Electrolysis of aqueous CuSO₄ with inert carbon electrodes

Steps:

• Ions present: Cu²⁺, SO₄²⁻, and H⁺/OH⁻ from water.
• Cathode reaction: Cu²⁺ + 2e⁻ → Cu (solid copper deposits).
• Anode reaction: 2H₂O → O₂ + 4H⁺ + 4e⁻ (oxygen gas evolves).
• Cu²⁺ depletion reduces blue color intensity.

Why C is correct:

• Cu²⁺ ions are selectively discharged at cathode, decreasing their concentration and fading the solution's blue color per Faraday's laws of electrolysis.

Why the others are wrong:

• A: Anode produces oxygen gas, not pink solid (pink relates to other ions like Co²⁺).
• B: Cathode deposits copper metal, no hydrogen bubbles (Cu²⁺ easier to reduce than H⁺).
• D: Cathode gains copper mass, increasing size.

Final answer: C