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Oxidation Number Method

In alkaline m...

Question

In alkaline medium, $C l O_{2}$ oxidises $H_{2} O_{2}$ to $O_{2}$ and is itself reduced to ${C l}^{⊖}$ . How many moles of $H_{2} O_{2}$ are oxidised by 1 mol of $C l O_{2}$ ?

1

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\frac{3}{2}

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\frac{5}{2}

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\frac{7}{2}

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Solution

The correct option is B $\frac{5}{2}$
The half cell reaction is given below:
$5 e^{-} + C l O_{2} ⟶ {C l}^{⊖}$
Eq. of $C l O_{2} =$ Eq. of $H_{2} O_{2}$
$(n = 5)$ $(n = 2)$
Therefore, $\frac{1}{5}$ mol $\equiv \frac{1}{2}$ mol.
or, 1 mol of $C l O_{2} \equiv \frac{5}{2}$ mol of $H_{2} O_{2}$ .

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