Dissolving CuSO4 in water will lead to dissociation of CuSO4 to Cu2+ and SO42-. 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 SO42- 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).
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