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

[tex]\[
2 Br^{-} \longrightarrow Br_{2} + 2 e^{-}
\][/tex]
[tex]\[
Cl_{2} + 2 e^{-} \longrightarrow 2 Cl^{-}
\][/tex]

What should he write as the final, balanced equation?

A. \( Cl_{2} + 2 Br^{-} \longrightarrow Br_{2} + 2 Cl^{-} \)

B. \( Cl_{2} + 2 Br^{-} + 2 e^{-} \longrightarrow Br_{2} + 2 Cl^{-} \)

C. \( Cl_{2} + 2 Br^{-} \longrightarrow Br_{2} + 2 Cl^{-} + 2 e^{-} \)

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

Sagot :

GinnyAnswer
To solve this problem using the half-reaction method, we need to follow these steps:

1. Determine the half-reactions:
- Oxidation half-reaction (loss of electrons):
[tex]\[ 2 Br^{-} \longrightarrow Br_{2} + 2 e^{-} \][/tex]
- Reduction half-reaction (gain of electrons):
[tex]\[ Cl_{2} + 2 e^{-} \longrightarrow 2 Cl^{-} \][/tex]

2. Balance the electrons in each half-reaction:
- In the given half-reactions, you can see that both have 2 electrons involved, so they are already balanced in terms of electron transfer.

3. Combine the half-reactions:
- The electrons lost in the oxidation half-reaction should equal the electrons gained in the reduction half-reaction. Hence, when combining, the electrons cancel each other out.
[tex]\[ Cl_{2} + 2 e^{-} + 2 Br^{-} \longrightarrow Br_{2} + 2 Cl^{-} + 2 e^{-} \][/tex]
- Simplify by canceling the electrons on both sides:
[tex]\[ Cl_{2} + 2 Br^{-} \longrightarrow Br_{2} + 2 Cl^{-} \][/tex]

This final equation is balanced with respect to both mass and charge.

The correct balanced equation is:
[tex]\[ Cl_{2} + 2 Br^{-} \longrightarrow Br_{2} + 2 Cl^{-} \][/tex]

Therefore, the student should write the final balanced equation as:
[tex]\[ Cl_{2} + 2 Br^{-} \longrightarrow Br_{2} + 2 Cl^{-} \][/tex]
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