MCQ
In alkaline medium $Cl{O_2}$ oxidize ${H_2}{O_2}$ in ${O_2}$ and reduced itself in $C{l^ - }$ then how many mole of ${H_2}{O_2}$ will oxidize by one mole of $Cl{O_2}$
  • A
    $1$
  • B
    $1.5$
  • $2.5$
  • D
    $3.5$

Answer

Correct option: C.
$2.5$
(c) $Cl{O_2} \to C{l^ - }$

$Cl{O_2} + 2{H_2}O + 5{e^ - } \to C{l^ - } + 4O{H^ - }$
${H_2}{O_2} \to {O_2}$
${H_2}{O_2} + 2O{H^ - } \to {O_2} + 2{H_2}O + 2{e^ - }$
$Cl{O_2} + 2{H_2}O + 5{e^ - } \to C{l^ - } + 4O{H^ - }] \times 2$
${H_2}{O_2} + 2O{H^ - } \to {O_2} + 2{H_2}O + 2e] \times 2$

-------------------------------------------------------------------------------------------------
$2Cl{O_2} + 5{H_2}{O_2} + 2O{H^ - } \to 2C{l^ - } + 5{O_2} + 5{H_2}O$
$2Cl{O_2} \equiv 5{H_2}{O_2}$
$\therefore $ $Cl{O_2} = 2.5{H_2}{O_2}$

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