since \(\Delta G=-n F E^{\circ}\)
Hence for \(\Delta G\) to be \(-v e \Delta E^{\circ}\) has to be positive.
Which is possible when \(X=Z n, Y=N i\)
\(Z n+N i^{++} \longrightarrow Z n^{++}+N i\)
\(E_{Z n / Z n^{+2}}^{\circ}+E_{N i^{2+} / N i}^{\circ}\)
\(=0.76+(-0.23)=+0.53(\text { positive })\)
${Zn}({s})+{Cu}^{2+}(0.02 {M}) \rightarrow {Zn}^{2+}(0.04 {M})+{Cu}({s})$
${E}_{\text {cell }}=...... \,\times 10^{-2} \,{~V}$ { (નજીકના પૂર્ણાંકમાં) }
${\left[ {E}_{{Cu} / {Cu}^{2+}}^{0}=-0.34\, {~V}, {E}_{2 {n} / {Zn}^{2+}}^{0}=+0.76 \,{~V}\right.}$
$\left.\frac{2.303 {RT}}{{F}}=0.059\, {~V}\right]$
$\left(E_{A g^{+} / A g}^{0}=0.80\, V, E_{A n^{+} / A u}^{0}=1.69\, V\right)$
$Zn = Z{n^{2 + }} + 2{e^ - };\,\,{E^o} = + 0.76\,V$
$Fe = F{e^{2 + }} + 2{e^ - };\,\,{E^o} = + 0.41\,V$
નીચેના કોષ પ્રક્રિયા માટે $EMF$ ......... $\mathrm{V}$ છે
$F{e^{2 + }} + Zn\, \to \,Z{n^{2 + }} + Fe$