MHTCET 26 Apr 2025 Shift 1 Paper

Concept:For the reverse of a reduction half-cell, the standard oxidation potential is the negative of the standard reduction potential. The Nernst equation is then used for non-standard conditions.Explanation:Given reduction: $\mathrm{Cu}^{2+}_{(aq)} + 2e^- \rightarrow \mathrm{Cu}_{(s)}$, $E^\circ_{\text{red}} = +0.34\ \mathrm{V}$.For oxidation: $\mathrm{Cu}_{(s)} \rightarrow \mathrm{Cu}^{2+}_{(aq)} + 2e^-$, the standard potential is:$$E^\circ_{\text{ox}} = -E^\circ_{\text{red}} = -0.34\ \mathrm{V}$$For this oxidation, $n = 2$ and $[\mathrm{Cu}^{2+}] = 0.1\ \mathrm{M}$.Using the Nernst equation for oxidation:$$E = E^\circ_{\text{ox}} - \frac{0.0592}{n} \log[\mathrm{Cu}^{2+}]$$$$E = -0.34 - \frac{0.0592}{2} \log(0.1)$$$$E = -0.34 - 0.0296 \times (-1)$$$$E = -0.34 + 0.0296$$$$E = -0.3104\ \mathrm{V}$$Answer:The potential is $-0.3104\ \mathrm{V}$, which matches option D.