In electrolysis of NaCl when P t electrode… - NEET UG 2002 Chemistry PYQ

NEET UG 2002 — Chemistry Physical Chemistry

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In electrolysis of $\mathrm{NaCl}$ when $\mathrm{P} t$ electrode is taken then $\mathrm{H}_2$ is liberated at cathode while with $\mathrm{Hg}$ cathode it forms sodium amalgam because:

Official previous-year question

Held on 30 Apr 2002 · Verified 9 Jul 2026.

Options

A
$\mathrm{Hg}$ is more inert than $\mathrm{P} t$
B
more voltage is required to reduce $\mathrm{H}^{+}$ at $\mathrm{Hg}$ than at $\mathrm{P} t$
C
$\mathrm{Na}$ is dissolved in $\mathrm{Hg}$ while it does not dissolve in $\mathrm{P} t$
D
concentration of $\mathrm{H}^{+}$ ions is larger when $\mathrm{P} t$ electrode is taken

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Solution

Sodium chloride in water dissociates as: $\begin{aligned} \mathrm{NaCl} & \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} \\ \mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{H}^{+} \mathrm{OH}^{-} \end{aligned}$ When electric current is passed through this solution using platinum electrodes. $\mathrm{Na}^{+}$ and $\mathrm{H}^{+}$ move towards cathode whereas $\mathrm{Cl}^{-}$ and $\mathrm{OH}^{-}$ ions move towards anode. At cathode $\begin{aligned} \mathrm{H}^{+}+e^{-} & \rightarrow \mathrm{H} \\ \mathrm{H}+\mathrm{H} & \rightarrow \mathrm{H}_2 \end{aligned}$ An anode $\begin{aligned} \mathrm{Cl}^{-} & \rightarrow \mathrm{Cl}+e^{-} \\ \mathrm{Cl}+\mathrm{Cl} & \rightarrow \mathrm{Cl}_2 \end{aligned}$ If mercury is used as cathode $\mathrm{H}^{+}$ ions are not discharged at mercury cathode because mercury has a high hydrogen over voltage. $\mathrm{Na}^{+}$ ions are dischaged at cathode in preference of $\mathrm{H}^{+}$ ions yielding sodium, which dissolves in mercury to form amalgam

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