The charge required to convert three moles of $M n_{3} O_{4}$ to $M n O_{4}^{2}^{-}$ in alkaline medium is ______.

10F

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20F

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30F

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40F

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Solution

The correct option is C 30F
Mn goes from an oxidation state of 8/3 to 6.

1 mole of $M n_{3} O_{4}$ contains 3 Mn and thus, the O.S. change comes out to be (3 X 6) - (3 X 8/3), equalling 10.

Therefore, for 3 moles of $M n_{3} O_{4}$ , the total moles of charge required will be $30 \times F$

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M n_{3} O_{4}

M n O_{4}^{-}

The charge required for the oxidation of one mole $M n_{3} O_{4}$ into $M n O {_{4}}^{2 -}$ in presence of alkaline medium is :

M n_{3} O_{4}

M n O_{4}^{2 -}

{M n}_{3} O_{4}

{M n O_{4}}^{2 -}

The charge required for the oxidation of one mole of $M n_{3} O_{4} \to M n O_{4}^{2 -}$ in alkaline medium is: (assume 100% current efficiency)

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