Q1(e). Aluminium–magnesium galvanic cell

Answer in brief

For 3Mg + 2Al³⁺ → 3Mg²⁺ + 2Al, E° = 0.71 V, ΔᵣG° = −411.1 kJ/mol of reaction and K ≃ 1.06 × 10⁷² at 298.15 K.

Key points

• Magnesium, with the more negative reduction potential, oxidises: Mg → Mg²⁺ + 2e⁻. Aluminium ions are reduced: Al³⁺ + 3e⁻ → Al.
• Multiply the half-reactions by three and two, respectively. Six electrons are transferred in the balanced reaction, so n = 6.
• E° = E°cathode − E°anode = −1.66 − (−2.37) = 0.71 V.
• ΔᵣG° = −nFE° = −6(96500)(0.71) = −411090 J/mol = −411.09 kJ/mol.
• From ΔᵣG° = −RT ln K, ln K = 411090/[8.314(298.15)] = 165.84, so K ≃ 1.06 × 10⁷².