Statement I: When acidified solution is electrolysed between zinc electrodes, zinc is deposited at c — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Assertion (I):} During electrolysis of aqueous ZnSO\textsubscript{4} solution, zinc deposits at the cathode. \textbf{Reason (II):} The discharge potential of Zn\textsuperscript{2+} is lower than the discharge potential of H\textsubscript{2}O (i.e., Zn\textsuperscript{2+} requires less energy to be reduced in the given conditions). \textbf{Analysis:} Standard reduction potentials: \[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad E^\circ = -0.76\ \text{V} \] \[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad E^\circ = 0.00\ \text{V} \] At first glance, H\textsuperscript{+} (from water) should be reduced preferentially since its E\textdegree is higher. However, at high concentrations of Zn\textsuperscript{2+} and due to overpotential effects on hydrogen evolution, Zn\textsuperscript{2+} is preferentially discharged at practical electrode potentials. This is analogous to why Cu deposits from CuSO\textsubscript{4} even though E\textdegree(Cu) = +0.34 V — concentration and overpotential factors can reverse simple E\textdegree predictions. Statement I is \textbf{correct}. Statement II, though phrased differently from standard Nernst reasoning, represents the operational "discharge potential" concept that explains Statement I. \textbf{Answer: A} (Both correct, II explains I)