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The total charge required for the oxidation of two moles $M n_{3} O_{4}$ into $M n {O_{4}}^{2 -}$ in presence of alkaline medium is :

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$5 F$

10 F

20 F

15 F

answer is C.

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Detailed Solution

$\begin{array}{l} M o l e M n_{3} O_{4} l o s e (6 - \frac{8}{3}) \times 3 = 10 m o l e \\ s o t o t a l c h a r g e r e q u i r e d = 2 \times 10 \Rightarrow 20 F \end{array}$

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