Electrochemistry
Electrochemistry questions combine cell EMF calculations, conductance relations, and applied cells such as lead storage and fuel cells.
Core relations
- E°cell = E°cathode − E°anode, both written as reduction potentials
- ΔrG° = −nFE°cell and log K = nE°cell/0.0591 at 298 K
- Nernst equation: E = E° − (0.0591/n)·log Q, where Q uses products/reactants of the cell reaction
Conductance
- κ = 1/ρ; molar conductivity Λm = 1000κ/c (S cm² mol⁻¹)
- strong electrolytes: Λm = Λ°m − A√c
- weak electrolytes: α = Λm/Λ°m and Ka = cα²/(1 − α); Λ°m comes from Kohlrausch's law of independent migration
Practical cells and corrosion
- Dry cell: zinc anode, MnO2/carbon cathode, paste of NH4Cl and ZnCl2; not rechargeable
- Lead storage: Pb anode and PbO2 cathode in ~38% H2SO4; both plates become PbSO4 on discharge and are regenerated on charging
- H2–O2 fuel cell: high efficiency, product is only water
- Rusting needs both oxygen and water; iron acts as the anode and is oxidised
Worked example
For Zn|Zn²⁺(0.10 M)||Cu²⁺(0.010 M)|Cu with E° = 1.10 V: n = 2, Q = [Zn²⁺]/[Cu²⁺] = 10. E = 1.10 − (0.0591/2)log 10 = 1.10 − 0.0296 = 1.07 V. Diluting the cathode relative to the anode lowers the EMF.
Traps that cost marks
- Reversing the anode and cathode in the Nernst expression.
- Multiplying a standard potential by n; only ΔG is multiplied.
- Extrapolating Λm of a weak electrolyte to get Λ°m.
- A scratched tin coating on iron accelerates corrosion because iron becomes the anode.