Iron(III) ions oxidise iodide ions: 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s)…

Step 1. The emf. Iron(III) is reduced and iodide is oxidised. $E^\circ = 0.77\ \text{V} - 0.54\ \text{V} = 0.23\ \text{V}$.

Step 2. The electrons. Each $\text{Fe}^{3+}$ takes one electron, and two of them take two. So $n = 2$.

Step 3. The Gibbs energy. $\Delta G^\circ = -2 \times 96.5\ \text{kJ/(V mol)} \times 0.23\ \text{V} = -44.4\ \text{kJ/mol}$.

Step 4. The equilibrium constant. $\log K = \dfrac{2 \times 0.23\ \text{V}}{0.0591\ \text{V}} = 7.78$, so $K \approx 6 \times 10^7$.