Question

Identify the correct representation of reaction occurring during the chlor-alkali process.
(A) $2NaC{{l}_{(l)}}+2{{H}_{2}}{{O}_{(l)}}\to 2NaO{{H}_{(l)}}+C{{l}_{2(g)}}+{{H}_{2(g)}}$
(B) $2NaC{{l}_{(s)}}+2{{H}_{2}}{{O}_{(aq)}}\to 2NaO{{H}_{(aq)}}+C{{l}_{2(g)}}+{{H}_{2(g)}}$
(C) $2NaC{{l}_{(aq)}}+2{{H}_{2}}{{O}_{(l)}}\to 2NaO{{H}_{(aq)}}+C{{l}_{2(aq)}}+{{H}_{2(aq)}}$
(D) $2NaC{{l}_{(aq)}}+2{{H}_{2}}{{O}_{(l)}}\to 2NaO{{H}_{(aq)}}+C{{l}_{2(g)}}+{{H}_{2(g)}}$

Answer 1

Hint: 

- When electricity is passed through an aqueous solution of sodium chloride, it decomposes to form sodium hydroxide.
-The process is called chlor-alkali because of the products formed, chlor for chlorine gas and alkali for sodium hydroxide.
-Chlorine gas is liberated at anode and hydrogen gas is liberated at the cathode. Sodium hydroxide solution is formed near the cathode.

Complete step by step solution:
Electrolysis of aqueous sodium chloride solution:
- NaCl decomposes in the presence of water to give sodium cation and chloride anion.
\[NaC{{l}_{(aq)}}\xrightarrow{{{H}_{2}}O}N{{a}^{+}}_{(aq)}+C{{l}^{-}}_{(aq)}\]  (Eq 1)
- ${{H}^{+}}$and $O{{H}^{-}}$ions are produced due to dissociation of ${{H}_{2}}O$
${{H}_{2}}{{O}_{(l)}}\to {{H}^{+}}_{(aq)}+O{{H}^{-}}_{(aq)}$ (Eq 2)
- At cathode there is a competition between the reduction of sodium ion and hydrogen ion.
$N{{a}^{+}}_{(aq)}+{{e}^{-}}\to N{{a}_{(s)}}$ $\implies {{E}^{o}}_{(cell)}$ = -2.71 V
$2{{H}^{+}}_{(aq)}+2{{e}^{-}}\to {{H}_{2(g)}}$ $\implies {{E}^{o}}_{(cell)}$ = 0.00 V (Eq 3)
- The reaction which is having higher ${{E}^{o}}$ is preferred and Hence reduction of hydrogen ion takes place at the cathode.
- Combining reaction (2) and (3), the cathode reaction can be written as
$2{{H}_{2}}{{O}_{(l)}}+2{{e}^{-}}\to {{H}_{2(g)}}+2O{{H}^{-}}_{(aq)}$ (Eq 4)
- At the anode, there is a competition between oxidation of chlorine and hydroxide ion. Even if the ${{E}^{o}}_{(cell)}$ of oxidation of hydroxide ion is greater the oxidation of chloride ion takes place at anode due to overpotential of Oxygen. Hence the reaction at the anode is as follows:
$2C{{l}^{-}}_{(aq)}\to C{{l}_{2(g)}}+2{{e}^{-}}$  (Eq 5)
The net reaction is given by combining (1), (4) and (5)
$2NaC{{l}_{(aq)}}+2{{H}_{2}}{{O}_{(l)}}\to 2NaO{{H}_{(aq)}}+C{{l}_{2(g)}}+{{H}_{2(g)}}$

Hence, the correct option is (D) $2NaC{{l}_{(aq)}}+2{{H}_{2}}{{O}_{(l)}}\to 2NaO{{H}_{(aq)}}+C{{l}_{2(g)}}+{{H}_{2(g)}}$.

Note: In the electrolysis of molten NaCl, the products are sodium metal and chlorine gas. Sodium cation is reduced to sodium atom at cathode and Chlorine anion is oxidized to chlorine at the anode.