To determine the final balanced equation for the given redox reaction, we need to follow a step-by-step approach to balance both the atoms and the charges.
### Step 1: Write the Half-Reactions
The given redox reaction can be split into two half-reactions:
1. The oxidation half-reaction for chlorine:
[tex]\[
2 Cl^{-}(aq) \longrightarrow Cl_2(g) + 2 e^-
\][/tex]
2. The reduction half-reaction for chromium:
[tex]\[
Cr^{3+}(aq) + 3 e^- \longrightarrow Cr(s)
\][/tex]
### Step 2: Balance the Electrons
To ensure electron balance between the oxidation and reduction half-reactions, we need to match the number of electrons lost and gained in both processes. In the oxidation half-reaction, 2 electrons are released, and in the reduction half-reaction, 3 electrons are gained. To balance the electrons, the common multiple of 2 and 3 is 6. Thus:
- Multiply the chlorine oxidation half-reaction by 3:
[tex]\[
3(2 Cl^{-}(aq) \longrightarrow Cl_2(g) + 2 e^-) \Rightarrow 6 Cl^{-}(aq) \longrightarrow 3 Cl_2(g) + 6 e^-
\][/tex]
- Multiply the chromium reduction half-reaction by 2:
[tex]\[
2(Cr^{3+}(aq) + 3 e^- \longrightarrow Cr(s)) \Rightarrow 2 Cr^{3+}(aq) + 6 e^- \longrightarrow 2 Cr(s)
\][/tex]
### Step 3: Combine the Half-Reactions
Now, we combine the two balanced half-reactions, ensuring to cancel out the electrons:
[tex]\[
6 Cl^{-}(aq) \longrightarrow 3 Cl_2(g) + 6 e^-
\][/tex]
[tex]\[
2 Cr^{3+}(aq) + 6 e^- \longrightarrow 2 Cr(s)
\][/tex]
When combined:
[tex]\[
6 Cl^{-}(aq) + 2 Cr^{3+}(aq) \longrightarrow 3 Cl_2(g) + 2 Cr(s)
\][/tex]
### Step 4: Write the Balanced Equation
The final balanced redox reaction equation becomes:
[tex]\[
2 Cr^{3+}(aq) + 6 Cl^{-}(aq) \longrightarrow 2 Cr(s) + 3 Cl_2(g)
\][/tex]
So, the correct balanced equation for the given redox reaction is:
[tex]\[
2 Cr^{3+}(aq) + 6 Cl^{-}(aq) \longrightarrow 2 Cr(s) + 3 Cl_2(g)
\][/tex]
This confirms that among the provided options, the first one is the correct balanced equation:
[tex]\[2 Cr^{3+}(aq) + 6 Cl^{-}(aq) \longrightarrow 2 Cr(s) + 3 Cl_2(g)\][/tex]
