The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is_____.

Answer

To determine the number of electrons involved in the reduction of permanganate (KMnO4) to manganese dioxide (MnO2) in acidic medium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the oxidation states**:
- In KMnO4, the oxidation state of manganese (Mn) is +7. This can be calculated as follows:
- Let the oxidation state of Mn be \( x \).
- The oxidation state of potassium (K) is +1 and that of oxygen (O) is -2.
- The overall charge of the compound is 0, so we set up the equation:
\[ x + 1 + (-2 \times 4) = 0 \]
\[ x + 1 - 8 = 0 \implies x - 7 = 0 \implies x = +7 \]

2. **Determine the oxidation state of manganese in MnO2**:
- In MnO2, the oxidation state of manganese is +4. This can be calculated similarly:
- Let the oxidation state of Mn be \( y \).
- The oxidation state of oxygen is -2.
- The overall charge of the compound is 0, so we set up the equation:
\[ y + (-2 \times 2) = 0 \]
\[ y - 4 = 0 \implies y = +4 \]

3. **Calculate the change in oxidation state**:
- The change in oxidation state from +7 (in KMnO4) to +4 (in MnO2) is:
\[ +7 \to +4 \implies \text{Change} = 7 - 4 = 3 \]

4. **Determine the number of electrons involved**:
- In acidic medium, the reduction of permanganate to manganese dioxide involves a total change of 3 units in oxidation state. Each unit change corresponds to the transfer of one electron. Therefore, the total number of electrons involved in this reduction process is:
\[ 3 \text{ electrons} \]

### Final Answer:
The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is **3**.