Question

During the electrolysis of copper (II) sulphate solution using platinum as cathode and carbon as anode: What do you observe at the cathode and at the anode.

Answer 1

Electrolysis is the process of dissociation of a compound into its ions bypassing the electric current.

Electrolysis of copper sulphate solution with platinum as cathode and carbon as anode gives copper ion and sulphate ion in the solution.

$$\begin{gathered} {\mathrm{CuSO}}_4\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right) \\ \mathrm{copper}\mathrm{copper}\mathrm{sulphate} \\ \mathrm{sulphate}\mathrm{ion}\mathrm{ion} \end{gathered}$$

During this electrolysis reaction at the anode, oxygen gas is liberated and the copper metal gets deposited at the cathode.

At the anode: hydroxide ions will undergo oxidation.

$$\begin{gathered} 4{\mathrm{OH}}^{-}\to 2{\mathrm{O}}_2\uparrow +2{\mathrm{H}}_2+4{\mathrm{e}}^{-} \\ \mathrm{hydroxide}\mathrm{oxygen}\mathrm{water} \\ \mathrm{ion}\mathrm{gas} \end{gathered}$$

At the cathode: copper ions will undergo reduction.

${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$