Question 12
State the reason for difference in product formed at the anode during electrolysis of aq. CuSO4 using :
(i) active electrode — copper anode
(ii) inert electrode — platinum anode.
(i) Electrolysis of aq. CuSO4 using copper anode — Product at anode is nil. [Copper ions are formed]
Reaction at anode: Cu - 2e- ⟶ Cu2+
SO42- and OH- ions migrate to anode but neither of them are discharged due to the nature of anode, (copper loses electrons more easily than SO42- and OH-) because copper anode itself ionises to give Cu2+ ions.
(ii) Electrolysis of aq. CuSO4 using inert platinum anode — Product at anode is oxygen gas.
SO42- and OH- ions migrate to anode but OH- ions are discharged since they are lower in the electrochemical series.
Oxidation of the OH- ions gives unstable hydroxyl radical which forms water with the liberation of oxygen.