Equivalent weights of `KMnO_(4)` in acidic medium, alkaline medium and neutral (dilute alkaline) medium respectively are `M/5, M/1 , M/3` . Reduced products can be :

Explanation

To solve the problem regarding the reduced products of KMnO4 in different media, we will analyze the equivalent weights provided and determine the products formed in each case.

### Step-by-Step Solution:

1. **Understanding Equivalent Weight**:
The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \]
where \( n \) is the number of electrons gained or lost in the reaction.

2. **Given Equivalent Weights**:
- In acidic medium: \( \frac{M}{5} \)
- In alkaline medium: \( \frac{M}{1} \)
- In neutral (dilute alkaline) medium: \( \frac{M}{3} \)

3. **Finding the Oxidation State of Manganese in KMnO4**:
The oxidation state of manganese (Mn) in KMnO4 can be calculated as follows:
\[ K + \text{Mn} + 4 \times O = 0 \]
Substituting the known values:
\[ 1 + x + 4 \times (-2) = 0 \implies x - 8 + 1 = 0 \implies x = +7 \]
Thus, the oxidation state of Mn in KMnO4 is +7.

4. **Acidic Medium**:
- Equivalent weight = \( \frac{M}{5} \)
- This implies \( n = 5 \) (number of electrons gained).
- Mn changes from +7 to +2 (gaining 5 electrons).
- The reduced product in acidic medium is \( \text{Mn}^{2+} \).

5. **Alkaline Medium**:
- Equivalent weight = \( \frac{M}{1} \)
- This implies \( n = 1 \) (number of electrons gained).
- Mn changes from +7 to +6 (gaining 1 electron).
- The reduced product in alkaline medium is \( \text{MnO}_4^{2-} \) (manganese in +6 oxidation state).

6. **Neutral (Dilute Alkaline) Medium**:
- Equivalent weight = \( \frac{M}{3} \)
- This implies \( n = 3 \) (number of electrons gained).
- Mn changes from +7 to +4 (gaining 3 electrons).
- The reduced product in neutral medium is \( \text{MnO}_2 \) (manganese in +4 oxidation state).

### Final Answer:
The reduced products of KMnO4 in different media are:
- Acidic medium: \( \text{Mn}^{2+} \)
- Alkaline medium: \( \text{MnO}_4^{2-} \)
- Neutral medium: \( \text{MnO}_2 \)

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