Answered

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A student is asked to balance an equation using the half-reaction method. He determines the two half-reactions as shown below:

[tex]\[
\begin{array}{l}
2 \text{Br} ^{-} \longrightarrow \text{Br}_2 + 2 e^{-} \\
\text{Cl}_2 + 2 e^{-} \longrightarrow 2 \text{Cl} ^{-}
\end{array}
\][/tex]

What should he write as the final, balanced equation?

A. [tex]\(\text{Cl}_2 + 2 \text{Br} ^{-} \longrightarrow \text{Br}_2 + 2 \text{Cl} ^{-}\)[/tex]

B. [tex]\(\text{Cl}_2 + 2 \text{Br} ^{-} + 2 e^{-} \longrightarrow \text{Br}_2 + 2 \text{Cl} ^{-}\)[/tex]

C. [tex]\(\text{Cl}_2 + 2 \text{Br} ^{-} \longrightarrow \text{Br}_2 + 2 \text{Cl} ^{-} + 2 e^{-}\)[/tex]

D. [tex]\(\text{Cl}_2 + \text{Br} ^{-} + 2 e^{-} \longrightarrow \text{Br}_2 + \text{Cl} ^{-} + 2 e^{-}\)[/tex]

Sagot :

GinnyAnswer
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]
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