Electrolysis of $X$ gives $Y$ at anode. Vacuum distillation of $Y$ gives $\mathrm{H}_2 \mathrm{O}_2$. The…
Electrolysis of $X$ gives $Y$ at anode. Vacuum distillation of $Y$ gives $\mathrm{H}_2 \mathrm{O}_2$. The number of peroxy (O-O) bonds present in $X$ and $Y$ respectively are :
1,1
1,2
zero, 1
zero, zero
Solution
A $30 \%$ solution of hydrogen peroxide can be obtained by the electrolysis of $50 \%$ sulphuric acid followed by vacuum distillation. The first product of electrolysis is perdisulphuric acid $\left(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8\right)$ which reacts with water during distillation to form $\mathrm{H}_2 \mathrm{O}_2$.
$2 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{H}^{+}+2 \mathrm{HSO}_4^{-}$
$2 \mathrm{HSO}_4^{-} \longrightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8+2 e^{-}$(At anode)
$\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2$
' $X$ ' is $\mathrm{H}_2 \mathrm{SO}_4$ and ' $Y$ ' is $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$. So ' $X$ ' and ' $Y$ ' contains zero and one peroxy bond respectively.