Question

Choose the correct option regarding the substances which are liberated at the cathode and anode during the electrolysis of aqueous NaCl.

• A. Anode: $\text{Na (s)}$ and Cathode: ${\text{H}}_2\text{O (l)}$
• B. Anode: ${\text{H}}_2\text{(s)}$ and Cathode: $\text{Na (s)}$
• C. Anode: ${\text{Cl}}_2\text{(g)}$ and Cathode: ${\text{H}}_2\text{(g)}$
• D. Anode: $\text{Na (s)}$ and Cathode: ${\text{Cl}}_2\text{(g)}$

Answer 1

The correct option is C Anode: ${\text{Cl}}_2\text{(g)}$ and Cathode: ${\text{H}}_2\text{(g)}$

Water plays a role during the electrolysis of $\text{NaCl}$ solution. Since water can be both oxidized and reduced, it competes with the dissolved ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions. Having a greater tendency to get reduced than sodium ions, water gets reduced and produces hydrogen. Reduction takes place at cathode so hydrogen gas is evolved at the cathode. Similarly, ${\text{Cl}}^{-}$ has a greater tendency than water to get oxidized so it gets oxidized and produces chlorine gas. Oxidation takes place at anode so chlorine gas will be produced at the anode. The reactions taking place at the electrodes are as follows:

At Cathode: ${\text{H}}_2\text{O (l)}+2e^{-}\to {\text{H}}_2\text{(g)}+2{\text{OH}}^{-}$

At Anode: $2{\text{Cl}}^{-}\to {\text{Cl}}_2\text{(g)}+2e^{-}$