Question

During the electrolysis of fused NaCl at anode
(A) $C{{l}^{-}}$ Ions are oxidised
(B) $C{{l}^{-}}$ Ions are reduced
(C) $N{{a}^{-}}$ Ions are oxidised
(D) $N{{a}^{+}}$ Ions are reduced

Answer 1

The electrolysis of fused NaCl is used for the production of metallic Na. Oxidation takes place at anode and reduction takes place at cathode.

Complete Solution:
The electrolysis of molten or fused NaCl is done above the temperature of 801$^{o}C$. Electric current is allowed to pass through the molten salt when the two electrodes are introduced. Then the chemical reaction takes place at the electrodes.
- The solution of NaCl or sodium chloride contains $N{{a}^{+}}$ and $C{{l}^{-}}$ ions but besides these ions there is also ${{H}^{+}}$ and $O{{H}^{-}}$ ions due to the ionization of water. But if we compare the number of ions of water molecules to the number of ions of NaCl, the number of ions of water molecules is small because water is a weak electrolyte. When electric current is passed through the solution a potential difference is established across the two electrodes, $N{{a}^{+}}$ and ${{H}^{+}}$ ions moves towards the cathode and $C{{l}^{-}}$ and $O{{H}^{-}}$ ions moves towards the anode.
- At cathode, reduction of $N{{a}^{+}}$ ion takes place and reduces $N{{a}^{+}}$ to sodium metal.
Reaction at cathode, sodium ion is reduced to sodium metal-
$N{{a}^{+}}+{{e}^{-}}\to Na$

At cathode, oxidation of $C{{l}^{-}}$ ion takes place and oxidizes $C{{l}^{-}}$ to chlorine gas.
Reaction at anode production of chlorine gas,
$C{{l}^{-}}\to \dfrac{1}{2}C{{l}_{2}}+{{e}^{-}}$

So, the correct answer is “Option A”.

Note: Pay attention to the question whether molten or aqueous NaCl is mentioned because for the electrolysis of Aqueous NaCl, Hydrogen gas is produced instead or sodium and reduction of water takes place at cathode.