Question

Explain why electrolysis of an aqueous solution of NaCl gives \( H_2 \) gas at cathode and \( Cl_2 \) gas at anode? Write the overall reaction.
(Given: \( E^\circ_{\text{Na}^+/Na} = -2.71 \, \text{V}, \, E^\circ_{\text{H}_2/\text{H}_2O} = -0.83 \, \text{V}, \, E^\circ_{\text{Cl}_2/\text{Cl}^-} = +1.36 \, \text{V} \))

Hint:

The electrolysis of aqueous sodium chloride is a classic example of water and salt splitting, where water is reduced to form hydrogen gas and chloride ions are oxidized to form chlorine gas.

Answer 1

During electrolysis of aqueous NaCl, the reaction at the cathode and anode involves the reduction and oxidation processes respectively. At the cathode, the reduction of \(\text{H}_2\text{O}\) occurs because the reduction potential of \( \text{H}_2/\text{H}_2O \) is higher than that of \( \text{Na}^+/Na \). Hence, the reaction at the cathode is: \[ \text{2H}_2\text{O(l)} + 2e^- \longrightarrow \text{H}_2(g) + 2\text{OH}^-(aq) \] At the anode, the oxidation of chloride ions takes place as chlorine has a more positive electrode potential than hydroxide ions. Hence, the reaction at the anode is: \[ \text{2Cl}^-(aq) \longrightarrow \text{Cl}_2(g) + 2e^- \] The overall reaction is: \[ 2\text{NaCl(aq)} + 2\text{H}_2\text{O(l)} \longrightarrow \text{H}_2(g) + \text{Cl}_2(g) + 2\text{NaOH(aq)} \] Thus, during the electrolysis of NaCl solution, hydrogen gas is liberated at the cathode and chlorine gas at the anode.