In electrolysis of $\mathrm{NaCl}$ when $\mathrm{P} t$ electrode is taken then $\mathrm{H}_2$ is liberated…

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:
  1. $\mathrm{Hg}$ is more inert than $\mathrm{P} t$
  2. more voltage is required to reduce $\mathrm{H}^{+}$at $\mathrm{Hg}$ than at $\mathrm{P} t$
  3. $\mathrm{Na}$ is dissolved in $\mathrm{Hg}$ while it does not dissolve in $\mathrm{P} t$
  4. concentration of $\mathrm{H}^{+}$ions is larger when $\mathrm{P} t$ electrode is taken

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

Asked in: NEET 2002

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