Question

Give the reactions that occur at cathode and anode in the electrolysis of Copper sulphate solution using platinum anode and copper or platinum cathode.

Answer 1

Part 1: Electrolysis of Copper sulphate:-

When aqueous Copper sulphate is electrolysed using platinum anode and copper or platinum cathode, the Copper sulphate gets dissociated as follows:-

$$\begin{gathered} \underset{\mathrm{Copper}\mathrm{sulphate}}{{\mathrm{CuSO}}_4\left(\mathrm{aq}\right)}\rightleftharpoons \underset{\mathrm{Copper}\mathrm{ion}}{{\mathrm{Cu}}^{2+}}+\underset{\mathrm{Sulphate}\mathrm{ion}}{{{\mathrm{SO}}_4}^{2-}} \\ \underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\rightleftharpoons \underset{\mathrm{Hydrogen}\mathrm{ion}}{{\mathrm{H}}^{+}}+\underset{\mathrm{Hydroxide}\mathrm{ion}}{{\mathrm{OH}}^{-}} \end{gathered}$$

Part 2: Reaction at anode:-

• The sulphate $\left({{\mathrm{SO}}_4}^{2-}\right)$ and the hydroxide $\left({\mathrm{OH}}^{-}\right)$ ions both migrate toward the anode
• But since the hydroxide ions$\left({\mathrm{OH}}^{-}\right)$ are slower than sulphate ions$\left({{\mathrm{SO}}_4}^{2-}\right)$ in the electrochemical series, only the hydroxide ions$\left({\mathrm{OH}}^{-}\right)$ get discharged.

$$\begin{gathered} [\underset{\mathrm{Hydroxide}\mathrm{ion}}{{\mathrm{OH}}^{-}}-\underset{\mathrm{Electron}}{{\mathrm{e}}^{-}}\to \underset{\mathrm{Hydroxide}}{\mathrm{OH}}]\times 2 \\ \underset{\mathrm{Hydroxide}}{2\mathrm{OH}}\to \underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{\left(\mathrm{O}\right)} \\ \underset{\mathrm{Hydroxide}}{4\mathrm{OH}}\to 2\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2} \end{gathered}$$

Part 3: Reaction at cathode:-

• Copper and the hydrogen ions both migrate toward the cathode.
• Since the position of copper is lower than hydrogen in the electrochemical series, as a result, only copper gets discharged at the cathode, by accepting two electrons and forming a neutral copper atom

$\underset{\mathrm{Copper}\mathrm{ion}}{{\mathrm{Cu}}^{+}}+\underset{\mathrm{Electron}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Copper}}{\mathrm{Cu}}$

Therefore, reaction at the cathode$\underset{\mathrm{Copper}\mathrm{ion}}{{\mathrm{Cu}}^{+}}+\underset{\mathrm{Electron}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Copper}}{\mathrm{Cu}}$,

reaction at the anode $$\begin{gathered} [\underset{\mathrm{Hydroxide}\mathrm{ion}}{{\mathrm{OH}}^{-}}-\underset{\mathrm{Electron}}{{\mathrm{e}}^{-}}\to \underset{\mathrm{Hydroxide}}{\mathrm{OH}}]\times 2 \\ \underset{\mathrm{Hydroxide}}{2\mathrm{OH}}\to \underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{\left(\mathrm{O}\right)} \\ \underset{\mathrm{Hydroxide}}{4\mathrm{OH}}\to 2\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2} \end{gathered}$$