The quantity of charge required to obtain 2 mol of \(Mn^{2+}\) from \(MnO_4^-\) is:

10 F

The correct answer is option 2. 10 F.

To solve this problem, we need to determine the balanced chemical equation for the reduction of \(MnO_4^-\) to \(Mn^{2+}\), and then calculate the quantity of charge required.

The balanced chemical equation for the reduction of \(MnO_4^-\) to \(Mn^{2+}\) in acidic solution is:

\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

In this reaction, 5 moles of electrons (\(5\text{e}^-\)) are required to produce 1 mole of \(Mn^{2+}\).

Given that we want to obtain 2 moles of \(Mn^{2+}\), we will need \(5 \times 2 = 10\) moles of electrons.

Now, we know that 1 mole of electrons corresponds to 1 Faraday (F). Therefore, 10 moles of electrons will correspond to 10 Faradays.

So, the correct answer is option (2) 10 F.