Determine the values of equilibrium constant $\left({\mathbf{K}}_{\mathrm{c}}\right)$ and ${\mathrm{\Delta G}}^{\circ }$ or the following reaction :
$\begin{array}{l}\mathrm{Ni}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{Ag}\left(\mathrm{s}\right)\\ {\mathrm{E}}^{\circ }=1.05\mathrm{V}\\ \left(1\mathrm{F}=96500\mathrm{C}{\mathrm{mol}}^{-1}\right)\end{array}$

According to the formula
$$\begin{gathered} {\mathrm{\Delta G}}^{\circ }=-{\mathrm{nFE}}^{\circ }=-2\times 96500\times 1.05 \\ \mathrm{or} {\mathrm{\Delta G}}^{\circ }=-202650\mathrm{J}{\mathrm{mol}}^{-1}=-202.65\mathrm{KJ}{\mathrm{mol}}^{-1} \\ \mathrm{Now} {\mathrm{\Delta G}}^{\circ }\Rightarrow -202650\mathrm{J}{\mathrm{Mol}}^{-1} \\ \mathrm{R}=8.314\mathrm{J}/\mathrm{Mol}/\mathrm{K},\mathrm{T}=298\mathrm{K} \end{gathered}$$
$\log \mathrm{K}=\frac{{\mathrm{\Delta G}}^{\circ }}{2.303\mathrm{RT}}$
or $\log \mathrm{K}=\frac{-202650}{2.303\times 8.314\times 298}$
$$\begin{gathered} \log \mathrm{K}=\frac{-202650}{5705.84831}=35.52 \\ \mathrm{K}=\text{ Antilog of }35.52\therefore \mathrm{K}=0.35\times {10}^7 \end{gathered}$$