Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Balancing Redox Reactions in Acidic Solutions**
**Problem:**
Balance the following redox reaction in acidic solution:
\[ \text{ClO}_3^- (\text{aq}) + \text{Cl}^- (\text{aq}) \rightarrow \text{Cl}_2 (\text{aq}) \]
**Explanation:**
In redox reactions, the number of electrons lost in oxidation must equal the number of electrons gained in reduction. Here’s how to balance this reaction in an acidic solution:
1. **Separate the reaction into two half-reactions:**
- **Reduction Half-Reaction:**
\[ \text{ClO}_3^- \rightarrow \text{Cl}_2 \]
- **Oxidation Half-Reaction:**
\[ \text{Cl}^- \rightarrow \text{Cl}_2 \]
2. **Balance each half-reaction for mass and charge:**
- **Reduction Half-Reaction:**
Identify the oxidation state changes and balance electrons.
\[ \text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Cl}_2 + 3\text{H}_2\text{O} \]
- **Oxidation Half-Reaction:**
\[ 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \]
3. **Combine the balanced half-reactions:**
Ensure that the electrons cancel each other out. Multiply the oxidation half-reaction by 3 and add to the reduction half-reaction:
\[
\begin{align*}
&\text{Reduction: } \text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Cl}_2 + 3\text{H}_2\text{O} \\
&\text{Oxidation: } 6\text{Cl}^- \rightarrow 3\text{Cl}_2 + 6\text{e}^- \\
&\text{Overall: } \text{ClO}_3^- + 6\text{Cl}^- + 6\text{H}^+ \rightarrow 3\text{Cl}_2 +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3b4aa68-1696-4204-931e-d52a90d9be65%2Fb0363761-7d60-42d3-a86b-bfd091c14856%2Fzaxjhop_processed.png&w=3840&q=75)
Transcribed Image Text:**Balancing Redox Reactions in Acidic Solutions**
**Problem:**
Balance the following redox reaction in acidic solution:
\[ \text{ClO}_3^- (\text{aq}) + \text{Cl}^- (\text{aq}) \rightarrow \text{Cl}_2 (\text{aq}) \]
**Explanation:**
In redox reactions, the number of electrons lost in oxidation must equal the number of electrons gained in reduction. Here’s how to balance this reaction in an acidic solution:
1. **Separate the reaction into two half-reactions:**
- **Reduction Half-Reaction:**
\[ \text{ClO}_3^- \rightarrow \text{Cl}_2 \]
- **Oxidation Half-Reaction:**
\[ \text{Cl}^- \rightarrow \text{Cl}_2 \]
2. **Balance each half-reaction for mass and charge:**
- **Reduction Half-Reaction:**
Identify the oxidation state changes and balance electrons.
\[ \text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Cl}_2 + 3\text{H}_2\text{O} \]
- **Oxidation Half-Reaction:**
\[ 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \]
3. **Combine the balanced half-reactions:**
Ensure that the electrons cancel each other out. Multiply the oxidation half-reaction by 3 and add to the reduction half-reaction:
\[
\begin{align*}
&\text{Reduction: } \text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Cl}_2 + 3\text{H}_2\text{O} \\
&\text{Oxidation: } 6\text{Cl}^- \rightarrow 3\text{Cl}_2 + 6\text{e}^- \\
&\text{Overall: } \text{ClO}_3^- + 6\text{Cl}^- + 6\text{H}^+ \rightarrow 3\text{Cl}_2 +
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