The cell in which the following reaction occurs: 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s) has E°cell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given: 1 F = 96,500 C mol⁻¹)
Solution:
Standard Gibbs free energy of the cell is given by the formula: ΔG° = -nFE° = -2 × 96500 × 0.236 = -45540 J = -45.54 KJ