How many \(C\, \ mol^{-1}\) are required to convert \(1\, \ mol\) of \(MnO_4^-\) to \(Mn^{+2}\)?

[1F = 96500 C]

482500 \(C\, \ mol^{-1}\)

The correct answer is option 4. 482500 \(C\, \ mol^{-1}\).

To solve this problem, we need to determine the number of electrons required to convert \(1\, \text{mol}\) of \(MnO_4^-\) to \(Mn^{+2}\). Then, we'll use Faraday's constant to convert the number of electrons to coulombs.

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

\[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{+2} + 4H_2O \]

From this equation, we can see that \(5\) moles of electrons are required to convert \(1\) mole of \(MnO_4^-\) to \(Mn^{+2}\).

Given that \(1\) Faraday (\(F\)) is equivalent to \(96500\) coulombs (\(C\)):

\[ \text{Number of coulombs required} = \text{Number of moles of electrons} \times F \]

\[ = 5 \times 96500 \]

\[ = 482500 \, C\, \text{mol}^{-1} \]

So, the correct answer is option 4: \(482500 \, C\, \text{mol}^{-1}\).