Question

Which of the following(s) is/are true for electrolysis of dilute $LiOH$ solution using platinum electrodes?

• A. $H_2$ is evolved at anode
• B. $H_2$ is evolved at cathode
• C. $O_2$ is produced at anode
• D. $Li$ metal is obtained at cathode

Answer 1

Answer: (B), (C)

$O_2$ is produced at anode

On electrolysis of aqueous $LiOH$ solution, $H^{+}$ is reduce to $H_2$ and gets liberated at cathode.

$L{i}^{+}\left(aq\right)+e^{-}\rightleftharpoons Li\left(s\right)$
$E_{L{i}^{+}/Li}^0=-3.04V$

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

This is because of the higher reduction potential of $H^{+}$ than $L{i}^{+}$.
Cell reaction:

$\text{At cathode:}2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$
$\text{At anode:}2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}\left(aq\right)+O_2\left(g\right)+4e^{-}$
$E_{H_{2}O/O_2}^0=-1.23V$