Total charge required for the oxidation of two moles Mn₃O₄ into MnO₄^2- in presence of alkaline medium is :

$\stackrel{+\frac{8}{3}}{{\mathrm{Mn}}_3}{\mathrm{O}}_4 \to \stackrel{+6}{\mathrm{Mn}}{\mathrm{O}}_4^{2-}$

1 mole Mn₃O₄ lose $\left(6-\frac{8}{3}\right)\times 3=10 \mathrm{moles} \mathrm{of} \mathrm{electrons}$.
Total charge required for two moles of Mn₃O₄ = 2 x 10 F $\Rightarrow$ 20 F ;