Electrolysis of $X$ gives $Y$ at anode. Vacuum distillation of $Y$ gives ${{\rm{H}}_2}{{\rm{O}}_2}$. The number of peroxy (O - O) bonds present in $X$ and $Y$ respectively are

NEET UG CHEMISTRY MCQ: HYDROGEN

1 Medium

1 of 1

Analyzing +4 | -1

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

View answer and explanation

Correct option: C

Zero, 1

Option A (

2 \mathrm{KBr}+\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{Br}_2+2 \mathrm{KOH}

) is correct because in this reaction,

\mathrm{H}_2 \mathrm{O}_2

acts as an oxidizing agent, not a reductant.

Practice All Chapter MCQs →

\mathrm{CaC}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OH})_2+' \mathrm{X}\,'

\mathrm{Mg}_2 \mathrm{C}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Mg}(\mathrm{OH})_2+'\mathrm{Y}\,'

\mathrm{X}

\mathrm{Y}

{H_2}{O_2} \to 2{H^ + } + {O_2} + 2{e^ - };{E^0} = - 0.68V

{\mathrm{H}_\mathrm{2}}{\mathrm{O}_\mathrm{2}}

{\mathrm{H}_2}{\mathrm{O}_2}

{SiH}_4

AlH_3

{NH}_3