To balance the given redox reaction using the half-reaction method, we need to follow these steps:
1. Write the oxidation and reduction half-reactions:
- Oxidation: [tex]\(2 Br^{-} \rightarrow Br_{2} + 2 e^{-}\)[/tex]
- Reduction: [tex]\(Cl_{2} + 2 e^{-} \rightarrow 2 Cl^{-}\)[/tex]
2. Equalize the number of electrons in both half-reactions:
- Both half-reactions already have 2 electrons, so they are balanced in terms of electrons.
3. Add the half-reactions together to cancel out the electrons:
[tex]\[
(2 Br^{-} \rightarrow Br_{2} + 2 e^{-}) + (Cl_{2} + 2 e^{-} \rightarrow 2 Cl^{-})
\][/tex]
When we add these two reactions, the electrons [tex]\(2 e^{-}\)[/tex] on each side will cancel out:
[tex]\[
2 Br^{-} + Cl_{2} + 2 e^{-} \rightarrow Br_{2} + 2 e^{-} + 2 Cl^{-}
\][/tex]
4. Simplify the equation to get the final balanced reaction:
[tex]\[
Cl_{2} + 2 Br^{-} \rightarrow Br_{2} + 2 Cl^{-}
\][/tex]
Thus, the final balanced equation is:
[tex]\[
Cl_{2} + 2 Br^{-} \rightarrow Br_{2} + 2 Cl^{-}
\][/tex]
So, the correct answer is:
[tex]\[ Cl_{2} + 2 Br^{-} \longrightarrow Br_{2} + 2 Cl^{-} \][/tex]
