Question

A direct current of electricity is passed through a dilute solution of copper sulphate using platinum electrodes. Describe and explain all that happens during electrolysis

Answer 1

Electrolysis of Aqueous copper sulfate solution.

Electrolyte

• Aqueous copper sulfate. A few drops of diluted sulphuric acid are added as it increases the electrical conductivity of the bath and also prevents the hydrolysis of the electrolyte.

Electrolytic cell - Iron crucible or glass voltameter

Temperature - Room temperature

Current - 12 volts or 3 Ampere

Electrodes

• Cathode - Copper
• Anode - Copper or platinum

1. Dissociation of ${\mathrm{CuSO}}_4$

• Copper sulfate is dissociating into copper ions as positive ions and sulfate ions as negative ions

${\mathrm{CuSO}}_4\rightleftharpoons \mathbf{}{\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}+\mathbf{}{{\mathbf{SO}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$

2. Dissociation of ${\mathrm{H}}_2\mathrm{O}$

• The water is dissociated into hydrogen ions as positive ions and hydroxyl ions as negative ions

${\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathbf{H}}^{\mathbf{+}}\mathbf{}+\mathbf{}{\mathbf{OH}}^{\mathbf{-}}$

3. Electrode reaction

• The reduction half-reaction occurring at the cathode is given in this reaction
• Both ${\mathrm{Cu}}^{+2}$ and ${\mathrm{H}}^{+}$ ions migrate to the cathode and ${\mathrm{Cu}}^{+2}$ ions are discharged at the cathode as neutral atoms by accepting electrons. product at the cathode is a reddish-brown copper deposit

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

4. Electrode reaction

• 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 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.