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Valency and Oxidation State

A 20.0 mL sol...

Question

$A 20.0 m L$ solution containing $0.2 g$ impure $H_{2} O_{2}$ reacts completely with $0.316 g$ of $K M n O_{4}$ in acid solution. The purity of $H_{2} O_{2}$ (in%) is (Nearest integer) $(mol.wt.of H_{2} O_{2} = 34 mole.wt.of.KM n O_{4} = 158)$

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Solution

Step 1:
The redox change are:
$M n^{7 +} + 5 e^{-} \to M n^{2 +}$
$O_{2}^{-} \to O_{2}^{0} + 2 e [E o_{2} = \frac{M}{2}]$
$∵ M e q . of H_{2} O_{2} = M e q . of K M n O_{4}$
Step 2:

$\frac{w \times 10000}{\frac{34}{2}} = \frac{0.316}{\frac{158}{5}} \times 1000$
$∴^{W} H_{2} O_{2} = 0.17 g$
Step 3:
$∴$ % purity of sample $H_{2} O_{2} = \frac{017}{0.2} \times 100 = 85 %$

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