Electrode reactions in aqueous electrolysis

Steps:

• At the cathode, reduce the species with the highest reduction potential (least negative E°).
• At the anode, oxidize the species with the lowest oxidation potential.
• For NaCl(aq) concentrated, cathode favors water reduction over Na+ (E° Na+ = -2.71 V vs. water = -0.83 V).
• Apply rules to verify each statement's products or changes.

Why A is correct:

• In concentrated NaCl(aq), water is reduced at the cathode (2H₂O + 2e⁻ → H₂ + 2OH⁻) because Na⁺ has a more negative reduction potential than water, producing hydrogen gas.

Why the others are wrong:

• B: Dilute H₂SO₄ produces H₂ at the cathode from H⁺ reduction (E° = 0 V); oxygen forms at the anode from water oxidation.
• C: With copper electrodes, Cu²⁺ reduction occurs at the cathode, but the net process transfers Cu from anode to cathode without depleting solution ions; the half-reaction is balanced by anode dissolution.
• D: The anode (copper) undergoes oxidation (Cu → Cu²⁺ + 2e⁻), dissolving metal and decreasing its mass.

Final answer: A