20 " mL of " `(M)/(60)` `KBrO_3` was reacted with a sample of `SeO_(3)^(2-)` 20 The `Br_(2)` thus evolved was removed and the excess of `NaAsO_(2)` The reaction involved are
`SeO_(3)^(2-)+BrO_(3)^(ɵ)+H^(o+)toSeO_(4)^(2-)+Br_(2)+H_(2)O` .. (i)
`BrO_(3)^(ɵ)+ASO_(2)^(ɵ)+H_(2)OtoBr^(ɵ)+AsO_(4)^(3-)+H^(o+)` ..(ii)
`[Mw(SeO_(3)^(2-))=79+48=127gmol^(-1)]`
Q. n-factors of `BrO_(3)^(ɵ)` ion in equations (i) and (ii) respectively are

Options

A. 10,6
B. 5,6
C. 6,10
D. 6,5

Answer

To find the n-factors of the \( \text{BrO}_3^{-} \) ion in the given reactions, we need to analyze each reaction step by step.

### Step 1: Analyze the first reaction
The first reaction is:
\[ \text{SeO}_3^{2-} + \text{BrO}_3^{-} + \text{H}^{+} \rightarrow \text{SeO}_4^{2-} + \text{Br}_2 + \text{H}_2\text{O} \]

**Oxidation states:**
- In \( \text{BrO}_3^{-} \), the oxidation state of Br is +5.
- In \( \text{Br}_2 \), the oxidation state of Br is 0.

**Change in oxidation state:**
- The change in oxidation state for Br is from +5 to 0, which involves a change of 5 units.

**Electrons involved:**
- To convert 1 mole of \( \text{BrO}_3^{-} \) to \( \text{Br}_2 \), we need to balance the reaction. Since 2 moles of \( \text{BrO}_3^{-} \) produce 2 moles of \( \text{Br}_2 \), we need 10 electrons (5 for each Br). Thus, for 1 mole of \( \text{BrO}_3^{-} \), it will be:
\[ \text{n-factor} = \frac{10 \text{ electrons}}{2} = 5 \]

### Step 2: Analyze the second reaction
The second reaction is:
\[ \text{BrO}_3^{-} + \text{AsO}_2^{3-} + \text{H}_2\text{O} \rightarrow \text{Br}^{-} + \text{AsO}_4^{3-} + \text{H}^{+} \]

**Oxidation states:**
- In \( \text{BrO}_3^{-} \), the oxidation state of Br is +5.
- In \( \text{Br}^{-} \), the oxidation state of Br is -1.

**Change in oxidation state:**
- The change in oxidation state for Br is from +5 to -1, which involves a change of 6 units.

**Electrons involved:**
- To convert 1 mole of \( \text{BrO}_3^{-} \) to \( \text{Br}^{-} \), we need to add 6 electrons to balance the charge:
\[ \text{n-factor} = 6 \]

### Final Result
- The n-factor of \( \text{BrO}_3^{-} \) in the first reaction is **5**.
- The n-factor of \( \text{BrO}_3^{-} \) in the second reaction is **6**.

### Summary
Thus, the n-factors of \( \text{BrO}_3^{-} \) in equations (i) and (ii) respectively are:
- **5** and **6**.