Which of the following statements about the electrolysis of CuSO4(aq) using copper cathode and platinum anode is not correct?

Analyze the question: We are asked to find the statement that is NOT correct about electrolysis of CuSO₄(aq) with a copper cathode and a platinum (inert) anode.

Concept: The products depend on the electrodes and the ions present.

• Ions in solution: Cu²⁺, H⁺, SO₄²⁻, OH⁻
• Platinum is inert, so it does not dissolve.

Step-by-step reactions:

• Cathode (reduction): Cu²⁺ is preferentially discharged over H⁺. Cu²⁺ + 2e⁻ → Cu So copper is deposited at the cathode. A is correct.

• Anode (oxidation): Since Pt is inert, OH⁻ is discharged in preference to SO₄²⁻. 4OH⁻ → 2H₂O + O₂ + 4e⁻ So oxygen is liberated. B is correct.

• Solution: Cu²⁺ is removed and OH⁻ is removed (from water), leaving excess H⁺ with SO₄²⁻. This forms H₂SO₄, so the solution becomes acidic. D is correct.

• Purification of copper: This requires an impure copper anode that dissolves, and a pure copper cathode. Here the anode is platinum, which does not dissolve, so copper is not being transferred from impure to pure. The solution just gets depleted of Cu²⁺. So C is not correct for this setup.

Tip: Remember the difference:

• Inert electrodes (Pt or C): The electrolyte is decomposed, the blue colour fades, and the solution becomes acidic.
• Copper anode (active): The anode dissolves, the cathode grows, the blue colour stays constant, and this is the purification/refining setup.

Whenever an exam question mentions "purification" or "refining", look for an impure active metal anode.