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 because:

• 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 dissolved in $Pt$
• D. Concentration of $H^{+}$ ions is larger when $Pt$ electrode is taken

Answer 1

Answer: (B)

More voltage is required to reduce $H^{+}$ at $Hg$ than at $Pt$

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