Question
Correct Answer: B
Solution:
Sodium chloride in water dissociates as
[
N a C l rightleftharpoons N a^{+}+C l^{-} quad H_{2} O rightleftharpoons H^{+}+O H^{-}
]
when electric current is passed through this solution using platinum electrode, ( N a^{+} ) and ( H^{+} ) move towards cathode.
Whereas ( C l^{-} ) and ( O H^{-} ) ions move towards anode.
At cathode:
( H^{+}+e^{-} rightarrow H )
[
H+H rightarrow H_{2}
]
At anode:-
[
C l^{-}-e^{-} rightarrow C l
]
( C l+C l rightarrow C l_{2} quad ) If mercury is used as cathode, ( H^{+} )
tons are not discharged at mercury cathode because mercury
has a high hydrogen over voltage. ( N a^{+} ) ions are discharged at
cathode in preference of ( mathrm{H}+ ) ions yielding sodium, which
dissolves in mercury to form sodium amalgam.

ETECLULUIGJ 0 (55) In electrolysis of NaCl when Pt electrode is taken then His liberated at cathode while with Hg cathode it forms sodium amalgam (1) Hg is more inert than Pt (2) More voltage is required to reduce H+ at Hg than at Pt (3) Na is dissolved in Hg while it does not dissolve in Pt (4) Conc. of H+ ions is larger when Pt electrode is taken
Solution
