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)

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

Answer 2

Sodium chloride in water dissociates as
$NaCl \rightleftharpoons Na^++Cl^-$
$H_2O \rightleftharpoons H^++OH^-$
When electric current is passed through this solution using platinum electrodes, $Na^+ \, and \, H^+$ move towards cathode whereas $Cl^- \, and \, OH^-$ ions move towards anode.
At cathode
$H^++e^- \rightarrow H$
$H+H \rightarrow H_2$
At anode
$Cl^- \rightarrow Cl+e^-$
$Cl+Cl \rightarrow Cl_2$
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.