${H_{2} O}_{2} (a q)$ reduces $C l O_{4}^{-} (a q)$ in acidic medium. If 1 mole of ${H_{2} O}_{2}$ reduces $0.5$ mole of $C l O_{4}^{-} (a q)$ . What is the oxidation number of reduced species?
(Answer is the modulus of the oxidation number)

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Solution

Let the reduced species be $Z$ and its oxidation number be $x$ .
The reaction for reduction of $C l O_{4}^{-}$ by $H_{2} O_{2}$ is as follows:

${H_{2} O}_{2} - 2 + C l O_{4}^{-} + 7 ⟶ O_{2} 0 + r e d u c e d s p e c i e s (Z) O . N = x$

Now balancing the oxidation number,
$(7 - x) {H_{2} O}_{2} + 2 C l O_{4} ⟶ (7 - x) O_{2} + 2 Z$

Hence, $(7 - x)$ moles of ${H_{2} O}_{2}$ will reduce 2 mols of $C l O_{4}^{-}$

1 mole of ${H_{2} O}_{2}$ will reduce $\frac{2}{7 - x}$ moles of $C l O_{4}^{-}$ .
Hence,
$\frac{2}{7 - x} = 0.5 m o l C l O_{4}^{-} (g i v e n)$
$4 = 7 - x$
$x = 3$
Thus, the oxidation number of $C l$ in $Z$ is $= + 3$ , $Z$ is $C l O_{2}$ .
So answer is $| + 3 | = 3$ .

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