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 insert than `Pt`
B. More valtage 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

B

Explanation

`NaCl(aq.) rarr Na^(+)(aq.)+Ci^(-)`
At cathode
`Na^(+)(aq.)+e^(-) rarr Na(s)`
or `2H_(2)O(l) rarr O_(2)(g)+4H^(+)(aq.)+4e^(-)`
When `Pt` electrode is used, `H_(2)O` is reduced as its reduction potential is higher than that of `Na^(+)` ions. However, due to over-voltage for the liberation of `H_(2)(g)` at `Hg` electrode, the `H^(+)` ions arc not discharged at `Hg` cathode. Thus, `Na^(+)` ions are discharged at cathode (in perference of `H^(+)` ions) yielding `Na` metal, which dissolves in `Hg` to form sodium amalgam.