1. Chapter 20 Electrochemistry
Lecture Presentation
Chapter 20 Electrochemistry
James F. Kirby
Quinnipiac University
Hamden, CT

2. 20.1 Oxidation States and Redox Reactions
Electrochemistry and Redox Reactions
Oxidation–reduction (redox) reaction – involves a transfer of electrons from one species to another species. +2 −2
2Mg(s) + O2(g) MgO(s)
Oxidation half-reactions
Reduction half-reactions −2
2Mg Mg2+
O2 + 4e− O2−
+ 4e−
+ 4e−
Oxidation – loss of electrons
Reduction – gain of electrons
Mg is the reducing agent
O2 is the oxidizing agent
Reducing agent – electron donor
Oxidizing agent – electron acceptor
Reducing agent – itself is oxidized
Oxidizing agent – itself is reduced

3. 20.1 Oxidation States and Redox Reactions
Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation number of zero.
Mg, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to the charge on the ion.
Li+ = +1; Fe3+ = +3; O2− = −2
3. The oxidation number of oxygen is usually –2. In H2O2 and O22−, it is –1.

4. 20.1 Oxidation States and Redox Reactions
4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.
5. Fluorine is always –1.
6. Group IA metals are +1. Group IIA metals are +2.
7. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.
HPO42−
Oxidation numbers of all the atoms in HPO42− ?
O = −2
H = +1
4(−2) ? = −2
P = +5

5. 20.2 Balancing Redox Equations
EXERCISE #1
Balance the following redox reaction that occurs in acidic and basic solution, respectively.
Cr2O72−(aq) + SO32−(aq) Cr3+(aq) + SO42−(aq)
Step 1: Write redox half-reactions.
Step 2: Balance atoms other than oxygens and hydrogens.
Step 3a: Balance electrons (e−) in each half-reaction.
Step 3b: Balance oxygens by adding H2O in each half-reaction.
Step 3c: Balance hydrogens by adding H+ in each half-reaction.
Step 4: Add up the half-reactions. Balance the electrons between oxidation and reduction half-reactions.
Step 5: Basic solutions: Add OH- to both sides of the equation.

6. Color code: Step 1, Step 2, Step 3a, Step 3b, Step 3c, Step 4, Step 5
EXERCISE #1 +6 +4 +3 +6
Cr2O72−(aq) + SO32−(aq) Cr3+(aq) + SO42−(aq)
Step 3b: Balance oxygens by adding H2O in each half-reaction.
Step 5: Basic solutions: Add OH− to both sides of the equation.
Step 3a: Balance electrons (e−) in each half-reaction.
Step 3c: Balance hydrogens by adding H+ in each half-reaction.
Step 4: Add up the half-reactions.
Step 1: Write redox half-reactions.
Step 2: Balance atoms other than oxygens and hydrogens.
Ox: ( )
SO32−
SO42−
H2O +
2e− +
2H+ +
x 3
Red:
Cr2O72−
Cr3+
6e− +
14H+ + 2
7H2O +
Acidic soln:
8OH− +
3SO32−
+ Cr2O72−
+ 8H+
3SO42−
+ 2Cr3+
+ 4H2O
+ 8OH−
+ 8H2O 4
Basic soln:
3SO32−
+ Cr2O72−
+ 4H2O
3SO42−
+ 2Cr3+
+ 8OH−
Color code: Step 1, Step 2, Step 3a, Step 3b, Step 3c, Step 4, Step 5

7. 20.3 Voltaic/Galvanic Cells
Galvanic cell: a device in which chemical energy is converted to electrical energy through spontaneous redox reaction.
anode
oxidation
cathode
reduction
Oxidation half-reaction
Reduction half-reaction
spontaneous
redox reaction

8. 20.3 Voltaic/Galvanic Cells
Cell Diagram
oxidation
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
reduction
[Cu2+] = 1 M and [Zn2+] = 1 M
Zn (s) |
Zn2+ (1 M) ||
Cu2+ (1 M) |
Cu (s)
anode
cathode
Cell Potentials (Ecell)
The difference in electrical potential between the anode and cathode is called cell voltage or electromotive force (emf) or cell potential (Ecell).
The unit of electrical potential is the volt (V)
1V = 1 J/C

9. 20.4 Cell Potentials under Standard Conditions
Standard Reduction Potential (E0)
Standard conditions:
Gas 1 atm
Solution 1 M
Temperature 25oC
The voltage associated with a reduction reaction at an electrode under standard-state conditions.
Standard hydrogen electrode (SHE)
2H+ (1 M) + 2e− H2 (1 atm)
E = 0 V
H+/H2
Standard reduction potential of a substance can be determined using SHE as reference electrode.

10. 20.4 Cell Potentials under Standard Conditions
Standard Emf (E0 )
cell
E cell 0 = E cathode 0 − E anode 0
Cell voltage under standard-state conditions.
E cell 0 =
E H + / H 2 0 −
E Zn 2+ /Zn 0
SHE
0.76 V =
0 V −
E Zn 2+ /Zn 0
E Zn 2+ /Zn 0
= −0.76 V
Zn2+ (1 M) + 2e− Zn (s)
E0 = −0.76 V
Zn (s) | Zn2+ (1 M) ||
H+ (1 M) | H2 (1 atm) | Pt (s)
anode
cathode
Anode (oxidation):
Zn (s) Zn2+ (1 M) + 2e−
Cathode (reduction):
2H+ (1 M) + 2e− H2 (1 atm)
Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

11. 20.4 Cell Potentials under Standard Conditions
Standard Emf (E0 )
cell
E cell 0 =
E cathode 0 −
E anode 0
E0 = 0.34 V
cell
E cell 0 =
E Cu 2+ /Cu 0 −
E H + / H 2 0
SHE
0.34 V =
E Cu 2+ /Cu 0 −
0 V
E Cu 2+ /Cu 0 =
0.34 V
Cu2+ (1 M) + 2e− Cu (s)
E0 =0.34 V
Pt (s) | H2 (1 atm) | H+ (1 M) ||
Cu2+ (1 M) | Cu (s)
anode
cathode
Anode (oxidation):
H2 (1 atm) H+ (1 M) + 2e−
Cathode (reduction):
Cu2+ (1 M) + 2e− Cu (s)
H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)

12. 20.4 Cell Potentials under Standard Conditions
E0 is for the reaction as written. The sign of E0 changes when the reaction is reversed.

13. 20.4 Cell Potentials under Standard Conditions
The half-cell reactions are reversible. Thus, an electrode can act as either anode or cathode depending on the conditions.
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)

14. 20.4 Cell Potentials under Standard Conditions
Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0.
2e− + Cu2+ (1 M) Cu (s)
E0 = 0.34 V
4e− + 2Cu2+ (1 M) Cu (s)
E0 = 0.34 V

15. 20.4 Cell Potentials under Standard Conditions
The more positive E0 the greater the tendency for the substance to be reduced (stronger oxidizing agent).
F2(1 atm) + 2e− F−(1 M)
E0 = V
strongest oxidizing agent
Li+(1 M) + e− Li(s)
E0 = −3.05 V
strongest reducing agent

16. 20.4 Cell Potentials under Standard Conditions
Diagonal rule: Under standard conditions, any species on the left of a half-reaction reacts spontaneously with any species on the right of another half-reaction below it.
Cu2+ (1 M) + 2e− Cu (s)
E0 = V
Zn2+ (1 M) + 2e− Zn (s)
E0 = −0.76 V
Zn (s) +
Cu2+ (aq)
Cu (s) +
Zn2+ (aq)
red. agent
ox. agent

17. Fe3+(aq) + Cu(s) Cu2+(aq) + Fe2+(aq)
EXERCISE #2
Calculate the standard cell potential of the following galvanic cell under standard-state conditions at 25oC. Identify also the reducing and oxidizing agents.
E Cu 2+ /Cu 0 = 0.34 V
Fe3+(aq) + Cu(s) Cu2+(aq) + Fe2+(aq)
E Fe 3+ / Fe = 0.77 V
E cell 0 = E cathode 0 − E anode 0
Anode (ox):
Cu(s) Cu2+ + 2e–
Cathode (red):
( ) x 2
Fe3+ + e– Fe2+
Cu(s) + 2Fe3+(aq) Cu2+(aq) + 2Fe2+(aq) =
E cathode 0 −
E anode 0
E cell 0
Oxidizing agent:
Reducing agent:
Fe3+
E cell 0 =
E Fe 3+ / Fe 2+ 0 −
E Cu 2+ /Cu 0
Cu(s)
E cell 0 =
0.77 V –
(0.34 V)
= 0.43 V

18. 20.5 Free Energy and Redox Reactions
Spontaneity of Reversible Redox Reactions
DG = −nFEcell
n = # of moles of electron in the reaction
F = C/mol e− = J/V∙mol e−
Cell Emf and Spontaneity of Redox Reactions DG
Spontaneity
Ecell
< 0
> 0
Spontaneous in the forward direction.
> 0
< 0
The reaction is nonspontaneous as written.
The reaction is at equilibrium.

19. 20.5 Free Energy and Redox Reactions
Redox Reaction and Equilibrium Constant
At 25oC:
E cell 0 = V n ln 𝐾
E cell 0 = V n log 𝐾
DG0 = − nFE cell 0
DG0 = −RT ln K
DG0 = −RT ln K = − nFE cell 0
E cell 0 = RT nF ln 𝐾
Relationship among ∆G0, K, and E0cell
∆Go K
E0cell
Reaction under standard-state conditions
< 0
> 1 +
Products favored over reactants. 1
Products and reactants equally favored.
> 0
< 1 –
Reactants favored over products.

20. EXERCISE #3
Calculate the standard cell potential, standard Gibbs energy, and equilibrium constant for the following redox reaction at 25oC:
Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)
E Sn 2+ /Sn 0 = −0.14 V
E cell 0 = E cathode 0 − E anode 0
E Cu 2+ / Cu = 0.15 V
Anode (ox):
Sn(s) Sn2+ + 2e−
n = 2 mol e−
Cathode (red):
2Cu2+ + 2e− Cu+
Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)
E cell 0
E cathode 0 −
E anode 0 =
ΔG0= −nFE0cell =
E Cu 2+ / Cu + 0 −
E Sn 2+ /Sn 0 = −
(2)
(96500 J/V·mol)
(0.29 V)
= 0.29 V
= −56.0 kJ =
0.15 V –
(−0.14 V)
= −55970 J

21. Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)
EXERCISE #3
Continued
Calculate the standard cell potential, standard Gibbs energy, and equilibrium constant for the following redox reaction at 25oC.
Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)
E cell 0 = 0.29 V
n = 2 mol e−
E cell 0 = V n ln K
(n) (E 0 ) V
(2)(0.29 V) V
ln K = =
= 22.6
K =
e22.6
= 7 x 109
E cell 0 > 0, ∆G0 < 0, and K > 1
Products favored over reactants at equilibrium.

22. 20.6 Cell Potentials under Nonstandard Conditions
The Effect of Concentration on Cell Emf
When the conditions of the cell are not standard-state conditions:
At 25oC:
Ecell = E cell 0 – V n ln Q
Ecell = E cell 0 – V n log Q
Ecell = E cell 0 – RT nF ln Q
Nernst equation
Cell Emf and Spontaneity of Redox Reactions
Ecell > 0
Spontaneous in the forward direction.
Ecell < 0
The reaction is nonspontaneous as written.
Ecell = 0
The reaction is at equilibrium.

23. Fe2+(0.60 M) + Cd(s) Fe(s) + Cd2+(0.010 M)
EXERCISE #4
Will the following redox reaction occur spontaneously at 25oC?
Fe2+(0.60 M) + Cd(s) Fe(s) + Cd2+(0.010 M)
Is Ecell greater than zero?
E Fe 2+ /Fe 0 = −0.44 V
E Cd 2+ /Cd 0 = −0.40 V
Anode (ox):
Cd (s) Cd2+ + 2e−
n = 2 mol e−
Cathode (red):
Fe2+ + 2e− Fe (s)
Cd(s) + Fe2+(aq) Cd2+(aq) + Fe(s)
ln [Cd 2+ ]0 [Fe 2+ ]0
Ecell = E cell 0 – V n ln Q ln
ln Q = =
E cell 0
E cathode 0 −
E anode 0 = =
E Fe 2+ /Fe 0 −
E Cd 2+ /Cd 0 =
−0.44 V –
(−0.40 V)
= −0.04 V
Ecell = E cell 0 – V n ln
[Cd 2+ ]0 [Fe 2+ ]0
V 2 ln
> 0 =
−0.04 V –
= V
Spontaneous

24. 20.9 Electrolysis
Electrolysis: the process in which electrical energy is used to cause a nonspontaneous (Ecell < 0) chemical reaction to occur.
Electrolytic cell
Electroplating

25. 20.9 Electrolysis Stoichiometry of Electrolysis
How much chemical change occurs with the flow of a given current for a specified time?
Faraday constant
metal C
(current)(time) =
Coulombs of charge =
C/s s
Faraday constant
F = C/mol e−
mol metal mol e −
mol e− = x =
mol metal x
(molar mass metal) =
mass metal
(current)(time) Faraday constant
mol metal mol e −
mass metal =
(molar mass metal)

26. Ca2+(l) + 2Cl−(l) Ca(s) + Cl2(g)
EXERCISE #5
How many grams calcium will be produced in an electrolytic cell of molten CaCl2 if a current of A is passed through the cell for 1.5 hours? +2 −1
CaCl Ca(s) + Cl2(g)
Anode:
2Cl−(l) Cl2(g) + 2e−
Cathode:
Ca2+(l) + 2e− Ca(s)
2 mole e− = 1 mole Ca
Ca2+(l) + 2Cl−(l) Ca(s) + Cl2(g)
F = C/mol e−
molar mass Ca =
g/mol
mass Ca =
(current)(time) Faraday constant
mol Ca mol e −
(molar mass Ca)
(0.452 C/s)(1.5 hrs)(60 min)(60 s)
1 mol Ca 2 mol e − =
mass Ca
( g/mol)
= 0.51 g
96500 C/mol e−

27. End of Chapter 20

28. 20.9 Electrolysis
Electrolysis: the process in which electrical energy is used to cause a nonspontaneous (Ecell < 0) chemical reaction to occur.
2Cl− + 2Na Cl2(g) + 2Na(l)
2H2O(l) O2(g) + 2H2(g)