Question

Which of the following(s) not true about the electrolysis of concentrated $NaCl\left(aq\right)$?

• A. When $Pt$ electrode is used, $H_2$ is liberated at cathode
• B. When $Pt$ electrode is used, $Na$ is produced at cathode
• C. When $Hg$ electrode is used, sodium amalgam is produced at cathode
• D. All of the above.

Answer 1

The correct option is B When $Pt$ electrode is used, $Na$ is produced at cathode

Aqueous $NaCl$ will dissociate as $N{a}^{+}$ and $C{l}^{-}$ in the solution.
Ions available in the electrolyte are $N{a}^{+},C{l}^{-},H^{+}\text{and}O{H}^{-}$

(i) Using platinum electrodes:
Platinum electrode is an inert electrode.
This metal electrode doesn’t involve in the redox reaction of the cell.
Hence, this electrode can conduct electrons into or out of the cell but cannot take part in the half-reactions.
Due to higher reduction potential of $H^{+}$ than $N{a}^{+}$, $H^{+}$ get reduced to $H_2$ at cathode.
Also, $C{l}_2$ is liberated at anode since it is a concentrated aqueous $NaCl$ solution.

(ii) Using mercury electrode:
At anode:
When concentrated aqueous $NaCl$ is used $C{l}^{-}$ get oxidised to $C{l}_2$ due to higher concentration of $C{l}^{-}$ ions.
$2C{l}^{-}\left(aq\right)\rightleftharpoons C{l}_2+2e^{-}$

At cathode, we have two possibilities

1. $N{a}^{+}\left(aq\right)+e^{-}\rightleftharpoons Na\left(s\right)$
$E_{N{a}^{+}/Na}^0=-2.71V$

2. $2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$
$E_{H_{2}O/H_2}^0=-0.83V$

$\because$
$E_{H_{2}O/H_2}^0>E_{N{a}^{+}/Na}^0$
$H^{+}$ should get reduced to $H_2$ and liberated at cathode but this is not occuring.
Due to the high solubilty $Na$ in $Hg$, $N{a}^{+}$ is getting reduced to $Na$
Solubility of $Na$ in $Hg$ is a spontaneous reaction.
Hence, reaction at cathode is,
$N{a}^{+}\left(aq\right)+e^{-}\rightleftharpoons Na\left(s\right)Na\left(s\right)+Hg\rightleftharpoons Na-Hg$
Thus, sodium amalgam is formed at cathode.

Hence, statement (b) is wrong.