Which of the following will give $H_{2} (g)$ at cathode and $O_{2} (g)$ at anode on electrolysis using platinum electrodes ?

Concentrated aqueous solution of

N a C l

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Molten

N a C l

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Dilute aqueous solution of

N a C l

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Solid

N a C l

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Solution

The correct option is C Dilute aqueous solution of $N a C l$
When a dilute aqueous solution of $N a C l$ is electrolysed, hydrogen and oxygen gases are released at cathode and anode respectively. This is because of the high reduction potential of $N a^{+}$ as compared to $H^{+}$ and high oxidation potential of $C l^{-}$ as compared to $O H^{-}$ .

$N a C l (A q u e o u s s o l u t i o n) ⟶ N a^{+} + C l^{-}$

$H_{2} O p r e s e n t i n a q u e o u s s o l u t i o n o f N a c l ⟶ H^{+} + O H^{-}$
At anode:
$2 C l^{-} \to C l_{2} + 2 e^{-}; E_{o x i}^{o} = + 1.36 V$
$4 O H^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}; E_{o x i}^{0} = + 1.23 V$
At cathode:
$N a^{+} + e^{-} \to N a; E_{r e d}^{o} = - 2.71 V$
$2 H^{+} + 2 e^{-} \to H_{2}; E_{r e d}^{o} = - 0.83 V$

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Which of the following will give H2(g) at cathode and O2(g) at anode on electrolysis using platinum electrodes ?

Which of the following will give $H_{2} (g)$ at cathode and $O_{2} (g)$ at anode on electrolysis using platinum electrodes ?