In electrolysis of `NaCl` when `Pt` electrode is taken `H_(2)` is liberated at cathode while `Hg` cathode it forms sodium amalgam because

Options

A. `Hg` is more inert than `Pt`
B. more voltage is required to reduce `H^(+)` at Hg than at `Pt`
C. `Na` is dissovled in `Hg` while it does not dissolved in `Pt`
D. concentration of `H^(+)` ion is larger when `Pt` electrode is taken

Answer

B

Explanation

(b) Sodium Chloride in water dissociates as
`NaCl iff Na^(+) +Cl^(-)`
`H_(2)O iff H^(+) +OH^(-)`
When electric current is passed through this solution using platinum eletrodes , `Na^(+) and H^(+)` move towards cathode whereas `Cl^(-) and OH^(-)` ions move towards anode.
At cathode
`H^(+)+e^(-) to H`
`H+H to H_(2)`
At anode
`Cl^(-) to Cl+e^(-)`
`Cl+Cl to Cl_(2)`
If mercury is used as cathode, `H^(+)` ions are not discharged at mercury cathode because mercury has high hydrogen over voltage. `Na^(+)` ions are discharged at cathode in preference of `H^(+)` ions yielding sodium, which dissolves in mercury to form sodium amalgum.