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 prod

Vedclass · Accepted Answer

zero,$1$

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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 $H_2S_2O_8$.$H_2SO_4$ (sulphuric acid) contains zero peroxy bonds.$H_2S_2O_8$ (Marshall's acid) contains one peroxy bond $(-O-O-)$.Therefore,the number of peroxy bonds in $X$ and $Y$ are zero and $1$ respectively.

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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