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 dissolved in Hg while it does not dissolve in Pt
D. `"Conc of" H^(+)` ions is larger when Pt electrode is taken

Answer

B