Electrolysis of X gives Y at anode. Vacuum distillation of Y gives H_2 O_2 . The number of peroxy (O-O) bonds present in X and Y respectively are

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Electrolysis of X gives Y at anode. Vacuum distillation of Y gives H_2 O_2 . The number of peroxy (O-O) bonds present in X and Y respectively are :

MARKS Content Team · Accepted Answer

Correct answer: (C) zero, 1. 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 (H_2 ~S_2 O_8 ) which reacts with water during distillation to form H_2 O_2 . 2 H_2 SO_4 longrightarrow 2 H^++2 HSO_4^- 2 HSO_4^- longrightarrow H_2 ~S_2 O_8+2 e^- (At anode) H_2 ~S_2 O_8+2 H_2 O longrightarrow 2 H_2 SO_4+H_2 O_2 ' X ' is H_2 SO_4 and ' Y ' is H_2 ~S_2 O_8 . So ' X ' and ' Y ' contains zero and one peroxy bond respectively.

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 :

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