Question

In electrolysis of $NaCl$ when $Pt$ electrode is taken then $H_2$​ is liberated at the 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 dissolve in $Pt$
• D. Concentration of $H^{+}$ ions is larger when $Pt$ electrode is taken
  

Answer 1

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

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 and $C{l}^{-}$ and $O{H}^{–}$ ions move towards anode.
If mercury is used as a cathode, $H^{+}$ ions are not discharged at mercury cathode because mercury has high hydrogen overvoltage. $N{a}^{+}$ ions are discharged at the cathode in preference of $H^{+}$ ions yielding sodium, which dissolves in mercury to form sodium amalgam.