Question

In alkaline medium, $Cl{O}_2$ oxidizes $H_2{O}_2$ in $O_2$ and reduce itself in $C{l}^{-}$. Then how many moles of $H_2{O}_2$ will be oxidized by one mole of $Cl{O}_2$?

• A. 1.5
• B. 2.5
• C. 3.5
• D. 5

Answer 1

The correct option is B 2.5

$Cl{O}_2\to C{l}^{-}$
$Cl{O}_2+2H_{2}O+5e^{-}\to C{l}^{-}+4O{H}^{-}$

$H_2{O}_2\to O_2$
$H_2{O}_2+2O{H}^{-}\to O_2+2H_{2}O+2e^{-}$

$Cl{O}_2+2H_{2}O+5e^{-}\to C{l}^{-}+4O{H}^{-}\times 2$
$H_2{O}_2+2O{H}^{-}\to O_2+2H_{2}O+2e^{-}\times 2$
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$2Cl{O}_2+5H_2{O}_2+2O{H}^{-}\to 2C{l}^{-}+5O_2+5H_{2}O$

$2Cl{O}_2\equiv 5H_2{O}_2$
$\therefore Cl{O}_2\equiv 2.5H_2{O}_2$