MHTCET 23 Apr 2025 Shift 2 Paper

Concept:In a galvanic cell, the left electrode is the anode where oxidation occurs, and the right electrode is the cathode where reduction occurs.A positive emf means the cell reaction is spontaneous in the given direction.Explanation:The cell notation is:$$A_{(s)}|A^+_{(aq)}(1\,\mathrm{M})||B^{2+}_{(aq)}(1\,\mathrm{M})|B_{(s)}$$At the anode, oxidation takes place:$$A_{(s)} \rightarrow A^+_{(aq)} + e^-$$At the cathode, reduction takes place:$$B^{2+}_{(aq)} + 2e^- \rightarrow B_{(s)}$$To balance electrons, multiply the anode half-reaction by 2:$$2A_{(s)} \rightarrow 2A^+_{(aq)} + 2e^-$$Adding both half-reactions gives the net cell reaction:$$2A_{(s)} + B^{2+}_{(aq)} \rightarrow 2A^+_{(aq)} + B_{(s)}$$This is the possible spontaneous cell reaction.Answer:Option B: $2A_{(s)} + B^{2+}_{(aq)} \rightarrow 2A^+_{(aq)} + B_{(s)}$