Match Column I with Column II (Electrolysis products using inert electrodes). Column I: (A) Dilute s — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Match the Following — Electrolysis Products (Inert Electrodes)} \textbf{Column I (Solution)} \quad \textbf{Column II (Product at anode)} \textbf{Analysis of each:} \textbf{A. Dilute H\textsubscript{2}SO\textsubscript{4}:} Cathode: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ Anode: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ \quad (SO\textsubscript{4}\textsuperscript{2-} not discharged at low conc.) Products: H\textsubscript{2} at cathode, \textbf{O\textsubscript{2}} at anode \quad → \textbf{P, Q} \textbf{B. Dilute NaCl:} Cathode: $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$ Anode: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ \quad (Cl\textsuperscript{-} not preferentially discharged at low conc.) Products: H\textsubscript{2} at cathode, \textbf{O\textsubscript{2}} at anode \quad → \textbf{P, Q} \textbf{C. Concentrated NaCl (brine):} Cathode: $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$ Anode: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ \quad (Cl\textsuperscript{-} preferentially discharged at high conc.) Products: H\textsubscript{2} at cathode, \textbf{Cl\textsubscript{2}} at anode \quad → \textbf{Q, R} \textbf{D. CuSO\textsubscript{4} solution:} Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (Cu deposits) Anode: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ Products: \textbf{Cu} at cathode, \textbf{O\textsubscript{2}} at anode \quad → \textbf{P, S} \textbf{Answer: A→P,Q; B→P,Q; C→Q,R; D→P,S}