MHTCET 18 Apr 2026 Shift 1 Paper

Concept:In a galvanic cell, the half-reaction with the higher standard reduction potential undergoes reduction at the cathode, while the other half-reaction undergoes oxidation at the anode.Explanation:The given standard reduction potentials are:$$\mathrm{Cd}_{(aq)}^{2+} + 2e^- \rightarrow \mathrm{Cd}_{(s)}, \quad E^\circ = -0.403\ \mathrm{V}$$$$2\mathrm{H}^+_{(aq)} + 2e^- \rightarrow \mathrm{H}_{2(g)}, \quad E^\circ = 0.000\ \mathrm{V}$$Since $0.000 > -0.403$, the hydrogen electrode has the higher reduction potential.Therefore, $\mathrm{H}^+$ is reduced at the cathode:$$2\mathrm{H}^+_{(aq)} + 2e^- \rightarrow \mathrm{H}_{2(g)}$$Cadmium undergoes oxidation at the anode:$$\mathrm{Cd}_{(s)} \rightarrow \mathrm{Cd}^{2+}_{(aq)} + 2e^-$$Adding the two half-reactions gives the net cell reaction:$$\mathrm{Cd}_{(s)} + 2\mathrm{H}^+_{(aq)} \rightarrow \mathrm{Cd}^{2+}_{(aq)} + \mathrm{H}_{2(g)}$$The standard cell potential is:$$E^\circ_{\text{cell}} = 0.000 - (-0.403) = +0.403\ \mathrm{V}$$A positive $E^\circ_{\text{cell}}$ confirms that the reaction is spontaneous.Answer:The correct net cell reaction is option B:$$\mathrm{Cd}_{(s)} + 2\mathrm{H}^+_{(aq)} \rightarrow \mathrm{Cd}^{2+}_{(aq)} + \mathrm{H}_{2(g)}$$