Question

What is quantity of electricity in f required to convert 4Br- to 4Br2

Answer 1

To determine the quantity of electricity required to convert \( \text{4Br}^- \) (bromide ions) to \( \text{4Br}_2 \) (bromine gas), we first need to understand the oxidation process involved.

1. **Oxidation Reaction**: The conversion of bromide ions to bromine gas can be represented by the half-reaction:
\[
2 \text{Br}^- \rightarrow \text{Br}_2 + 2 \text{e}^-
\]
This shows that 2 moles of bromide ions lose 2 moles of electrons to form 1 mole of bromine gas.

2. **Scaling the Reaction**: Since we are interested in converting 4 moles of \( \text{Br}^- \) to \( \text{Br}_2 \), we need to scale the half-reaction:
\[
4 \text{Br}^- \rightarrow 2 \text{Br}_2 + 4 \text{e}^-
\]

3. **Calculating Total Charge**: To find the total quantity of electricity in coulombs, we can use the Faraday constant (\( F \)), which is approximately \( 96485 \, \text{C/mol} \). Since we are transferring 4 moles of electrons:
- Total moles of electrons = 4
- Total charge (Q) can be calculated using:
\[
Q = nF
\]
Where:
- \( n = 4 \, \text{mol of electrons} \)
- \( F = 96485 \, \text{C/mol} \)

Therefore, the total quantity of electricity \( Q \):
\[
Q = 4 \, \text{mol} \times 96485 \, \text{C/mol} = 385940 \, \text{C}
\]

Thus, the quantity of electricity required to convert \( 4 \, \text{Br}^- \) to \( 4 \, \text{Br}_2 \) is approximately \( 385940 \, \text{C} \).