Question

During the electrolysis of copper (II) sulphate solution using platinum as cathode and carbon as anode:
(i) What do you observe at the cathode and at the anode?
(ii) What change is noticed in the electrolyte?
(iii) Write the reactions at the cathode and at the anode.

Answer 1

(i) During this electrolysis reaction, copper metal is deposited at cathode and oxygen gas is liberated at anode.

(ii) In the electrolyte, the blue colour of copper sulphate becomes paler due to the formation of sulphuric acid.

(iii) Reaction at cathode: The ions migrating to cathode would be ${\mathrm{Cu}}^{\mathrm{2+}}$ and ${\mathrm{H}}^{+}$ ions. But the ion that is preferentially discharged at cathode is ${\mathrm{Cu}}^{\mathrm{2+}}$ as it is positioned lower than ${\mathrm{H}}^{+}$ ion in the electrochemical series, forming copper atoms at platinum cathode.
${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{Cu}$
Reaction at anode: Similarly, ions migrating to anode would be ${\mathrm{SO}}_4^{2-}$, ${\mathrm{OH}}^{-}$. But the ion that is preferentially discharged at anode is ${\mathrm{OH}}^{-}$, as it is positioned lower than ${\mathrm{SO}}_4^{2-}$ in the electrochemical series, producing neutral OH particles that react by itself to form water and oxygen. Oxygen gas is liberated at carbon anode.
$$\begin{gathered} {\mathrm{OH}}^{-}-{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{O}\mathrm{H} \\ \mathrm{4(OH)}\stackrel{}{\to }{\mathrm{2H}}_2\mathrm{O}+{\mathrm{O}}_2 \end{gathered}$$