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. more voltage is required to reduce $H^{+}$ at Hg than at Pt
• B. Na is dissolved in Hg while it does not dissolve in Pt
• C. Hg is more inert than Pt
• D. concentration of $H^{+}$ ions is larger when Pt electrode is taken

Answer 1

Answer: (A)

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.