Dissolving CuSO₄ in water will lead to dissociation of CuSO₄ to Cu²⁺ and SO₄^2-. Water will dissociate as H⁺ and OH^-. In this case, copper electrodes are used instead of platinum inert electrodes. So the reactions will be different. Electrolysis of CuSO₄ using Cu electrodes will deposit copper metal at the cathode and the copper at the anode dissolves to form cations in solution.

At the anode, the negatively charged ions SO₄^2- and OH^- are attracted but they are stable to exist on their own and do not get oxidized, so the copper on the anode gets oxidized to form Cu²⁺ ions. The reaction is Cu(s) → Cu²⁺ + 2e^-.

At the cathode, the positively charged Cu²⁺ ions get attracted and get reduced to form Cu metal, which gets deposited on the copper cathode. It is favoured more than H⁺ to get reduced because of higher electrode potential. The reaction is given as Cu²⁺ + 2e^- → Cu(s).

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