Calculate the equilibrium constant at $298 \ K$ for the following reaction and also calculate the maximum work that can be obtained from this cell:
$Mg(s) \ | \ Mg^{2+}(aq) \ || \ Ag^{+}(aq) \ | \ Ag(s)$
Given: $E_{Mg^{2+} \mid Mg}^{o} = -2.37 \ V$ and $E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V$

$1$. Calculation of $E_{\text{cell}}^{o}$:
$E_{\text{cell}}^{o} = E_{\text{cathode}}^{o} - E_{\text{anode}}^{o} = E_{Ag^{+} \mid Ag}^{o} - E_{Mg^{2+} \mid Mg}^{o}$
$E_{\text{cell}}^{o} = 0.80 \ V - (-2.37 \ V) = 3.17 \ V$
$2$. Calculation of equilibrium constant $K_{C}$:
Using the formula $E_{\text{cell}}^{o} = \frac{0.0591}{n} \log K_{C}$,where $n = 2$:
$3.17 = \frac{0.0591}{2} \log K_{C}$
$\log K_{C} = \frac{3.17 \times 2}{0.0591} \approx 107.2758$
$K_{C} = \text{antilog}(107.2758) \approx 1.887 \times 10^{107}$
$3$. Calculation of maximum work $(W_{\text{max}})$:
Maximum work is equal to the decrease in Gibbs free energy,$W_{\text{max}} = -\Delta G^{o} = nFE_{\text{cell}}^{o}$
$W_{\text{max}} = 2 \times 96487 \ C \ mol^{-1} \times 3.17 \ V$
$W_{\text{max}} = 611727.6 \ J \ mol^{-1} = 611.73 \ kJ \ mol^{-1}$