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 that Pt.
B. More voltage is required to reduce `H^(+)` at Hg then at Pt
C. Na is dissolved is Hg while it does not dissolve in Pt
D. Concentration of `H^(+)` ions is larger when Pt electrode is taken.

Answer

B

Explanation

When NaCl is dissolved in water, it ionise as
`NaCl Water also dissociate as:
`H_(2)O `H^(+)` ions gain electron and change into neutral atoms due to discharged potential
At cathode:
`H^(+) + e^(-) to H`
`H + H to H_(2)`
At anode:
`Cl^(+) + e^(-) to H`
`H + H to H_(2)`
At anode:
`Cl + e to Cl`
`Cl + Cl to Cl_(2)`
It mercury is used as cathode. `H^(+)` ions will not discharged mercury because mercury has a high voltage than hydrogen.