7. Identify the oxidizing agent in the following reaction: pracuollot self to one daid w trabag I₂ + Zn → Z1 + Zn bubong brie 2+ of spredlo palaritas adr Bilt to state sett fo
7. Identify the oxidizing agent in the following reaction: pracuollot self to one daid w trabag I₂ + Zn → Z1 + Zn bubong brie 2+ of spredlo palaritas adr Bilt to state sett fo
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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![**Question 7: Identifying the Oxidizing Agent in a Chemical Reaction**
**Problem Statement:**
Identify the oxidizing agent in the following reaction:
\[ \text{I}_2 + \text{Zn} \rightarrow 2\text{I}^- + \text{Zn}^{2+} \]
**Explanation:**
In this reaction, iodine (\( \text{I}_2 \)) and zinc (\( \text{Zn} \)) are the reactants that undergo a redox (reduction-oxidation) process.
- **Oxidizing Agent:** The substance that accepts electrons and gets reduced in a chemical reaction.
**Steps to Identify the Oxidizing Agent:**
1. **Determine Oxidation States:**
- \( \text{I}_2 \) is in the elemental form, so the oxidation state is 0.
- \( \text{Zn} \) is also in the elemental form with an oxidation state of 0.
- In the product, \( \text{I}^- \) has an oxidation state of -1.
- \( \text{Zn}^{2+} \) has an oxidation state of +2.
2. **Changes in Oxidation State:**
- \( \text{I}_2 \) goes from 0 to -1 (gains electrons, gets reduced).
- \( \text{Zn} \) goes from 0 to +2 (loses electrons, gets oxidized).
3. **Conclusion:**
- Since \( \text{I}_2 \) gains electrons and gets reduced, it acts as the oxidizing agent in this reaction.
This analysis helps in understanding how redox reactions work by identifying the transfer of electrons between substances.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F527d0796-f7b0-4877-ad88-5fb65d7b09e0%2F773de338-27a2-45a0-afc3-b07789cc5502%2Fqxfsu3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 7: Identifying the Oxidizing Agent in a Chemical Reaction**
**Problem Statement:**
Identify the oxidizing agent in the following reaction:
\[ \text{I}_2 + \text{Zn} \rightarrow 2\text{I}^- + \text{Zn}^{2+} \]
**Explanation:**
In this reaction, iodine (\( \text{I}_2 \)) and zinc (\( \text{Zn} \)) are the reactants that undergo a redox (reduction-oxidation) process.
- **Oxidizing Agent:** The substance that accepts electrons and gets reduced in a chemical reaction.
**Steps to Identify the Oxidizing Agent:**
1. **Determine Oxidation States:**
- \( \text{I}_2 \) is in the elemental form, so the oxidation state is 0.
- \( \text{Zn} \) is also in the elemental form with an oxidation state of 0.
- In the product, \( \text{I}^- \) has an oxidation state of -1.
- \( \text{Zn}^{2+} \) has an oxidation state of +2.
2. **Changes in Oxidation State:**
- \( \text{I}_2 \) goes from 0 to -1 (gains electrons, gets reduced).
- \( \text{Zn} \) goes from 0 to +2 (loses electrons, gets oxidized).
3. **Conclusion:**
- Since \( \text{I}_2 \) gains electrons and gets reduced, it acts as the oxidizing agent in this reaction.
This analysis helps in understanding how redox reactions work by identifying the transfer of electrons between substances.
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