Electrolysis of X gives Y at anode. Vacuum di | vedclass

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(C) $30\%$ solution of hydrogen peroxide can be obtained by the electrolysis of $50\%$ sulphuric acid followed by vacuum distillation. The first produ

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(C) $30\%$ 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 $(H_2S_2O_8)$,which reacts with water during distillation to form $H_2O_2$. $2H_2SO_4 \longrightarrow 2H^+ + 2HSO_4^-$ $2HSO_4^- \longrightarrow H_2S_2O_8 + 2e^-$ (At anode) $H_2S_2O_8 + 2H_2O \longrightarrow 2H_2SO_4 + H_2O_2$ Here,$X$ is $H_2SO_4$ and $Y$ is $H_2S_2O_8$. $H_2SO_4$ has $0$ peroxy bonds,while $H_2S_2O_8$ (Marshall's acid) has $1$ peroxy bond $(HO_3S-O-O-SO_3H)$.

Electrolysis of $X$ gives $Y$ at anode. Vacuum distillation of $Y$ gives $H_2O_2$. The number of peroxy $(O-O)$ bonds present in $X$ and $Y$ respectively are :

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