$\mathrm{Sn}$ (s) | $\mathrm{Sn}^{2+}(0.004\;\mathrm{M}) \parallel \mathrm{H}^{+}(0.020\;\mathrm{M}) \mid \mathrm{H}_2$ (g) (1 bar) | Pt (s) $\; \mathrm{Sn}(\mathrm{s}) \rightarrow \mathrm{Sn}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-}$ $2\mathrm{H}^{+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{g})$ $\text{────────────────────}$ $\mathrm{Sn}(\mathrm{s}) + 2\mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Sn}^{2+}(\mathrm{aq}) + \mathrm{H}_2(\mathrm{g})$ $\text{────────────────────}$ $E^{\circ}_{\;\text{cell}\;} = E^{\circ}_{(\mathrm{H}^{+}/\mathrm{H}_2)} - E^{\circ}_{(\mathrm{Sn}^{2+}/\mathrm{Sn})}$ $\; = 0.00 - (-0.14)$ $\; = +0.14\;\mathrm{V}$ $E_{\;\text{cell}\;} = E^{\circ}_{\;\text{cell}\;} - \frac{0.0591}{n} \log \frac{[\mathrm{Sn}^{2+}]}{[\mathrm{H}^{+}]^2}$ $\; = 0.14 - \frac{0.0591}{2} \log \frac{(4 \times 10^{-3})}{(2 \times 10^{-2})^2}$ $\; = 0.14 - 0.0295 \log 10$ $\; = 0.14 - 0.0295$ $\; = 0.1105\;\mathrm{V}$ (b) (i) $\; \mathrm{NaCl} \rightarrow \mathrm{Na}^{+} + \mathrm{Cl}^{-}$ $\; \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{-}$ The value of $E^{\circ}$ of $\mathrm{O}_2$ is higher than $\mathrm{Cl}_2$ but $\mathrm{O}_2$ is evolved from $\mathrm{H}_2\mathrm{O}$ only when the higher voltage is applied. So, because of this $\mathrm{Cl}_2$ is evolved instead of $\mathrm{O}_2$.(ii) Conductivity varies with the change in the concentration of the electrolyte. The number of ions per unit volume decreases on dilution. So, conductivity decreases with decrease in concentration. Therefore, conductivity of $\mathrm{CH}_3\mathrm{COOH}$ decreases on dilution. OR $\;\; \text{(a)}\; \Delta G^{\circ} = -n F E^{\circ}$ $\; -43600 = -2 \times 96500 \times E^{\circ}$ $\; E^{\circ} = 0.226\;\mathrm{V}$ $\; E = E^{o} - \frac{0.059}{2} \log$ $\left( \frac{[\mathrm{H}^{+}]^2 [\mathrm{Cl}^{-}]^2}{[\mathrm{H}_2]} \right)$ $\; = 0.226 - \frac{0.059}{2} \log \left[ (0.1)^2 \right]$ $\; \times (0.1)^2 ] / 1$ $\; = 0.226 - \frac{0.059}{2} \log 10^{-4}$ $\; = 0.226 + 0.118 = 0.344\;\mathrm{V}$ (b) Cells that convert the energy of combustion of fuels (like hydrogen, methane, methanol etc.) directly into electrical energy are called fuel cells. Advantages : High efficiency, non polluting (or any other suitable advantage)