For the reaction
`3Br_(2)+6OH^(ө)-ltBr^(ө)+BrO_(3)^(ө)+3H_(2)O`
Equivalent weight of `Br_(2)` (molecular weitht M) is

Options

A. `(M)/(2)`
B. `(M)/(10)`
C. `((M)/(2)+(M)/(10))`
D. `((M)/(6))`

Explanation

To find the equivalent weight of \( Br_2 \) in the given reaction:

### Step 1: Write the balanced reaction
The balanced reaction is:
\[ 3Br_2 + 6OH^- \rightarrow 5Br^- + BrO_3^- + 3H_2O \]

### Step 2: Identify the changes in oxidation states
In this reaction:
- \( Br_2 \) is being reduced to \( Br^- \) (oxidation state changes from 0 to -1).
- \( Br_2 \) is also being oxidized to \( BrO_3^- \) (oxidation state changes from 0 to +5).

### Step 3: Calculate the change in oxidation states
- For the reduction to \( Br^- \):
- The change in oxidation state (\( \Delta R \)) for \( Br_2 \) to \( Br^- \) is \( 2 \) (since each \( Br \) atom goes from 0 to -1).
- For the oxidation to \( BrO_3^- \):
- The change in oxidation state (\( \Delta R \)) for \( Br_2 \) to \( BrO_3^- \) is \( 5 \) (since each \( Br \) atom goes from 0 to +5).

### Step 4: Determine the total change in oxidation states
- Since there are 3 moles of \( Br_2 \):
- Total change for reduction: \( 3 \times 2 = 6 \)
- Total change for oxidation: \( 3 \times 5 = 15 \)

### Step 5: Calculate the total change in equivalents
- The total change in equivalents for \( Br_2 \) is:
\[ \text{Total change} = 6 + 15 = 21 \]

### Step 6: Relate the total change to equivalent weight
- The equivalent weight (EW) is given by:
\[ EW = \frac{\text{Molecular weight (M)}}{\text{Total change in equivalents}} \]
Thus, we have:
\[ EW = \frac{M}{21} \]

### Conclusion
The equivalent weight of \( Br_2 \) is:
\[ \text{Equivalent weight of } Br_2 = \frac{M}{21} \]

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