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 SO42-. 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, EӨcell = 1.23V


(ii) 2SO42- (l) S2O82- + 2e-, EӨcell = 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), EӨcell = 0.34V


(ii) H+ + e- H2(g), Eocell = 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).

1