State the reason for difference in product formed at the anode during

1.	State the reason for difference in product formed at the anode during electrolysis of aq. CuSO4 using1. active electrode – copper anode2. inert electrode – platinum anode.
Answer» 1. Electrolysis of aq. CuSO₄ using copper anode: As copper can easily lose electron, copper from anode will dissolve as Cu²⁺ ions. Cu (s) – 2e^–→ Cu²⁺ (aq) 2. Electrolysis of aq. CuSO₄ using inert platinum anode: Due to very low tendency of platinum to lose electron platinum anode does not take part in electrolytic reaction. Further tendency of SO₄^2- to lose electron is much less than that of OH^– (from feebly ionised water). Thus OH^– ions get oxidised in preference to SO₄^2- ions to give O₂ 4OH – 4e → 4OH 4OH→ 2H₂O + O₂↑

State the reason for difference in product formed at the anode during electrolysis of aq. CuSO4 using1. active electrode – copper anode2. inert electrode – platinum anode.

Answer»

1. Electrolysis of aq. CuSO₄ using copper anode: As copper can easily lose electron, copper from anode will dissolve as Cu²⁺ ions.

Cu (s) – 2e^–→ Cu²⁺ (aq)

2. Electrolysis of aq. CuSO₄ using inert platinum anode: Due to very low tendency of platinum to lose electron platinum anode does not take part in electrolytic reaction. Further tendency of SO₄^2- to lose electron is much less than that of OH^– (from feebly ionised water). Thus OH^– ions get oxidised in preference to SO₄^2- ions to give O₂

4OH – 4e → 4OH

4OH→ 2H₂O + O₂↑

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