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Consider the following reaction $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O, E^{0} = 1.51 V .$ The quantity of electricity required in Faraday to reduce five moles of $M n O_{4}^{-}$ is ____(Integer answer)

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answer is 25.

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Detailed Solution

For the balanced equation, $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
We can say that 1 mole of $M n O_{4}^{-}$ require 5F electricity. Hence, 5moles of $M n O_{4}^{-}$ will require 25F of electricity.

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