Determine the equivalent weight of the following oxidising and reducing agents :
(a) `KMnO_(4)` (reacting in acidic medium `MnO_(4)^(-)rarrMn^(2+)`)
(b) `KMnO_(4)` (reacting in neutral medium `MnO_(4)^(-)rarr MnO_(2)`)

Answer

To determine the equivalent weight of potassium permanganate (KMnO₄) in both acidic and neutral mediums, we will follow these steps:

### Step 1: Understand the Reactions
- In acidic medium, KMnO₄ is reduced from MnO₄⁻ to Mn²⁺.
- In neutral medium, KMnO₄ is reduced from MnO₄⁻ to MnO₂.

### Step 2: Calculate the Change in Oxidation Number
#### (a) Acidic Medium Reaction
1. Identify the oxidation states:
- In MnO₄⁻, the oxidation state of Mn is +7.
- In Mn²⁺, the oxidation state of Mn is +2.

2. Calculate the change in oxidation number:
- Change = Final oxidation state - Initial oxidation state
- Change = +2 - (+7) = -5 (The oxidation state decreases by 5).

#### (b) Neutral Medium Reaction
1. Identify the oxidation states:
- In MnO₄⁻, the oxidation state of Mn is +7.
- In MnO₂, the oxidation state of Mn is +4.

2. Calculate the change in oxidation number:
- Change = Final oxidation state - Initial oxidation state
- Change = +4 - (+7) = -3 (The oxidation state decreases by 3).

### Step 3: Calculate the Molecular Weight of KMnO₄
- The molecular weight of KMnO₄ is approximately 158.034 g/mol.

### Step 4: Calculate the Equivalent Weight
The equivalent weight is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Change in Oxidation Number}} \]

#### (a) Acidic Medium
1. Using the change in oxidation number of -5:
\[ \text{Equivalent Weight} = \frac{158.034}{5} = 31.6068 \, \text{g/equiv} \]

#### (b) Neutral Medium
1. Using the change in oxidation number of -3:
\[ \text{Equivalent Weight} = \frac{158.034}{3} = 52.678 \, \text{g/equiv} \]

### Final Answers
- (a) Equivalent weight of KMnO₄ in acidic medium: **31.6068 g/equiv**
- (b) Equivalent weight of KMnO₄ in neutral medium: **52.678 g/equiv**

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