1. Redox Reactions and Electrochemistry
Galvanic cells
Standard reduction potential
Cell potential
Free energy
Effect of concentration
Battery
Corrosion
Electrolysis

2. I. Oxidization–Reduction (Redox) Reaction
Electrochemical processes are redox reactions in which:
energy released by a spontaneous reaction is converted to electricity
electrical energy is used to cause a nonspontaneous reaction 2+ 2-
2Fe (s) + O2 (g) FeO (s)
Oxidation half-reaction (lose e-):
Fe is oxidized and Fe is the reducing agent
2Fe Fe2+ + 4e-
Reduction half-reaction (gain e-):
O2 is reduced and O2 is the oxidizing agent
O2 + 4e O2-

3. A. Oxidation number/state
Free elements have an oxidation number (ON) =0
Monatomic ions, the ON is equal to the charge on the ion.
The ON of oxygen is usually –2. In H2O2 and O22-:–1, in O2- : -1/2
The ON of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.
Fluorine is always –1, Group IA metals are +1, IIA metals are +2 in compounds.
The sum of the ON of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.
Ex. Oxidation numbers of all the atoms in HSO3–
1 +3x (-2)+S= -1
O =-2
H= +1
S=?
S=4

4. B. Balance redox reaction
By inspection
Cu+ H Cu2+ + H2 2
Ion-electron method (half-reaction method)
By balancing the number of electrons lost and gained
during the reaction
Sn2+ (aq) + Fe3+ (aq) Sn4+(aq) + Fe2+(aq) 2
Oxidation half reaction
Sn2+ (aq) Sn4+(aq) + 2 e–
Reduction half reaction
( ) x 2
Fe3+ (aq) + e– Fe2+(aq)
2 Fe3+ (aq) + 2e– Fe2+(aq)

5. Ion-electron method
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
Write the unbalanced equation for the reaction ion ionic form.
Fe2+ + Cr2O Fe3+ + Cr3+
Separate the equation into two half-reactions.
Fe Fe3+ +2 +3
Oxidation:
Cr2O Cr3+ +6 +3
Reduction:
Balance the atoms other than O and H in each half-reaction.
Cr2O Cr3+

6. Ion-electron method
For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.
Cr2O Cr3+ + 7H2O
14H+ + Cr2O Cr3+ + 7H2O
Add electrons to one side of each half-reaction to balance the charges on the half-reaction.
Fe Fe3+ + 1e-
6e- + 14H+ + Cr2O Cr3+ + 7H2O
If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.
6Fe Fe3+ + 6e-
6e- + 14H+ + Cr2O Cr3+ + 7H2O

7. Ion-electron method
Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.
Oxidation:
6Fe Fe3+ + 6e-
Reduction:
6e- + 14H+ + Cr2O Cr3+ + 7H2O
14H+ + Cr2O Fe Fe3+ + 2Cr3+ + 7H2O
Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.

8. Summary of balancing redox reaction in acidic solutions
1. Divide reaction into two incomplete half-reactions
2. Balance each half-reaction by doing the following:
a. Balance all elements, except O and H
b. Balance O by adding H2O
c. Balance H by adding H+
d. Balance charge by adding e– as needed
3. If the electrons in one half-reaction do not balance those in the other, then multiply each half-reaction to get a common multiple
4. The overall reaction is the sum of the half-reactions
5. The oxidation half-reaction is the one that produces electrons as products, and the reduction half-reaction is the one that uses electrons as reactants

9. Summary of balancing redox reaction in basic solutions
Steps 1-3 are the same as the reaction in acidic solution
4. Since we are under basic conditions, we neutralize any H+ ions in solution by adding an equal number of OH– to both sides
5. The overall reaction is the sum of the half-reactions
Ex. Balance I– + OCl– I2 + Cl– in basic solution?
Oxidation
I– I2
2 I– I2
+ 2 e–
Reduction
Add up
2 OH– +
2 H+ +
2e– +
OCl– Cl–
+ H2O
+ 2 OH–
2 H2O+2 e– + OCl– Cl– + H2O + 2 OH–
2 e– + H2O + OCl– Cl– + 2 OH–
2 I– + H2O + OCl– I2 + Cl– + 2 OH–

10. II. Galvanic Cells
Consider reaction Zn(s) + Cu(NO3)2 --> Cu(s) + Zn(NO3)2
Spontaneous (activity series, chapter 4)
half reaction: Zn(s) --> Zn2+ + 2e– Cu2+ +2 e– --> Cu(s)
Galvanic cell: a device uses the spontaneous redox reaction to produce an electric current
e– flow e–
Wire
Oxidation
Zn --> Zn2+ + 2e–
Anode
Reduction
Cu2+ +2 e– --> Cu
Cathode
Flow of cation
Flow of anion
Salt bridge
Electrode: anode (oxidation rxn) and cathode (reduction rxn)
Salt bridge: contains conc. soln of strong electrolyte (KCl)

11. II. Galvanic Cells
anode
oxidation
cathode
reduction

12. II. Galvanic cell
Cell diagram : a conventional notation to represent galvanic cell
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M and [Zn2+] = 1 M
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode
cathode
Anode is on the left, cathode on the right
| : represent boundary between phases
|| : represent salt bridge
Ex. Draw cell diagram for the following reaction:
Al (s) + Ag+ (aq) Al3+ (s) + Ag (s)
Al (s) | Al3+ || Ag+ | Ag (s)

13. III. Standard reduction potential
A. Cell voltage, cell potential, electromotive force (emf)
The potential energy of the e– is higher at the anode than at the cathode
The difference in electrical potential between the anode and cathode is called as emf
Unit: V (volt) = 1 J/C
C: coulomb, unit of charge
4. Standard cell potential, Eocell
Cell potential under standard condition
Zn(s)| Zn2+(1 M) || H+ (1 M)| H2 (1 atm)| Pt (s)
Anode (oxidation):
Zn (s) Zn2+ (1 M) + 2e-
Cathode (reduction):
2e- + 2H+ (1 M) H2 (1 atm)
Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

14. B. Standard reduction potential
Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm.
Reduction Reaction
2e- + 2H+ (1 M) H2 (1 atm)
E0 = 0 V
Standard hydrogen electrode (SHE) : half-cell contain 1M H+ and 1 atm H2, used as a reference to determine Eo for other species

15. B. Standard reduction potential
E0 = 0.76 V
cell
Standard emf (E0cell )
E0 = Ecathode - Eanode
cell
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
E0 = EH /H - EZn /Zn
cell + 2+ 2
0.76 V = EZn /Zn 2+
EZn /Zn = V 2+
Zn2+ (1 M) + 2e Zn E0 = V

16. B. Standard reduction potential
E0 = 0.34 V
cell
E0 = Ecathode - Eanode
cell
Ecell = ECu /Cu – EH /H 2+ + 2
0.34 = ECu /Cu - 0 2+
ECu /Cu = 0.34 V 2+
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
Anode (oxidation):
H2 (1 atm) H+ (1 M) + 2e-
Cathode (reduction):
2e- + Cu2+ (1 M) Cu (s)
H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)

17. E0 is for the reaction as written
The more positive E0 the greater the tendency for the substance to be reduced
The half-cell reactions are reversible
The sign of E0 changes when the reaction is reversed
Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

18. Cd is the stronger oxidizer
Ex.
What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e Cd (s) E0 = V
Cd is the stronger oxidizer
Cd will oxidize Cr
Cr3+ (aq) + 3e Cr (s) E0 = V
Anode (oxidation):
Cr (s) Cr3+ (1 M) + 3e-
x 2
Cathode (reduction):
2e- + Cd2+ (1 M) Cd (s)
x 3
2Cr (s) + 3Cd2+ (1 M) Cd (s) + 2Cr3+ (1 M)
E0 = Ecathode - Eanode
cell
E0 = – (-0.74)
cell
E0 = 0.34 V
cell

19. Ex. Calculate Eocell for the following reactions:
(a) 2 Al (s) + 6 H+ -> 2 Al H2 (g)
(b) H2O2 + 2 H+ + Cu (s) -> Cu H2O
E0 = Ecathode - Eanode
cell
(a) 2 Al (s) + 6 H+ -> 2 Al H2 (g)
Anode (oxidation):
2 Al(s) Al3+ + 6e-
Cathode (reduction):
6 H+(aq) + 6e H2(g)
E0 = EH /H - EAl /Al
cell + 3+ 2
= 1.66 V
(b) H2O2 + 2 H+ + Cu (s) -> Cu H2O
Anode (oxidation):
Cu(s) Cu2+ + 2e-
Cathode (reduction):
H2O2+2 H+ +2e H2O
E0 = E ECu /Cu
cell 2+
H2O2/H2 O
= 1.77 V V = 1.43 V

20. IV. Cell Potential Free energy : spontaneity of redox reactions
DG = -nFEcell
n = number of moles of electrons in reaction
F = 96,500 J
V • mol
DG0 = -nFEcell
= 96,500 C/mol
DG0 = -RT ln K
= -nFEcell
Ecell = RT nF
ln K
(8.314 J/K•mol)(298 K)
n (96,500 J/V•mol)
ln K = = V n
ln K
Ecell = V n
log K
Ecell

21. A. Free energy and spontaneity of redox reaction= V n
ln K
Ecell
DG = -nFEcell = V n
log K
Ecell
DG0 = -nFEcell

22. Ex.
What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq) = V n
ln K
Ecell
Oxidation:
2Ag Ag+ + 2e-
n = 2
Reduction:
2e- + Fe Fe
E0 = EFe /Fe – EAg /Ag 2+ +
E0 = – (0.80)
E0 = V V
x n E0
cell
exp
K = V
x 2
-1.24 V
= exp
K = 1.23 x 10-42

23. B. Concentration effect on cell potential
DG = DG0 + RT ln Q
DG = -nFE
DG0 = -nFE
-nFE = -nFE0 + RT ln Q
Nernst equation
E = E0 -
ln Q RT nF
At 298 K - V n
ln Q E
E = - V n
log Q E
E =

24. Ex. Will the following reaction occur spontaneously at 250C if
[Fe2+] = 0.60 M and [Cd2+] = M?
Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
Ex.
Oxidation:
Cd Cd2+ + 2e-
n = 2
Reduction:
2e- + Fe Fe
= – (-0.40) V
= V
E0 = EFe /Fe – ECd /Cd 2+
[Cd2+]
[Fe2+]
Q = - V n
ln Q E
E = - V 2 ln
-0.04 V
E =
0.010
0.60
E = 0.013
E > 0
Spontaneous

25. Ex.
A redox reaction for the oxidation of nickel by silver ion is set up with an initial concentration of 5.0 M silver ion and M nickel ion. What is the cell EMF at 298K?
Oxidation:
Ni Ni e-
n = 2
Reduction:
2e- +2 Ag Ag
Ni + 2 Ag Ni Ag
E0 = EAg /Ag – E Ni /Ni + 2+
= 0.80 – (-0.25) V
= 1.05V - V n
ln Q E
E =
[Ni2+]
[Ag+]2
Q = - V 2 ln
1.05 V
E =
0.050
5.02
E = 1.13

26. V. Battery Dry cell Leclanché cell About 1.5 V Anode:
Zn (s) Zn2+ (aq) + 2e-
Cathode:
2NH4 (aq) + 2MnO2 (s) + 2e Mn2O3 (s) + 2NH3 (aq) + H2O (l) +
Zn (s) + 2NH4+ (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

27. V. Battery Mercury Battery 1.35 V Anode:
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-
Cathode:
HgO (s) + H2O (l) + 2e Hg (l) + 2OH- (aq)
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)

28. V. Battery Lead storage battery Total 12V Anode:
Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4
Cathode:
PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e PbSO4 (s) + 2H2O (l) 4
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) PbSO4 (s) + 2H2O (l) 4

29. VI. Corrosion
Deterioration of metal by an electrochemical process

30. Cathodic Protection of an Iron Storage Tank
VI. Corrosion
Cathodic Protection of an Iron Storage Tank

31. VII. Electrolysis
Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

32. A. Electrolysis of water

33. B. Quantitative aspect of electrolysis
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
Ex. It will take the greates amount electricity to produce 2 mol of the ____ metal by electrolysis.
(1) potassium (2) calcium (3) aluminum (4) silver

34. Ex.
How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of A is passed through the cell for 1.5 hours?
Anode:
2Cl- (l) Cl2 (g) + 2e-
Cathode:
2 mole e- = 1 mole Ca
Ca2+ (l) + 2e Ca (s)
Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)
mol Ca = 0.452 C s
x 1.5 hr x 3600 s hr
96,500 C
1 mol e- x
2 mol e-
1 mol Ca x
= mol Ca
= 0.50 g Ca