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Q. Total charge required for the oxidation of two moles $Mn _{3} O _{4}$ into $MnO _{4}^{2}$ in presence of alkaline medium is-

Electrochemistry

Solution:

$16 OH ^{-}+ Mn _{3} O _{4} \rightarrow 3 MnO _{4}^{2-}+8 H _{2} O +10 e ^{-}$

1 mol electron required is 10

$2 mol =2 \times 10=20$

Charge required

$=20 \times F =20 F$