Electrochemistry - Result Question 23
####25. In electrolysis of $NaCl$, when $Pt$ electrode is taken, then $H_2$ is liberated at cathode while with $Hg$ cathode, it forms sodium amalgam. This is because
[2002] (a) $Hg$ is more inert than $Pt$
(b) More voltage is required to reduce $H^{+}$at $Hg$ than at $Pt$
(c) $Na$ is dissolved in $Hg$ while it does not dissolve in $Pt$
(d) Conc. of $H^{+}$ions is larger when Pt electrode is taken
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Solution:
- (b) If mercury is used as cathode, $H^{+}$ions are not discharged at mercury cathode because mercury has a high over voltage for hydrogen.
In electrolysis of $NaCl$, following ionisations take place:
$NaCl \rightarrow Na^{+}+Cl^{-}$
$H_2 O \rightarrow H^{+}+OH^{-}$
$Na^{+}$and $H^{+}$ions move towards cathode.
However, only $H^{+}$ions are discharged more readily than $Na^{+}$ions because in electrochemical series, hydrogen is lower than sodium.
At cathode : $2 H^{+}+2 e^{-} \longrightarrow H_2(g)$
If mercury is used as cathode, then $Na^{+}$ions are discharged at cathode in preference to $H^{+}$ions, yielding sodium, which dissolves in mercury to form sodium amalgam.
At cathode : $Na^{+}+e^{-} \longrightarrow Na(s)$
$ Na(s)+Hg(s) \longrightarrow Na-Hg(s) $
