In alkaline medium, ${\mathrm{ClO}}_2$ oxidises $H_2{O}_2$ to $O_2$ and itself reduced to ${\text{Cl}}^{-}$. How many moles of ${\mathrm{H}}_2{\mathrm{O}}_2$ are oxidised by 1 $\mathrm{mole}$ of ${\text{ClO}}_2 ?$

${\mathrm{ClO}}_2⟶{\mathrm{Cl}}^{-}$

${\mathrm{ClO}}_2+4{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{Cl}}^{-}+2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{OH}}^{-}$

 ${\mathrm{ClO}}_2+2{\mathrm{H}}_2\mathrm{O}+5{\mathrm{e}}^{-}⟶{\mathrm{Cl}}^{-}+4{\mathrm{OH}}^{-}$   …(1)

(b)      ${\mathrm{H}}_2{\mathrm{O}}_2⟶{\mathrm{O}}_2$

${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{OH}}^{-}⟶{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}$    …(2)

Thus : ${\mathrm{ClO}}_2+2{\mathrm{H}}_2\mathrm{O}+5{\mathrm{e}}^{-}⟶{\mathrm{Cl}}^{-}+4{\mathrm{OH}}^{-}\times 2$

       ${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{OH}}^{-}⟶{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\times 5$

On addition,

 $2{\mathrm{ClO}}_2+5{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{OH}}^{-}⟶2{\mathrm{Cl}}^{-}+5{\mathrm{O}}_2+6{\mathrm{H}}_2\mathrm{O}$

$2\mathrm{mol}5\mathrm{mol}$

$\therefore 2\mathrm{mol} {\mathrm{ClO}}_2=5\mathrm{mol} {\mathrm{H}}_2{\mathrm{O}}_2$

$1\mathrm{mol} {\mathrm{ClO}}_2=\frac{5}{2}$ or $2.5 {\mathrm{molH}}_2{\mathrm{O}}_2$