In acidic medium one mole of `MnO_(4)^(-)` accepts how many moles of electrons in a redox process?

Options

A. 1
B. 3
C. 5
D. 6

Answer

To determine how many moles of electrons one mole of \( \text{MnO}_4^- \) accepts in an acidic medium during a redox process, we can follow these steps:

### Step 1: Identify the half-reaction
In acidic medium, the half-reaction for the reduction of permanganate ion (\( \text{MnO}_4^- \)) is:
\[ \text{MnO}_4^- \rightarrow \text{Mn}^{2+} \]

### Step 2: Balance the manganese atoms
In the reaction, we have one manganese atom on both sides, so it is already balanced.

### Step 3: Balance the oxygen atoms
The \( \text{MnO}_4^- \) has 4 oxygen atoms. To balance the oxygen atoms, we add water molecules (\( \text{H}_2\text{O} \)) to the product side:
\[ \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 4: Balance the hydrogen atoms
Now, we have 8 hydrogen atoms on the product side (from 4 water molecules). To balance the hydrogen atoms, we add hydrogen ions (\( \text{H}^+ \)) to the reactant side:
\[ \text{MnO}_4^- + 8 \text{H}^+ \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 5: Balance the charge
Now, we need to balance the charges. The left side has a total charge of \( +7 \) (from \( 8 \text{H}^+ \) and \( -1 \) from \( \text{MnO}_4^- \)), while the right side has a charge of \( +2 \) (from \( \text{Mn}^{2+} \)). To balance the charges, we add 5 electrons (\( 5e^- \)) to the left side:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 e^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Conclusion
From the balanced equation, we can see that 1 mole of \( \text{MnO}_4^- \) accepts 5 moles of electrons in the redox process in acidic medium.

Thus, the answer is:
**5 moles of electrons.**

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