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Question

Electrolysis of dil. $${\text{H}}_{\text{2}} {\text{SO}}_{\text{4}} $$ liberates gases at anode and cathode respectively


A
O2,SO3
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B
O2,SO2
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C
O2,H2
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D
H2,SO2
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Solution

The correct option is C $$
{\text{O}}_{\text{2}} ,{\text{H}}_{\text{2}}
$$
in electrolysis of dil. $$H_2SO_4$$ we have,
at cathode: $$2H^+ + 2e^- \rightarrow H_2$$
at anode: $$4OH^-  \rightarrow 2H_2O + O_2 + 4e^-$$
because $$OH^-$$ is having lower reduction potential than $$SO_4^{2-}$$.
so, oxidation of $$OH^-$$ ion takes palce.


Chemistry

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