Question

Electrolysis of $NaCl\left(aq\right)$ is carried out using inert electrodes, choose the correct option:

• A. 1 and 2
• B. 2 and 3
• C. 3 and 4
• D. 1 and 4

Answer 1

The correct option is C 3 and 4

Electrolysis of concentrated $NaCl\left(aq\right)$ carried out using inert electrodes:

Since ,
$E_{H_{2}O/H_2}^0>E_{N{a}^{+}/Na}^0$

Reaction at cathode:
$2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$

Reaction at anode:
Even if $E_{C{l}^{-}/C{l}_2}^0<E_{H_{2}O/O_2}^0$
Kinetically rate of oxidation of water is very slow. To increase its rate, the greater potential difference is applied called as overpotential.Because of this overpotential, oxidation of $C{l}^{–}$ ions becomes more feasible
The reaction that occurs at anode is :
$2C{l}^{-}\rightleftharpoons C{l}_2\left(g\right)+2e^{-}$
So, chroine gas is involved at anode.

Electrolysis of dilute $NaCl\left(aq\right)$ carried out using inert electrodes:

Since ,
$E_{H_{2}O/H_2}^0>E_{N{a}^{+}/Na}^0$

Reaction at cathode:
$2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$

Reaction at anode: If concentration of $C{l}^{-}$ is low, oxidation of $H_{2}O$ to give $O_2$ is favoured.

$2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}\left(aq\right)+O_2\left(g\right)+4e^{-}$

When dilute $NaCl$ is used, oxygen gas is liberated at anode.
Hence (c) is correct.