (electrolytic)

18. II. Voltaic (Galvanic) Cells
B. Cell potential
3. cell potential and free energy
e.g., Zn | Zn2+(1 M) || Fe3+(1 M) | Fe
2Fe3+ + 3Zn  2Fe + 3Zn2+ Eºcell = 0.72 V
What is DGº for the cell?

19. II. Voltaic (Galvanic) Cells
C. Effect of concentration on cell potential
1. Nernst equation
DG = DGº + RTlnQ
–nFE = –nFEº + RTlnQ
E = Eº –
RTlnQ nF
at 25ºC: E = Eº –
0.0592 n
logQ
Nernst
equation

20. II. Voltaic (Galvanic) Cells
C. Effect of concentration on cell potential
1. Nernst equation
e.g., Ni | Ni2+ (0.05 M) || Ag+ (0.01 M) | Ag
Ni + 2Ag+  Ni2+ + 2Ag Eº = 1.05 V

21. II. Voltaic (Galvanic) Cells
C. Effect of concentration on cell potential
2. applications
a. measuring Ksp
e.g., AgCl(s)
Ag+ + Cl– Ksp = [Ag+][Cl–]
DVM Ag Ni
AgCl(s)
0.10 M
Cl–
[Ag+] = ?
1.0 M
Ni2+
Find: E = 0.53 V; What is Ksp for AgCl?
Ni + 2Ag+  Ni2+ + 2Ag Eº = 1.05 V

22. II. Voltaic (Galvanic) Cells
C. Effect of concentration on cell potential
2. applications
b. measuring pH
Find E = 0.94 V; What is pH? Ag
1.0 M
Ag+
1 atm H2
lemon juice
[H+] = ?
DVM
H2 + 2Ag+  2H+ + 2Ag Eº = 0.80 V

23. II. Voltaic (Galvanic) Cells
C. Effect of concentration on cell potential
2. applications
A cell was constructed using the standard hydrogen electrode ([H+] = 1.0 M) in one compartment and a lead electrode in a 0.10 M K2CrO4 solution in contact with undissolved PbCrO4 in the other. The potential of the cell was measured to be 0.51 V with the Pb electrode as the anode. Determine the Ksp of PbCrO4 from this data. (Pb2+ + 2e–  Pb Eº = –0.13 V)

24. II. Voltaic (Galvanic) Cells
C. Effect of concentration on cell potential
2. applications
A galvanic cell was constructed with a Cu electrode in a solution of 1.0 M Cu2+ in one compartment and a hydrogen electrode immersed in a sample of a soft drink. The cell potential was measured to be V. What was the pH of the soft drink? (Cu2+ + 2e–  Cu Eº = V)

25. II. Voltaic (Galvanic) Cells
D. Commercial voltaic cells
anode: Pb + SO42–  PbSO4 + 2e–
cathode: PbO2 + 4H+ + SO42– + 2e–  PbSO4 + H2O
discharge
charge
cell: Pb + PbO2 + 4H+ + 2SO42–
2PbSO4 + 2H2O E ~ 2 V (6 cells in series)
discharge: H2SO4 consumed, H2O produced
dilutes electrolyte solution
can measure with densitometer

26. II. Voltaic (Galvanic) Cells
D. Commercial voltaic cells
Zinc cup
anode
anode: Zn + 2OH–  ZnO + H2O + 2e–
cathode: 2MnO2+ H2O + 2e– 
Mn2O3 + 2OH–
cell: Zn + 2MnO2  ZnO + Mn2O3
E ~ 1.5 V
Graphite
cathode
Moist paste
of MnO2,
KOH and
H2O
Porous
partition

27. II. Voltaic (Galvanic) Cells
D. Commercial voltaic cells
Anode
cap
Partition
Cathode:
Ag2O paste
Anode:
Zn and KOH
Gasket
Cell
can
anode: Zn + 2OH–  ZnO + H2O + 2e–
cathode: Ag2O + H2O + 2e–  2Ag + 2OH–
cell: Zn + Ag2O  ZnO + 2Ag
E ~ 1.5 V

28. III. Electrolytic cells
A. Electrolysis
electrical energy  chemical energy
e.g., NaCl(l)  Na (l) + Cl2(g) DG >> 0
cell: 2Na+ + 2Cl–  2Na + Cl2
Eº = Eº(Na+) - Eº(Cl2)
= (-2.71) - (1.36)
= V
DGº = kJ/mol

29. III. Electrolytic cells
A. Electrolysis E
e.g., NaCl(aq)  ? Eº
cathode: 2H2O + 2e–  H2 + OH– V
(reduction) Na+ + e–  Na V
H2O more easily
reduced than Na+ Eº
anode: Cl2 + 2e–  2Cl– V
(oxidation) O2 + 4H+ + 4e–  2H2O V
H2O more easily
oxidized than Cl–
net: (2H2O + 2e–  H2 + OH–) x 2
2H2O  O2 + 4H+ + 4e–
2H2O  2H2 + O2

30. III. Electrolytic cells
A. Electrolysis E
e.g., CuCl2(aq)  ? Eº
cathode: Cu2+ + 2e–  Cu V
(reduction) 2H2O + 2e–  H2 + OH– V
Cu2+ more easily
reduced than H2O Eº
anode: Cl2 + 2e–  2Cl– V
(oxidation) O2 + 4H+ + 4e–  2H2O V
H2O more easily
oxidized than Cl–
net: (Cu2+ + 2e–  Cu) x 2
2H2O  O2 + 4H+ + 4e–
2Cu2+ + 2H2O  2Cu + O2 + 4H+

31. IV. Electrolytic cells B. Quantitative aspects of electrolysis
Units of charge: 1 faraday (F) = 1 mol e–s
1 coulomb = 1 amp ·1 sec (A·s)
experimentally: 1 F = 96,500 C
e.g., How many moles of Na and Cl2 are produced in the electrolysis of NaCl(l) when a current of 25 A is applied for 8.0 hours?

32. IV. Electrolytic cells B. Quantitative aspects of electrolysis
e.g., How long would it take to deposit 21.4 g of Ag from a solution of AgNO3 using a current of 10.0 A?