Electrochemical Cells

Reference Sheet Practice MCQs

Cell Voltage

+1.100 V

spontaneous as drawn

E°cell+1.100 V

n2 e⁻

Q1.000

ΔG-212 kJ/mol

E = E° − (RT / nF) ln Q= 1.100 − (1.28e-2) ln 1.000

Half-Reactions

Anode · Oxidationleft · Zn

Zn(s) → Zn²⁺ + 2e⁻

E°red = −0.760 V

Cathode · Reductionright · Cu

Cu²⁺ + 2e⁻ → Cu(s)

E°red = +0.340 V

Net cell reactionZn(s) + Cu²⁺ → Zn²⁺ + Cu(s)Q = [Zn²⁺] / [Cu²⁺]

Standard Reduction Potentials

Half-reaction	E° (V)
F₂ + 2e⁻ → 2F⁻	+2.87
O₂ + 4H⁺ + 4e⁻ → 2H₂O	+1.23
Ag⁺ + e⁻ → Ag	+0.80
CuCu²⁺ + 2e⁻ → Cu	+0.34
2H⁺ + 2e⁻ → H₂	+0.00
Fe²⁺ + 2e⁻ → Fe	−0.44
ZnZn²⁺ + 2e⁻ → Zn	−0.76
Al³⁺ + 3e⁻ → Al	−1.66
Na⁺ + e⁻ → Na	−2.71
Li⁺ + e⁻ → Li	−3.04

The stronger oxidising agent sits higher. E°cell = E°cathode − E°anode, and both values come from this table unchanged, however many electrons the balanced equation needs.

Key Concepts

Galvanic cell

ΔG < 0E > 0chem → elec

The reaction runs on its own, converting chemical energy to electrical work. Zn is oxidised at the anode and Cu is reduced at the cathode.

Anode vs cathode

Anode

Oxidation, loses e⁻. Negative (−) terminal, on the left here.

Cathode

Reduction, gains e⁻. Positive (+) terminal, on the right here.

Oxidation is always the anode. The sign changes with the cell type, the role does not.

Nernst equation

E = E° − (RT / nF) ln Q

Building up products raises Q and pulls E down. At equilibrium E = 0 and Q = K, which is a dead battery. Here log₁₀K = 37.2 at 298 K.

Free energy

ΔG = −nFE

With n = 2 and E = 1.100 V, ΔG = -212 kJ/mol. A positive cell potential is a negative ΔG, so the sign of E alone tells you whether the reaction is thermodynamically favoured.

Salt bridge

K⁺ migrates toward the cathode and NO₃⁻ toward the anode. Without that ion path, charge piles up within seconds and current stops.