Question

How many Faraday charge is required for ${KMnO}_4\to {Mn}^{2+}$?

• A. 3 F
• B. 2 F
• C. 5 F
• D. 1 F

Answer 1

The correct option is C 5 F

$Mn{O}_4^{-}+8H^{+}+5e^{-}⟶M{n}^{2+}+6H_{2}O$

1 mole 5mole 1mole

$M{n}^{7+}+5e^{-}⟶M{n}^{2+}$

Required charge = $n\times F$

Where, $n$ = difference of charge on ions

$F$= constant= $96487C$

Oxidation state of Mn is changing from +7 to +2 therefore 5 moles of electron are needed for reduction of 1 mole of $Mn{O}_4^{-}toM{n}^{2+}$.

5 moles of electrons = $5F$.

Hence,Required charge = $5\times F$

Therefore option C is correct answer.