State the reason for difference in product formed at the anode during electrolysis of aq. CuSO4 using :
(i) active electrode — copper anode
(ii) inert electrode — platinum anode.

(i) Electrolysis of aq. CuSO₄ using copper anode — Product at anode is nil. [Copper ions are formed]

Reaction at anode: Cu – 2e^– ⟶ Cu²⁺

SO₄^2- and OH^– ions migrate to anode but neither of them are discharged due to the nature of anode, (copper loses electrons more easily than SO₄^2- and OH^–) because copper anode itself ionises to give Cu²⁺ ions.

(ii) Electrolysis of aq. CuSO₄ using inert platinum anode — Product at anode is oxygen gas.

SO₄^2- and OH^– ions migrate to anode but OH^– ions are discharged since they are lower in the electrochemical series.

Oxidation of the OH^– ions gives unstable hydroxyl radical which forms water with the liberation of oxygen.