Which of the following is true for electrolysis of dilute $L i O H$ solution using platinum electrodes?

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Solution

On electrolysis of aqueous $L i O H$ solution, $H^{+}$ is reduce to $H_{2}$ and gets liberated at cathode.

$L i^{+} (a q) + e^{-} ⇌ L i (s)$
$E_{L i^{+} / L i}^{0} = - 3.04 V$

$2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 O H^{-} (a q)$
$E_{H_{2} O / H_{2}}^{0} = - 0.83 V$

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

$At cathode: 2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 O H^{-} (a q)$
$At anode: 2 H_{2} O (l) ⇌ 4 H^{+} (a q) + O_{2} (g) + 4 e^{-}$
$E_{H_{2} O / O_{2}}^{0} = - 1.23 V$

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L i O H

L i O H

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