Electrolysis of aqueous CuSO₄ with inert electrodes

Steps:

• Electrolyte contains Cu²⁺, SO₄²⁻, and H₂O.
• Cathode (reduction): Cu²⁺ + 2e⁻ → Cu (preferred over 2H₂O + 2e⁻ → H₂ + 2OH⁻ due to higher E°).
• Anode (oxidation): 2H₂O → O₂ + 4H⁺ + 4e⁻ (preferred over SO₄²⁻ oxidation).
• Products: Cu at cathode, O₂ at anode.

Why C is correct:

• Oxygen forms at anode via water oxidation (2H₂O → O₂ + 4H⁺ + 4e⁻), as sulfate ions require higher potential.

Why the others are wrong:

• A: Copper deposits at cathode via reduction, not anode.
• B: Copper, not hydrogen, forms at cathode due to Cu²⁺ easier reduction.
• D: No sulfur present; solution is CuSO₄, no S production.

Final answer: C