Question

When a dilute aqueous solution of lithium chloride is electrolysed using graphite electrodes, which of the following is/are true?

• A. both 1 and 2
• B. only 1
• C. only 2
• D. 1, 2, 3, 4

Answer 1

The correct option is A both 1 and 2

In the given electrolysis method, the electrodes used is graphite.
Graphite is an inert/inactive 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.
In such cell reactions, no reactants or products are capable of serving as electrodes.

In general, electrolysis of lithium chloride yield the product of $C{l}_2$ gas at anode because eventhough the oxidation potential of $O{H}^{-}$ in $H_{2}O$ is higher, the rate of oxidation of water is lower than the rate of production of $C{l}_2$.
Thus, to increase the rate of the reaction, an extra potential is applied which is known as overpotential.
Because of this overpotential, oxidation of $C{l}^{-}$ becomes more feasible.
So,
Anode reaction:
$2C{l}^{-}\left(aq\right)\rightleftharpoons C{l}_2\left(g\right)+2e^{-}$

But in dilute $LiCl$, the oxidation of $C{l}^{-}$ does not occurs.
At low concentration of $C{l}^{-}$, oxidation of $H_{2}O$ to give $O_2$ is favoured.
Hence, oxygen is liberated at anode.
Anode reaction:
$2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}\left(aq\right)+O_2\left(g\right)+4e^{-}$

At cathode,
$E_{L{i}^{+}/Li}^0=-3.04V$
$E_{H_{2}O/H_2}^0=-0.83V$

Since, reduction potential of $H^{+}$ is higher than the reduction potential of $L{i}^{+}$. $H_{2}O$ is reduced to give $H_2$ at cathode.
Cathode reaction:
$2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$