In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of $C O_{2}$ is:

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Solution

The correct option is C 1
The balanced reaction of oxalate with permanganate in an acidic medium is

$2 K M n O_{4} + 5 H_{2} C_{2} O_{4} + 3 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C O_{2} + 8 H_{2} O$

The number of electrons involved in producing one molecule of $C O_{2}$ are :

$= \frac{No.of electron involved in balanced reaction}{No.of moles of C O_{2} produced} = \frac{10}{10} = 1$

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