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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. \(2Cr^{3+}(aq) + 6Cl^{-}(aq) \longrightarrow 2Cr(s) + 3Cl_2(g)\)

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

C. \(Cr^{3+}(aq) + BCl^{-}(aq) + 3e^{-} \longrightarrow 2Cr(s) + 3Cl_2(g)\)

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

Sagot :

GinnyAnswer
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]
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