To find the balanced redox reaction, we need to balance the given half-reactions for both the reduction and oxidation processes and then combine them.
Step-by-step solution:
1. Identify the given half-reactions:
- Oxidation half-reaction:
[tex]\[
2 Cl^- (aq) \rightarrow Cl_2 (g) + 2 e^-
\][/tex]
- Reduction half-reaction:
[tex]\[
Cr^{3+} (aq) + 3 e^- \rightarrow Cr (s)
\][/tex]
2. Balance the electrons in each half-reaction:
- The oxidation half-reaction produces 2 electrons. To balance this with the reduction half-reaction (which requires 3 electrons), we need a common multiple of electrons. Therefore:
- Multiply the oxidation half-reaction by 3:
[tex]\[
3 \times \left(2 Cl^- (aq) \rightarrow Cl_2 (g) + 2 e^- \right)
\][/tex]
This gives us:
[tex]\[
6 Cl^- (aq) \rightarrow 3 Cl_2 (g) + 6 e^-
\][/tex]
- Multiply the reduction half-reaction by 2:
[tex]\[
2 \times \left(Cr^{3+} (aq) + 3 e^- \rightarrow Cr (s)\right)
\][/tex]
This gives us:
[tex]\[
2 Cr^{3+} (aq) + 6 e^- \rightarrow 2 Cr (s)
\][/tex]
3. Combine the balanced half-reactions:
- When we add these two balanced half-reactions, the electrons on both sides will cancel out:
[tex]\[
\left(6 Cl^- (aq) \rightarrow 3 Cl_2 (g) + 6 e^- \right) + \left(2 Cr^{3+} (aq) + 6 e^- \rightarrow 2 Cr (s)\right)
\][/tex]
This results in:
[tex]\[
2 Cr^{3+} (aq) + 6 Cl^- (aq) \rightarrow 2 Cr (s) + 3 Cl_2 (g)
\][/tex]
Thus, the final balanced equation for the given redox reaction is:
[tex]\[
2 Cr^{3+}(aq) + 6 Cl^{-}(aq) \rightarrow 2 Cr(s) + 3 Cl_2(g)
\][/tex]
