How many Faradays are required to reduce `1 mol of BrO_(3)^(c-)` to `Br^(c-)` in basic medium ?
1) 6
2) 5
3) 4
4) 3

Options

A. 6
B. 5
C. 4
D. 3

Explanation

To determine how many Faradays are required to reduce 1 mole of \( \text{BrO}_3^{-} \) to \( \text{Br}^{-} \) in a basic medium, we will follow these steps:

### Step 1: Write the half-reaction
The reduction half-reaction for the conversion of bromate ion \( \text{BrO}_3^{-} \) to bromide ion \( \text{Br}^{-} \) can be written as:
\[ \text{BrO}_3^{-} \rightarrow \text{Br}^{-} \]

### Step 2: Balance the oxygen atoms
In basic medium, we balance oxygen by adding hydroxide ions \( \text{OH}^{-} \) and water \( \text{H}_2\text{O} \). The bromate ion has 3 oxygen atoms, and we need to balance these by adding 6 hydroxide ions:
\[ \text{BrO}_3^{-} + 6 \text{OH}^{-} \rightarrow \text{Br}^{-} + 3 \text{H}_2\text{O} \]

### Step 3: Balance the hydrogen atoms
Now, we have 6 hydroxide ions on the left side, which gives us 6 hydrogen atoms. On the right side, we have 3 water molecules, contributing 6 hydrogen atoms. Thus, the hydrogen atoms are balanced.

### Step 4: Balance the charges
Next, we need to balance the charges. The left side has a charge of:
\[ -1 + 6(-1) = -1 - 6 = -7 \]
The right side has a charge of:
\[ -1 + 0 = -1 \]
To balance the charges, we need to add 6 electrons to the left side:
\[ \text{BrO}_3^{-} + 6 \text{OH}^{-} + 6 e^{-} \rightarrow \text{Br}^{-} + 3 \text{H}_2\text{O} \]

### Step 5: Final balanced half-reaction
The final balanced half-reaction is:
\[ \text{BrO}_3^{-} + 6 e^{-} + 6 \text{OH}^{-} \rightarrow \text{Br}^{-} + 3 \text{H}_2\text{O} \]

### Step 6: Determine the number of Faradays
Since 6 electrons are required to reduce 1 mole of \( \text{BrO}_3^{-} \) to \( \text{Br}^{-} \), it means that 6 Faradays are required for this reduction process.

### Conclusion
Thus, the number of Faradays required to reduce 1 mole of \( \text{BrO}_3^{-} \) to \( \text{Br}^{-} \) in basic medium is **6**.