Chemistry - Electrochemistry Question with Solution | TestHub

Apply Nernst equation to the reaction $\mathrm{Pb}+{\mathrm{Sn}}^{2+}\to {\mathrm{Pb}}^{2+}+\mathrm{Sn}$ or $E^{\circ }+\frac{0.059}{2}\log \frac{\left[{\mathrm{Sn}}^{2+}\right]}{\left[{\mathrm{Pb}}^{2+}\right]}=E_{\text{cell }}$ or $\log \frac{\left[{\mathrm{Sn}}^{2+}\right]}{\left[{\mathrm{Pb}}^{2+}\right]}=\frac{0.01\times 2}{0.059}=0.3\left(\because E_{\text{cell }}=0\right)$ or $\frac{\left[{\mathrm{Sn}}^{2+}\right]}{\left[{\mathrm{Pb}}^{2+}\right]}=$ antilog $(0.3)$