$H_{2} C_{2} O_{4} - ------- - + M n O_{2} + 2 H^{+} ⟶ {M n}^{2 +} + 2 H_{2} O + 2 C O_{2}$
The equivalent weight of the underlined species by taking its molecular weight as $M$ is :

E = \frac{M}{2}

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E = \frac{M}{4}

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E = \frac{M}{3}

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none of these

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Solution

The correct option is A $E = \frac{M}{2}$
The oxidation number of C in $H_{2} C_{2} O_{2}$ is +3.
The oxidation number of C in $C O_{2}$ is +4.
Thus, the oxidation number of one C changes by 1.
Total change in the oxidation number for 2 C atoms is 2.
$Equivalent weight = \frac{Molecular weight}{Change in the oxidation number}$
Hence, $E = \frac{M}{2}$

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