The half-reaction that occurs at the anode during the electrolysis of molten sodium bromide is :

Options

A. `2Br to Br_(2)+2e^(-)`
B. `Br_(2)+2e^(-) to 2Br^(-)`
C. `Na^(+)+e^(-) to Na`
D. `Na to Na^(+)+e^(-)`

Explanation

To determine the half-reaction that occurs at the anode during the electrolysis of molten sodium bromide (NaBr), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components in Molten Sodium Bromide**:
- When sodium bromide (NaBr) is melted, it dissociates into its constituent ions:
- Sodium ions (Na⁺)
- Bromide ions (Br⁻)

2. **Understand the Electrolysis Setup**:
- In an electrolytic cell, there are two electrodes: an anode (positive electrode) and a cathode (negative electrode).
- The anode is where oxidation occurs, and the cathode is where reduction takes place.

3. **Determine Ion Migration**:
- During electrolysis, the positive sodium ions (Na⁺) will migrate towards the cathode, while the negative bromide ions (Br⁻) will migrate towards the anode.

4. **Identify the Reaction at the Anode**:
- At the anode, the bromide ions (Br⁻) will undergo oxidation. Oxidation is defined as the loss of electrons.
- The half-reaction for the oxidation of bromide ions can be written as:
\[ 2 \text{Br}^- \rightarrow \text{Br}_2 + 2 \text{e}^- \]
- This equation shows that two bromide ions lose two electrons to form one molecule of bromine gas (Br₂).

5. **Conclusion**:
- Therefore, the half-reaction that occurs at the anode during the electrolysis of molten sodium bromide is:
\[ 2 \text{Br}^- \rightarrow \text{Br}_2 + 2 \text{e}^- \]