MHTCET 23 Apr 2025 Shift 2 Paper

Concept:For a cell reaction involving two ions, the Nernst equation relates emf to concentration ratio. When both ion concentrations change by the same factor, the ratio remains the same.Explanation:The cell reaction is:$\mathrm{Cd}_{(s)} + \mathrm{Cu}^{2+}_{(aq)} \longrightarrow \mathrm{Cd}^{2+}_{(aq)} + \mathrm{Cu}_{(s)}$Number of electrons transferred, $n = 2$.Using the Nernst equation:$E_{\text{cell}} = E_{\text{cell}}^{\ominus} - \frac{0.0592}{n} \log_{10} \frac{[\mathrm{Cd}^{2+}]}{[\mathrm{Cu}^{2+}]}$Initially, $[\mathrm{Cd}^{2+}] = [\mathrm{Cu}^{2+}] = 1\ \mathrm{M}$.So, $E_{\text{cell}} = 0.74 - \frac{0.0592}{2} \log_{10} \frac{1}{1} = 0.740\ \mathrm{V}$.When both concentrations decrease by 10 times:$[\mathrm{Cd}^{2+}] = [\mathrm{Cu}^{2+}] = 0.1\ \mathrm{M}$.The ratio remains $\frac{0.1}{0.1} = 1$, so the logarithmic term is zero.Hence, $E_{\text{cell}} = 0.740\ \mathrm{V}$.The emf is unchanged because the reaction quotient $Q$ stays the same.Answer:$E_{\text{cell}} = +0.740\ \mathrm{V}$Correct option: D. +0.740 V