For the following balanced redox reaction, `2MnO_(4)^(-) + 4H^(+) + Br_(2) hArr 2Mn^(2+) + 2BrO_(3)^(-) + 2H_(2)O`. If the molecular weight of `MnO_(4)^(-)` and `Br_(2)` are x & y respectively then

Options

A. Equivalent weight of `MnO_(4)^(-)` is `(x)/(5)`
B. Equivalent weight of `Br_(2) ` is `(y)/(5)`
C. Equivalent weight of `Br_(2)` is `(y)/(10)`
D. n-factor ratio of `MnO_(4)^(-)` and `Br_(2)` is 2:1

Answer

To solve the problem, we need to determine the equivalent weights of the species involved in the given redox reaction:

**Reaction:**
\[ 2 \text{MnO}_4^{-} + 4 \text{H}^{+} + \text{Br}_2 \rightleftharpoons 2 \text{Mn}^{2+} + 2 \text{BrO}_3^{-} + 2 \text{H}_2\text{O} \]

### Step 1: Determine the oxidation states
1. **For Mn in MnO4^-:**
- Let the oxidation state of Mn be \( x \).
- The oxidation state of oxygen is \(-2\).
- The equation becomes:
\[ x + 4(-2) = -1 \]
\[ x - 8 = -1 \]
\[ x = +7 \]

2. **For Mn in Mn2+:**
- The oxidation state is \( +2 \).

3. **For Br in Br2:**
- The oxidation state of Br in Br2 is \( 0 \).

4. **For Br in BrO3^-:**
- Let the oxidation state of Br be \( y \).
- The equation becomes:
\[ y + 3(-2) = -1 \]
\[ y - 6 = -1 \]
\[ y = +5 \]

### Step 2: Calculate the change in oxidation states (n-factor)
1. **For MnO4^-:**
- Change in oxidation state: \( 7 \to 2 \)
- Change = \( 7 - 2 = 5 \)
- Therefore, n-factor for MnO4^- = 5.

2. **For Br2:**
- Change in oxidation state: \( 0 \to 5 \)
- Change = \( 5 - 0 = 5 \)
- Since there are 2 Br atoms in Br2, the total change is \( 2 \times 5 = 10 \).
- Therefore, n-factor for Br2 = 10.

### Step 3: Calculate equivalent weights
1. **Equivalent weight of MnO4^-:**
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{x}{5} \]
Thus, equivalent weight of MnO4^- = \( \frac{x}{5} \).

2. **Equivalent weight of Br2:**
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{y}{10} \]
Thus, equivalent weight of Br2 = \( \frac{y}{10} \).

### Conclusion
- The equivalent weight of MnO4^- is \( \frac{x}{5} \).
- The equivalent weight of Br2 is \( \frac{y}{10} \).