Question

An electrolytic cell is set up using two platinum electrodes and an aqueous solution of copper[II] sulphate. write the equation for the reaction at anode

Answer 1

Reaction at Anode

• The oxidation half-reaction occurring at the anode is given in this reaction
• Here, copper electrodes are used. Copper loses electrons more easily than ${{\mathrm{SO}}_4}^{-2}$ and ${\mathrm{OH}}^{-}$ because copper anode itself ionizes to give ${\mathrm{Cu}}^{+2}$ions

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

• ${{\mathrm{SO}}_4}^{-2}$ and ${\mathrm{OH}}^{-}$ ions migrate to the anode but neither are discharged due to the nature of the anode. product at the anode is nil as ${\mathrm{Cu}}^{+2}$ ions are formed.
• If platinum electrode as the anode is used, then the oxidation half-reaction is:

$$\begin{gathered} {\mathrm{OH}}^{-}-{\mathrm{e}}^{-}\to \mathrm{OH} \\ 4\mathrm{OH}\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2 \end{gathered}$$

• Both ${{\mathrm{SO}}_4}^{-2}$ and ${\mathrm{OH}}^{-}$ ions migrate to the anode, but ${\mathrm{OH}}^{-}$ ions being lower in the electrochemical series are discharged.