If ${\mathrm{E}}_{\text{cell }}^0$ of a cell represented by

$\mathrm{Ni}/{\mathrm{Ni}}^{2+}(0.04\mathrm{M})$  $\parallel {\mathrm{Ag}}^{+}(0.002\mathrm{M})/\mathrm{Ag}\text{ is }1.05\mathrm{V}\text{, }{\mathrm{E}}_{\text{cell }}\text{ in }\mathrm{V}\text{ is : }$

$$\begin{gathered} Ni\hspace{0.17em} + 2 A{g}^{+}\to N{i}^{2+}+\hspace{0.17em}Ag ; E_{cell}^o= 1.05V \\ {E}_{cell}^{ }=E_{cell}^o-\frac{0.0591}{n}\log \frac{\left[N{i}^{2+}\right]}{{\left[A{g}^{+}\right]}^2} \\ = 1.05- \frac{0.0591}{2}\log \frac{(0.04)}{{(0.002)}^2} \\ on solving \\ {E}_{cell}= 0.932 V\hspace{0.17em} \end{gathered}$$