Question

Answer the following in brief.

Write electrode reactions for the electrolysis of aqueous NaCl.

Write gases liberated at cathode and anode during electrolysis of aqueous NaCl solution. 

Answer 1

Reduction half-reaction at cathode:

At cathode, two reduction reactions compete.

1. i. Reduction of sodium ions.
   \[\ce{Na{^+_{(aq)}} + e- -> Na_s, E^\circ = 2.17 V}\]
2. Reduction of water to hydrogen gas.
   

   \[\ce{2H2O_{(l)} + 2e- -> H2_{(g)} + 2OH{^{-}_{{(aq)}}}, E^\circ = - 0.83 V}\]

The standard potential for the reduction of water is higher than that for the reduction of Na⁺. Hence, water has a much greater tendency to get reduced than the Na⁺ ion. Therefore, reduction of water is the cathode reaction when the aqueous NaCl is electrolysed.

Oxidation half-reaction at anode: 

At anode, there will be competition between oxidation of Cl^– ion to Cl₂ gas as in the case of molten NaCl and the oxidation of water to O₂ gas.

1. Oxidation of Cl^- ions to chlorine gas
   

   \[\ce{2Cl{^-_{(aq)}} -> Cl2_{(g)} + 2e-, E^\circ_{oxd} = -1.36 V}\]

2. Oxidation of water to oxygen gas.
   

   \[\ce{2H2O_{(l)} -> O2_{(g)} + 4H{^+_{(aq)}} + 2e-, E^\circ_{oxd} = -0.4 V}\]

The standard electrode potential for the oxidation of water is greater than that of Cl^– ion or water has a greater tendency to undergo oxidation. Hence, an anode half-reaction would be oxidation of water. However, experiments have shown that the gas produced at the anode is Cl₂ and not O₂. This suggests that anode reaction is oxidation of Cl^– to Cl₂ gas. This is because of overvoltage.

Net cell reaction:

The net cell reaction is the sum of two electrode reactions.

Oxidation half reaction at anode:

\[\ce{2Cl{^-_{(aq)}} -> Cl2_{(g)} + 2e-}\]

Reduction half reaction at cathode:

\[\ce{2H2O_{(l)} + 2e- -> H2_{(g)} + 2OH{^-_{(aq)}}}\]

`2"H"_2"O"_(("l")) + 2 "e"^(-) -> "H"_(2("g")) + 2"OH"_(("aq"))^-`    

Overall cell reaction:

\[\ce{2Cl{^-_{(aq)}} + 2H2O_{(l)} -> Cl2_{(g)} + H2_{(g)} + 2OH{^-_{(aq)}}}\]