(આપેલ છે: $\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{0}=-0.14 \mathrm{\;V}$ $\left.\mathrm{E}_{\mathrm{Pb}^{+2}/{\mathrm{Pb}}}^{0}=-0.13 \;\mathrm{V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\right)$
$\mathrm{Sn}(\mathrm{s})+\mathrm{Pb}^{+2}(\mathrm{aq}) \rightarrow \mathrm{Sn}^{+2}(\mathrm{aq})+\mathrm{Pb}(\mathrm{s})\dots (1)$
Apply Nernst equation:
$\mathrm{E}_{\text {cell}}=\mathrm{E}_{\text {cel }}^{0}-\frac{0.06}{2} \log \frac{\left[\mathrm{Sn}^{+2}\right]}{\left[\mathrm{Pb}^{+2}\right]}$
$\mathrm{E}_{\mathrm{cell}}^{0}=-0.13+0.14=0.01 \mathrm{V}$
At equilibrium : $E_{\text {cell}}=0$
Substituting in $( 1)$
$0=0.01-\frac{0.06}{2} \log \frac{\left[\mathrm{Sn}^{+2}\right]}{\left[\mathrm{Pb}^{+2}\right]}$
$\Rightarrow \quad \frac{1}{3}=\log \frac{\left[\mathrm{Sn}^{+2}\right]}{\left[\mathrm{Pb}^{+2}\right]}$
$\Rightarrow\; \frac{\left[\mathrm{Sn}^{+2}\right]}{\left\lceil\mathrm{Pb}^{+2}\right\rceil}=2.15$
$Pt|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|0.1{\mkern 1mu} M{\mkern 1mu} HCl||{\mkern 1mu} {\mkern 1mu} 0.1{\mkern 1mu} M\,C{H_3}COOH|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|Pt$
$Sn ^{2+}+2 e ^{-} \rightarrow Sn$
$Sn ^{4+}+4 e ^{-} \rightarrow Sn$
ઈલેક્ટ્રોન (વિદ્યુતધ્રુવ) પોટેન્શિયલ ની $E _{ Sn ^{2+} / Sn }^{\circ}=-0.140 V$ અને $E _{ Sn ^{4+} / Sn }^{\circ}=0.010 V$ છે. $Sn ^{4+} / Sn ^{2+}$
$E^{o} _{ Sn ^{4+} / Sn ^{2+}}$માટે પ્રમાણિત ઈલેક્ટ્રોડ (વિદ્યુતધ્રુવ) પોંટેન્શિયલની માત્રા........ $\times 10^{-2} V$ છે. (નજીકનો પૂર્ણાક)