Number of coulombs required for the reduction of \(1 \text{ mol}\) of \(MnO_4^-\) to \(Mn^{2+}\) is :

\(4.825 × 10^5 C\)

The correct answer is (2) \(4.825 × 10^5 C\).

To calculate the number of coulombs required for the reduction of \(1 \text{ mol}\) of \(MnO_4^-\) to \(Mn^{2+}\), we need to consider the change in the number of electrons involved in the reduction reaction.
The reduction of \(MnO_4^-\) to \(Mn^{2+}\) can be represented by the following half-reaction:
\[MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\]
In this reaction, 5 moles of electrons (\(5e^-\)) are involved in the reduction of 1 mole of \(MnO_4^-\) to \(Mn^{2+}\).
Now, we know that 1 mole of electrons is equivalent to 1 faraday (F), and 1 faraday is equal to approximately \(96,485 \text{ C}\) (coulombs).
So, for 5 moles of electrons, we have:
\[5 \text{ moles of electrons} = 5 \text{ F} \times 96,485 \text{ C/F} = 482,425 \text{ C}\]
Therefore, the number of coulombs required for the reduction of \(1 \text{ mol}\) of \(MnO_4^-\) to \(Mn^{2+}\) is approximately \(4.82425 \times 10^5 \text{ C}\).
So, the correct option is: (2) \(4.825 \times 10^5 \text{ C}\)