Calculate the standard Gibb's energy and the equilibrium constant of the cell for the following reaction?

2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ …… E°(cell) = 0.236V
2e⁻ are involved in the above equation.

E°(cell) = [R T/(n F)] ln(K) …… [1]
ln(K) = n F E°(cell) / (R T)
K = e^[n F E°(cell) / (R T)]
Equilibrium constant, K = e^[2 × 96500 × 0.236 / (8.314 × 298)] = 9.64 × 10⁷

ΔG° = R T ln(K) …… [2]

[2]/[1] :
ΔG° / E°(cell) = n F
ΔG° = n F E°(cell)
Gibb's free energy, ΔG° = 2 × 96500 × 0.236 = 45500 J/mol = 45.5 kJ/mol

The Gibbs free energy equals -nFE where n = number of electrons, F + Faraday constant; E is the electrode potential; if you use the standard electrode potential you have the standard Gibbs free energy