Question

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

• A. ${\text{O}}_{\text{2}},{\text{SO}}_{\text{3}}$
• B. ${\text{O}}_{\text{2}},{\text{SO}}_{\text{2}}$
• C. ${\text{O}}_{\text{2}},{\text{H}}_{\text{2}}$
• D. ${\text{H}}_{\text{2}}{\text{,SO}}_{\text{2}}$

Answer 1

The correct option is C ${\text{O}}_{\text{2}},{\text{H}}_{\text{2}}$

in electrolysis of dil. $H_{2}S{O}_4$ we have,

at cathode: $2H^{+}+2e^{-}\to H_2$

at anode: $4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$

because $O{H}^{-}$ is having lower reduction potential than $S{O}_4^{2-}$.

so, oxidation of $O{H}^{-}$ ion takes palce.