To reduce MnO$_4^{-}$ to Mn$^{2+}$, the oxidation state of manganese changes from +7 in MnO$_4^{-}$ to +2 in Mn$^{2+}$, meaning 5 electrons are required for the reduction of 1 mol of MnO$_4^{-}$ to Mn$^{2+}$.
For 2 mol of Mn$^{2+}$, 5 electrons are required for each mole, so the total number of moles of electrons required is:
\[
\text{Electrons} = 5 \times 2 = 10 \text{ mol of electrons}
\]
To calculate the quantity of charge (Q), we use the relationship:
\[
Q = n \times F
\]
Where:
- \( n \) is the number of moles of electrons (10 mol),
- \( F \) is Faraday’s constant, \( F = 96500 \, \text{C/mol} \).
Thus, the total charge required is:
\[
Q = 10 \, \text{mol} \times 1 \, \text{F} = 10 \, \text{F}
\]
Thus, the correct answer is 10 F.