The cell in which the following reaction occurs:
$2Fe^{3+}_{(aq)} + 2I^{-}_{(aq)} \to 2Fe^{2+}_{(aq)} + I_{2(s)}$
has $E^{\Theta}_{cell} = 0.236 \ V$ at $298 \ K$. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Given: $n = 2$,$E^{\Theta}_{cell} = 0.236 \ V$,$T = 298 \ K$,$F = 96487 \ C \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$1$. Standard Gibbs energy $(\Delta_r G^{\Theta})$:
$\Delta_r G^{\Theta} = -nFE^{\Theta}_{cell}$
$\Delta_r G^{\Theta} = -2 \times 96487 \times 0.236 = -45541.864 \ J \ mol^{-1} \approx -45.54 \ kJ \ mol^{-1}$.
$2$. Equilibrium constant $(K_c)$:
$\Delta_r G^{\Theta} = -RT \ln K_c = -2.303 \ RT \log K_c$
$\log K_c = -\frac{\Delta_r G^{\Theta}}{2.303 \ RT}$
$\log K_c = -\frac{-45541.864}{2.303 \times 8.314 \times 298} \approx 7.981$
$K_c = \text{antilog}(7.981) \approx 9.57 \times 10^{7}$.