Question

In electrolysis of $NaCl$ when $Pt$ electrode is taken then $H_2$ is liberated at cathode while with $Hg$ cathode it forms sodium amalgam :-

• 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

Answer 1

Answer: (B)

Sodium chloride in water dissociates as
$NaCl\rightleftharpoons N{a}^{+}+C{l}^{-}$
$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$
When electric current is passed through this solution using platinum electrodes, $N{a}^{+}$and $H^{+}$ move towards cathode whereas $C{l}^{-}$and $O{H}^{-}$ions move towards anode.
At cathode
$H^{+}+e^{-}⟶H$
$H+H⟶H_2$
At anode
$C{l}^{-}⟶Cl+e^{-}$
$Cl+Cl⟶C{l}_2$
If mercury is used as cathode $H^{+}$ions are not discharged at mercury cathode because mercury has a high hydrogen over voltage. $N{a}^{+}$ions are discharged at cathode in preference of $H^{+}$ions yielding sodium, which dissolves in mercury to form sodium amalgam.