The information below describes a redox reaction.

[tex]\[
\begin{array}{l}
Cr^{3+}(aq) + 2Cl^{-}(aq) \longrightarrow Cr(s) + Cl_2(g) \\
2Cl^{-}(aq) \longrightarrow Cl_2(g) + 2e^{-} \\
Cr^{3+}(aq) + 3e^{-} \longrightarrow Cr(s)
\end{array}
\][/tex]

What is the final, balanced equation for this reaction?

A. [tex]\(2Cr^{3+}(aq) + 6Cl^{-}(aq) \longrightarrow 2Cr(s) + 3Cl_2(g)\)[/tex]

B. [tex]\(2Cr^{3+}(2aq) + 2Cl^{-}(aq) + 6e^{-} \longrightarrow Cl_2(g) + 2Cr(s)\)[/tex]

C. [tex]\(Cr^{3+}(aq) + 6Cl^{-}(aq) + 3e^{-} \longrightarrow 2Cr(s) + 3Cl_2(g)\)[/tex]

D. [tex]\(Cr^{3+}(aq) + 2Cl^{-}(aq) \longrightarrow Cr(s) + Cl_2(g)\)[/tex]



Answer :

To find the final, balanced equation for the given redox reaction, we begin by separately considering the oxidation and reduction half-reactions.

First, we identify the oxidation half-reaction and the reduction half-reaction:

1. Oxidation Half-Reaction:
[tex]\[ 2 \text{Cl}^- ( \text{aq} ) \rightarrow \text{Cl}_2 ( \text{g} ) + 2 \text{e}^- \][/tex]

2. Reduction Half-Reaction:
[tex]\[ \text{Cr}^{3+} ( \text{aq} ) + 3 \text{e}^- \rightarrow \text{Cr} ( \text{s} ) \][/tex]

Next, we need to balance the electrons in both half-reactions to combine them properly. The oxidation half-reaction produces 2 electrons, whereas the reduction half-reaction consumes 3 electrons. To balance the number of electrons, we need a common multiple of 2 and 3, which is 6. This means:

- Multiply the oxidation half-reaction by 3:
[tex]\[ 3 \times ( 2 \text{Cl}^- ( \text{aq} ) \rightarrow \text{Cl}_2 ( \text{g} ) + 2 \text{e}^- ) \][/tex]
[tex]\[ 6 \text{Cl}^- ( \text{aq} ) \rightarrow 3 \text{Cl}_2 ( \text{g} ) + 6 \text{e}^- \][/tex]

- Multiply the reduction half-reaction by 2:
[tex]\[ 2 \times ( \text{Cr}^{3+} ( \text{aq} ) + 3 \text{e}^- \rightarrow \text{Cr} ( \text{s} ) ) \][/tex]
[tex]\[ 2 \text{Cr}^{3+} ( \text{aq} ) + 6 \text{e}^- \rightarrow 2 \text{Cr} ( \text{s} ) \][/tex]

Now, combine the balanced half-reactions:

[tex]\[ 6 \text{Cl}^- ( \text{aq} ) \rightarrow 3 \text{Cl}_2 ( \text{g} ) + 6 \text{e}^- \][/tex]
[tex]\[ 2 \text{Cr}^{3+} ( \text{aq} ) + 6 \text{e}^- \rightarrow 2 \text{Cr} ( \text{s} ) \][/tex]

When these are added together, the electrons cancel out:

[tex]\[ 2 \text{Cr}^{3+} ( \text{aq} ) + 6 \text{Cl}^- ( \text{aq} ) \rightarrow 2 \text{Cr} ( \text{s} ) + 3 \text{Cl}_2 ( \text{g} ) \][/tex]

Thus, the final balanced equation for the redox reaction is:
[tex]\[ \boxed{2 \text{Cr}^{3+} (\text{aq}) + 6 \text{Cl}^- (\text{aq}) \longrightarrow 2 \text{Cr} (\text{s}) + 3 \text{Cl}_2 (\text{g})} \][/tex]