The reaction that occurs all the anode when CuSO4(aq) is electrolyzed using carbon electrode is

Analyze the question: We need the reaction at the anode (positive electrode, where oxidation occurs) when aqueous CuSO₄ is electrolyzed using inert carbon electrodes.

Concept:

• The anode is where oxidation happens (loss of electrons).
• Reduction (gain of electrons) happens at the cathode.
• In the electrolysis of an aqueous solution, the ions present are Cu²⁺ and SO₄²⁻ (from CuSO₄), plus H⁺ and OH⁻ (from water).
• At the anode, the ion that is preferentially discharged is OH⁻ rather than SO₄²⁻. This is because SO₄²⁻ is very hard to oxidize, and OH⁻ is lower in the electrochemical series. Carbon is inert, so it does not take part in the reaction.

Solution:

Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s). Copper is deposited, so options A and B describe reduction reactions and cannot be anode reactions.

Anode: OH⁻ is discharged:

4OH⁻(aq) − 4e⁻ → 2H₂O(l) + O₂(g)

Oxygen gas is given off, and the solution becomes more acidic because H⁺ and SO₄²⁻ remain, forming H₂SO₄.

Checking the options:

• A: H⁺ gains electrons, which is reduction, so it is not at the anode.
• B: Cu²⁺ is reduced, which happens at the cathode.
• C: OH⁻ is oxidized to water and oxygen. This is correct.
• D: Sulfate is not discharged, and the equation is also wrong (SO₄²⁻ does not give SO₂ and O₂ this way).

Tip: Remember "AN OX" (Anode = Oxidation) and "RED CAT" (Reduction = Cathode). For aqueous CuSO₄ with inert electrodes, you get copper at the cathode and oxygen at the anode. Note that with copper electrodes, the anode itself dissolves instead.