According to the equation, $\mathrm{Ni}+2\mathrm{Ag}^{+}\rightarrow\mathrm{Ni}^{2+}+2\mathrm{Ag}$ $\Delta G=-nFE^{\circ}$ $\text{ where }\Delta G=\text{Gibb}'\text{ ' free energy }$ $\Delta G=-2\times96500\times1.05$ $N=\text{ No. of electrons gain or lost }=2$ $\Delta G=-202.650\,\text{kJ}$ $F=\text{ Faraday's constant }=96500$ $E^{\circ}=\text{ Standard emf }=1.05\,\text{V}$ The relation between Gibb's free energy and Equilibrium constant is given by equation $E^{\circ}_{\text{cell}}=\frac{0.0591}{n}\log K_c$ $\log K_c=-\frac{1.05\times2}{0.0591}=35.53$ $K_c=3.39\times10^{35}$ OR According to the equation, $\mathrm{Fe}(s)+2\mathrm{H}^{+}(aq)\rightarrow\mathrm{Fe}^{2+}(aq)+\mathrm{H}_2(g)$ $E^{\circ}_{\text{cell}}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}$ $E^{\circ}_{\text{cell}}=0-(-0.44)\,\text{V}$ $E^{\circ}_{\text{cell}}=+0.44\,\text{V}$ By applying Nernst Equation, $E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.0591}{2}\log\frac{[\mathrm{Fe}^{2+}]}{[\mathrm{H}^{+}]^2}$ $E_{\text{cell}}=0.44-\frac{0.0591}{2}\log\frac{0.001}{(0.01)^2}$ $E_{\text{cell}}=0.44-\frac{0.0591}{2}\log10$ $E_{\text{cell}}=0.44-0.0295\times1$ $E_{\text{cell}}=+0.410\,\text{V}$