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In alkaline medium, ${ClO}_{2}$ oxidises $H_{2} O_{2}$ to $O_{2}$ and itself reduced to ${Cl}^{-}$ . How many moles of $H_{2} O_{2}$ are oxidised by 1 $mole$ of ${ClO}_{2} ?$

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$1.5$

$3.5$

$2.5$

$1.0$

answer is C.

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Detailed Solution

${ClO}_{2} ⟶ {Cl}^{-}$

${ClO}_{2} + 4 H_{2} O ⟶ {Cl}^{-} + 2 H_{2} O + 4 {OH}^{-}$

${ClO}_{2} + 2 H_{2} O + 5 e^{-} ⟶ {Cl}^{-} + 4 {OH}^{-}$ …(1)

(b) $H_{2} O_{2} ⟶ O_{2}$

$H_{2} O_{2} + 2 {OH}^{-} ⟶ O_{2} + 2 H_{2} O + 2 e^{-}$ …(2)

Thus : ${ClO}_{2} + 2 H_{2} O + 5 e^{-} ⟶ {Cl}^{-} + 4 {OH}^{-} \times 2$

$H_{2} O_{2} + 2 {OH}^{-} ⟶ O_{2} + 2 H_{2} O + 2 e^{-} \times 5$

On addition,

$2 {ClO}_{2} + 5 H_{2} O_{2} + 2 {OH}^{-} ⟶ 2 {Cl}^{-} + 5 O_{2} + 6 H_{2} O$

$2 mol 5 mol$

$∴ 2 mol {ClO}_{2} = 5 mol H_{2} O_{2}$

$1 mol {ClO}_{2} = \frac{5}{2}$ or $2.5 {molH}_{2} O_{2}$

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