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]
