COMEDK 2025 Evening Solved Paper

Concept:Electrolysis of aqueous NaCl involves competing reactions at electrodes. The actual product depends on both thermodynamic potentials and kinetic overpotential.Explanation:At the cathode, two reductions are possible:Reduction of Na⁺: $\mathrm{Na^+ + e^- \rightarrow Na(s)}$, $E^\circ = -2.71\ \mathrm{V}$Reduction of water: $\mathrm{2H_2O + 2e^- \rightarrow H_2 + 2OH^-}$, $E^\circ = -0.83\ \mathrm{V}$Since $-0.83\ \mathrm{V}$ is less negative (easier to reduce), water is reduced and $\mathrm{H_2}$ is liberated.At the anode, two oxidations are possible:Oxidation of water: $\mathrm{2H_2O \rightarrow O_2 + 4H^+ + 4e^-}$, $E^\circ = +1.23\ \mathrm{V}$Oxidation of chloride: $\mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}$, $E^\circ = +1.36\ \mathrm{V}$Thermodynamically, water oxidation (lower $E^\circ$) should occur first.However, oxygen evolution has a high kinetic overpotential on common electrode materials.This overpotential suppresses water oxidation, so chloride oxidizes preferentially to $\mathrm{Cl_2}$.Thus, Assertion is correct: $\mathrm{Cl_2}$ at anode and $\mathrm{H_2}$ at cathode.Reason is correct: the reaction with lower oxidation potential (water) is not preferred due to oxygen overpotential.Answer:Option B: Both Assertion and Reason are correct.