MHTCET 26 Apr 2025 Shift 2 Paper

Concept:In a galvanic cell, the half-cell with higher reduction potential undergoes reduction (cathode), and the one with lower reduction potential undergoes oxidation (anode).Explanation:Given: $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = +0.34\ \mathrm{V}$ and $E^\circ_{\mathrm{H}^{+}/\mathrm{H}_2} = 0.00\ \mathrm{V}$.Since copper has the higher reduction potential, $\mathrm{Cu}^{2+}$ is reduced at the cathode:$\mathrm{Cu}^{2+}_{(\mathrm{aq})} + 2e^- \longrightarrow \mathrm{Cu}_{(\mathrm{s})}$Hydrogen has the lower reduction potential, so $\mathrm{H}_2$ is oxidised at the anode:$\mathrm{H}_{2(\mathrm{g})} \longrightarrow 2\mathrm{H}^{+}_{(\mathrm{aq})} + 2e^-$Adding the two half-reactions gives the net cell reaction:$\mathrm{H}_{2(\mathrm{g})} + \mathrm{Cu}^{2+}_{(\mathrm{aq})} \longrightarrow 2\mathrm{H}^{+}_{(\mathrm{aq})} + \mathrm{Cu}_{(\mathrm{s})}$Answer:$\mathrm{H}_{2(\mathrm{g})} + \mathrm{Cu}^{2+}_{(\mathrm{aq})} \longrightarrow 2\mathrm{H}^{+}_{(\mathrm{aq})} + \mathrm{Cu}_{(\mathrm{s})}$Hence, option B is correct.