What will happen during the electrolysis of aqueous solution of CuSO₄ by using platinum electrodes?

Dissolving CuSO₄ in water will lead to dissociation of CuSO₄ to Cu²⁺ and SO₄^2-. Water will dissociate as H⁺ and OH^-. On electrolysis using inert platinum electrodes, the cathode and anode reactions along with their standard electrode potential are as follows:

At anode, two reactions are possible –

(i) 2H₂O (l) → 4H⁺ (aq) + O₂ (g) + 4e^– OR 4OH^– (aq) → 2H₂O (l) + O₂ (g) + 4e^–, E^Ө_cell = 1.23V

(ii) 2SO₄^2- (l) → S₂O₈^2- + 2e^-, E^Ө_cell = 1.96 V

The reaction with lower electrode potential will be favoured at the anode, so O₂ is released at anode.

At cathode, two reactions are possible –

(i) Cu²⁺ + 2e^- → Cu(s), E^Ө_cell = 0.34V

(ii) H⁺ + e^-→ H₂(g), E^o_cell = 0.00 V

The reaction with higher electrode potential is favoured at the cathode, so Cu metal is deposited at the anode.

The correct options are (i) and (iii).