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Redox Reactions

2 mole, equim...

Question

$2$ mole, equimolar mixture of $N a_{2} C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ required $V_{1} L$ of $0.1 M K M n O_{4}$ in acidic medium for complete oxidation. The same amount of the mixture required $V_{2} L$ of $0.2 M N a O H$ for neutralization. The ratio of $V_{1}$ and $V_{2}$ is:

1 : 2

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2 : 1

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4 : 5

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5 : 4

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Solution

The correct option is C $4 : 5$
Solution: (C) $4 : 5$
Case I:
2 mole, equimolar mixture of ${N a}_{2} C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ required $V_{1} L$ of $0.1 M K M n O_{4}$ in acidic medium.
As the mixture is equimolar, 1 mole of each, ${N a}_{2} C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ are present.
$∴ eq. of {N a}_{2} C_{2} O_{4} + eq. of H_{2} C_{2} O_{4} = eq. of K M n O_{4} (e q u i v a l e n t = m o l e \times v a l e n c e f a c t o r}$
$\Rightarrow 1 \times 2 + 1 \times 2 = V_{1} \times 0.1 \times 5 {∵ valence factor = Change in oxidation state per molecule}$
$\Rightarrow V_{1} = 8 L$
Case II:
2 mole, equimolar mixture of ${N a}_{2} C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ required $V_{2} L$ of $0.2 M N a O H$ for neutralization.
$eq. of {N a}_{2} C_{2} O_{4} + eq. of H_{2} C_{2} O_{4} = eq. of N a O H$
$\Rightarrow 1 \times 1 + 1 \times 1 = V_{2} \times 0.2 \times 1$
$\Rightarrow V_{2} = 10 L$
$∴ V_{1} : V_{2} = 8 : 10 = 4 : 5$

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