JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

For the given redox reaction, the half-cell reactions are: Anode: Red _1 Ox _1^ n_1^+ + n_1 e^- Cathode: Ox _2 + n_2 e^- Red _2^ n_2^- To obtain the overall balanced chemical equation, we multiply the anode reaction by n_2 and the cathode reaction by n_1 so that the number of electrons lost equals the number of electrons gained: n_2 Red _1 n_2 Ox _1^ n_1^+ + n_1 n_2 e^- n_1 Ox _2 + n_1 n_2 e^- n_1 Red _2^ n_2^- Adding these gives the overall reaction: n_2 Red _1 + n_1 Ox _2 n_2 Ox _1^ n_1^+ + n_1 Red _2^ n_2^- Thus, statement (A) is correct. In the overall reaction, electrons do not appear because the electrons produced at the anode are exactly consumed at the cathode. Thus, statement (B) is correct. From the Nernst equation: E = E^ - RT nF Q = E^ - 2.303 RT nF _ 10 Q E - E^ RT/F = - 2.303 n _ 10 Q The graph of E - E^ RT/F versus _ 10 Q is a straight line with a slope of - 2.303 n , where n is the number of electrons transferred in the balanced redox reaction. Thus, statement (C) is correct. The electrical work done by a galvanic cell is given by the product of the total charge transferred and the cell potential (potential difference), not their ratio. W_ electrical = Charge Potential difference = nF E_ cell Thus, statement (D) is incorrect. Answer: If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.