\(\Delta G\) is negative, if \(E _{\text {cell }}\) is positive
Anode : \(\quad Cu ( s ) \longrightarrow Cu ^{-2}\left( C _{1}\right)+2 e ^{-}: E ^{\circ}\)
\(\frac{\text { Cathode }: Cu ^{+2}\left( C _{2}\right)+2 e ^{-} \longrightarrow Cu ( S ):- E ^{\circ}}{\text { Cell reaction }: Cu ^{+2}\left( C _{2}\right) \longrightarrow Cu ^{+2}\left( C _{1}\right) E _{\text {cell }}^{\circ}=0}\)
\(E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{2.303 RT }{ nF } \log Q\)
\(E _{\text {cell }}=0-\frac{2.303 RT }{ nF } \log \left(\frac{ C _{1}}{ C _{2}}\right)\)
\(E _{\text {cell }}>0:\) if \(\frac{ C _{1}}{ C _{2}}<1 \Rightarrow C _{1}< C _{2}\)
${{\text{E}}^o }{\text{C}}{{\text{u}}^{{\text{2}} + }}{\text{/Cu = + 0}}{\text{.34 V, E}}_{{\text{F}}{{\text{e}}^{ + {\text{2}}}}/Fe}^o = \,\,{\text{ - 0}}{\text{.44 V}}$
$Cu ( s )+ Sn ^{2+}( aq ) \rightarrow Cu ^{2+}( aq )+ Sn ( s )$
$\left( E _{ Sn ^{2+} \mid Sn }^{0}=-0.16\, V , E _{ Cu ^{2+} \mid Cu }^{0}=0.34\, V \right.$ Take $F=96500\, C\, mol ^{-1}$ )
આપેલ $\left( E _{ Cu ^{2+} / Cu ^{+}}^{0}=0.16 V \right.$ $,E _{ Cu ^{+} / Cu }^{0}=0.52 V,$ $\left.\frac{ RT }{ F }=0.025\right)$