Answered

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

GinnyAnswer
To determine the final balanced equation for the redox reaction, we need to follow these steps:

### Given Half-Reactions:
1. Oxidation half-reaction:
[tex]\[ 2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2 e^- \][/tex]

2. Reduction half-reaction:
[tex]\[ \text{Cr}^{3+} + 3 e^- \rightarrow \text{Cr} \][/tex]

### Step-by-Step Solution:

#### Step 1: Balance the electrons in both half-reactions
- The first half-reaction (oxidation) involves [tex]\(2 e^-\)[/tex].
- The second half-reaction (reduction) involves [tex]\(3 e^-\)[/tex].

To combine these half-reactions, the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction. Find the least common multiple (LCM) of 2 and 3 to achieve this:

- The LCM of 2 and 3 is 6.

Therefore, we need to balance the half-reactions by multiplying them by appropriate factors to get 6 electrons in each half-reaction.

#### Step 2: Multiply each half-reaction by appropriate factors
- Multiply the oxidation half-reaction by 3:
[tex]\[ 3 \times (2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2 e^-) = 6 \text{Cl}^- \rightarrow 3 \text{Cl}_2 + 6 e^- \][/tex]

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

#### Step 3: Add the balanced half-reactions together
Now add the two half-reactions ensuring the electrons cancel each other out:
[tex]\[ 6 \text{Cl}^- \rightarrow 3 \text{Cl}_2 + 6 e^- \][/tex]
[tex]\[ 2 \text{Cr}^{3+} + 6 e^- \rightarrow 2 \text{Cr} \][/tex]

Combine the equations:
[tex]\[ 2 \text{Cr}^{3+} + 6 \text{Cl}^- \rightarrow 2 \text{Cr} + 3 \text{Cl}_2 \][/tex]

### Step 4: Write the final balanced equation
[tex]\[ 2 \text{Cr}^{3+}(aq) + 6 \text{Cl}^-(aq) \rightarrow 2 \text{Cr}(s) + 3 \text{Cl}_2(g) \][/tex]

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

Thus, the correct choice is:
[tex]\[ 2 \text{Cr}^{3+}(aq) + 6 \text{Cl}^-(aq) \rightarrow 2 \text{Cr}(s) + 3 \text{Cl}_2(g) \][/tex]