1. Electrochemistry The study of the interchange
of chemical and electrical energy.
Sample electrochemical processes:
1) Corrosion
4 Fe(s) + 3 O2(g) ⇌ 2 Fe2O3(s)
2) Biological processes
C6H12O6 + 6 O2 ⇌ 6 CO2 + 6 H2O
3) Batteries (Galvanic or Voltaic cells)
Electrochemical cells that produce a current (flow of electrons) as a result of a redox reaction
4) Electrolytic cells
Electrical energy is used to produce chemical change
Used to prepare or purify metals (such as sodium, aluminum, copper)

2. Chemical Change  Electron Flow
Copper: Cu(s), Cu2+(aq)
Cu(s)  Cu2+(aq) + 2e-
ΔG°rxn = ΔG°f(Cu2+) = 65.6 kJ
Silver: Ag(s), Ag+(aq)
Ag(s)  Ag+(aq) + e-
ΔG°rxn = ΔG°f(Ag+) = 77.2 kJ
Ag(s)
Cu2+ in solution 2(
Ag+(aq) + e-  Ag(s) ΔG° = kJ )
2( )
Cu(s)  Cu2+(aq) + 2e- ΔG° = kJ
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) ΔG° = kJ
Spontaneous
wmax = kJ

3. Harnessing the Energy − + Separate the half-reactions Red Cat Anode
Creates a galvanic or voltaic cell
Luigi Galvani
Alessandro
Volta e-
Red Cat
salt bridge −
(produces electrons) +
(attracts electrons) Cu
NO3- K+ Ag
KNO3(aq)
Anode
Cathode
“anode” and “oxidation” begin with vowels
“cathode” and “reduction” begin with consonants
Cu2+
SO42-
Ag+
NO3-
1 M CuSO4
Cu(s)  Cu2+(aq) + 2e-
1 M AgNO3
Ag+(aq) + e-  Ag(s)
Oxidation
Reduction

4. Line Notation for Galvanic CellsCu
NO3- K+ Ag
Anode
(−)
Cathode
(+)
Cu2+
SO42-
Ag+
NO3-
1 M CuSO4
Cu(s)  Cu2+(aq) + 2e-
1 M AgNO3
Ag+(aq) + e-  Ag(s)
Oxidation
Reduction
Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s)
Anode always on the left
Cathode always on the right

5. Chemical Change  Electrical Work
Chemical change produces electrical energy
Electrical energy can be used to do work!
ΔG = wmax
Electrical work: w = -nFℰ
n = # of moles e- transferred
F = charge on a mole of e-
ℰ = electrical potential
(electromotive force)
Cell Potential (ℰ) or Electromotive Force (emf): The driving force pushing the electrons from the anode to the cathode.
Units = Volts 1 Volt = 1 joule/coulomb

6. Standard Reduction Potentials
The cell potential ℰ°cell can be determined from the standard reduction potentials (ℰ°red) for the half-reactions:
Reduction potential = tendency for reduction to happen
Positive ℰ°red  spontaneous reduction reaction
Negative ℰ°red  non-spontaneous reduction or spontaneous oxidation (reverse reaction)
Standard (o) = standard conditions (1 M solutions, 1 atm gases)

7. Standard Reduction Potentials
Half-Reaction ℰ° (V)
F2 + 2e-  2F
Au e-  Au 1.50
Ag+ + e-  Ag 0.80
Cu2+ + 2e-  Cu 0.34
2H+ + 2e-  H
Ni2+ + 2e-  Ni -0.23
Zn2+ + 2e-  Zn -0.76
Al3+ + 3e-  Al -1.66
Li+ + e-  Li -3.05
ℰ° > 0
Spontaneous reduction
 ℰ° = 0 (SHE)
Standard Hydrogen Electrode
ℰ° < 0
Non-Spontaneous reduction
Spontaneous oxidation
(reverse rxn)
Reduction potential = tendency for reduction to happen
Standard = standard conditions (1 M solutions, 1 atm gases)

8. Standard Reduction Potentials
Half-Reaction ℰ° (V)
F2 + 2e-  2F
Au e-  Au 1.50
Ag+ + e-  Ag 0.80
Cu2+ + 2e-  Cu 0.34
2H+ + 2e-  H
Ni  Ni2+ + 2e
Zn  Zn2+ + 2e
Al  Al3+ + 3e
Li  Li+ + e
ℰ° > 0
Spontaneous reduction
Spontaneous oxidation
But remember, an oxidation CANNOT happen without a reduction

9. Standard Reduction Potentials
Half-Reaction ℰ° (V)
F2 + 2e-  2F
Au e-  Au 1.50
Ag+ + e-  Ag 0.80
Cu2+ + 2e-  Cu 0.34
2H+ + 2e-  H
Ni2+ + 2e-  Ni -0.23
Zn2+ + 2e-  Zn -0.76
Al3+ + 3e-  Al -1.66
Li+ + e-  Li -3.05
Strongest Oxidizing Agent
(most easily reduced)
Strongest Reducing Agent
(most easily oxidized)
Reduction potential = tendency for reduction to happen
Standard = standard conditions (1 M solutions, 1 atm gases)

10. Cell Potential ℰ°cell = ℰ°reduction + ℰ°oxidation
Ag+(aq) + e-  Ag(s) ℰ° = 0.80 V
Cu2+(aq) + 2 e-  Cu(s) ℰ° = 0.34 V
Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s)
Reduction reaction: 2(Ag+(aq) + e-  Ag(s)) ℰ° = V
Oxidation reaction: Cu(s)  Cu2+(aq) + 2 e- ℰ° = V
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) ℰ°cell = V
The ℰ°cell MUST be + and thus spontaneous for Galvanic cells
ℰ° is intensive, unlike ΔGo

11. Free Energy and Cell Potential
G = wmax = nFℰ
n = number of moles of electrons transferred
F = Faraday’s constant
= 96,485 coulombs per mole of electrons (C/mol e-)
ℰ° = standard cell potential (V or J/C)
Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s)
ℰ°cell = V
ΔG° = -nFℰ°cell
ΔG° = -(2 mol e-)(96485 C/mol e-)(0.46 V)
ΔG° = -88,800 J or kJ
Michael Faraday

12. Practice Time Given the following information, draw a galvanic cell.
Fe(s)Fe2+(1 M)Au3+(1 M)Au(s)
Be sure to include the following:
Anode/Cathode reactions
Balanced overall reaction
Complete circuit (external wire with e- flow direction, salt bridge)
Label all parts of the cell (solution, electrode, etc.)

13. Fe(s)Fe2+(1 M)Au3+(1 M)Au(s)
anions
cations Au
Anode
Cathode
Fe2+
Au3+
1 M Fe2+
Fe(s)  Fe2+(aq) + 2e-
1 M Au3+
Au3+(aq) + 3e-  Au(s)
Oxidation
Reduction
3Fe(s) + 2Au3+(aq) → 3Fe2+(aq) + 2Au(s)
ℰ°cell = V (Fe rxn) V (Au rxn) = 1.94 V

14. Reaction Quotient
The reaction quotient (Q) sets up a ratio of products and reactants
For a reaction, A + 2B → 3C + 4D
[C]3[D]4
[A]1[B]2
Only concentrations (aq) or pressures (g) are used to solve for Q
Solids (s) and liquids (l) are not included in the expression
Q =

15. Reaction Quotient practice
Write the Q expression for the following reaction
CH4(g) + O2(g) → CO2(g) + H2O(g)
Reaction must be balanced first
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
(CO2)(H2O)2
(CH4)(O2)2
Q =

16. Reaction Quotient practice
Write the Q expression for the following reaction
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
(Cu2+)(Ag)2
(Cu)(Ag+)2
Is this correct?
NO: Solids aren’t included in the equation!
(Cu2+)
(Ag+)2
Q =
Q =

17. Non-standard conditions: The Nernst Equation
We can calculate the potential of a cell in which some or all of the components are not in their standard states (not 1 M concentration or 1 atm pressure).
ΔG = ΔG° + RT lnQ
ΔG = -nFℰ
ΔG° = -nFℰ°
-nFℰ = -nFℰ° + RT lnQ
R = J/mol K
T = temperature
n = moles of e-
F = Faraday’s constant
96,485 C/mol e-
ℰ = ℰ° -
Walther Nernst

18. Practice with the Nernst Equation
What will be the cell potential ℰ of a Cu/Ag cell using 0.10 M Cu2+ and 1.0 M Ag+ solutions at 25°C?
Cu(s)Cu2+ (0.10 M)Ag+ (1.0 M)Ag(s)
Cu(s)  Cu2+(aq) + 2e-
Ag+(aq) + e-  Ag(s)
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
ℰ = ℰ° - ℰ Cu Ag
Cu2+
SO42-
Ag+
NO3-
ℰ = 0.46 V – (-0.03 V)
ℰ = 0.49 V

19. Brain Warmup What is ℰ° for each of the following reactions?
Half-Reaction ℰ° (V)
Ag+ + e-  Ag 0.80
Cu2+ + 2e-  Cu 0.34
Zn2+ + 2e-  Zn -0.76
Al3+ + 3e-  Al -1.66
What is ℰ° for each of the following reactions?
Which reaction(s) are spontaneous? ℰ°
2.46 V
1.10 V
-0.90 V
Spontaneous? Y N
3 Ag+(aq) + Al(s)  3 Ag(s) + Al3+(aq)
Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)
2 Al3+(aq) + 3 Zn(s)  2 Al(s) + 3 Zn2+(aq)
Zn can reduce Cu2+, but not Al3+