$\therefore \,\,\,{E_1} = {E^0} - \frac{{0.0592}}{2}\,\,\log \,\,\frac{{0.01}}{1}$
$\therefore \,\,{E_2} = {E^0} - \frac{{0.0592}}{2}\,\,\,\log \,\,\frac{{1.0}}{{0.01}}\,\,$
$\,\therefore \,\,{E_1} > {E_2}$
[આપેલ, $KCl$ નું મોલર દળ $74.5 \,g\, mol ^{-1}$ છે.]
${Cu}({s})\left|{Cu}^{2+}({aq})(0.01 {M}) \| {Ag}^{+}({aq})(0.001 {M})\right| {Ag}({s})$ કોષ માટે ,કોષનો પોટેન્શિયલ $=.....\times 10^{-2} {~V}$
[ઉપયોગ : $\frac{2.303 {RT}}{{F}}=0.059$ ]