Number of Faradays required to reduce `3 mol` of `MnO_(4)^(c-)` to `Mn^(2+)` is `……………………….F`

Explanation

To determine the number of Faradays required to reduce 3 moles of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \), we can follow these steps:

### Step 1: Write the half-reaction for the reduction of \( \text{MnO}_4^{-} \)

The reduction of permanganate ion \( \text{MnO}_4^{-} \) to manganese ion \( \text{Mn}^{2+} \) can be represented by the following half-reaction:

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

### Step 2: Determine the number of electrons transferred

From the half-reaction, we see that 1 mole of \( \text{MnO}_4^{-} \) requires 5 moles of electrons (5 Faradays) to be reduced to \( \text{Mn}^{2+} \).

### Step 3: Calculate the total number of electrons for 3 moles

Since we need to reduce 3 moles of \( \text{MnO}_4^{-} \), we multiply the number of electrons required for 1 mole by 3:

\[ \text{Total electrons} = 3 \text{ moles} \times 5 \text{ moles of e}^- = 15 \text{ moles of e}^- \]

### Step 4: Convert moles of electrons to Faradays

Since 1 mole of electrons corresponds to 1 Faraday, the total number of Faradays required for 15 moles of electrons is:

\[ \text{Total Faradays} = 15 \text{ Faradays} \]

### Final Answer

Thus, the number of Faradays required to reduce 3 moles of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \) is:

\[ \boxed{15 \text{ F}} \]

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