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 25**
*Which is the oxidizing agent in the reaction below?*
\[ \text{Cr}_2\text{O}_7^{2-} + 6 \text{S}_2\text{O}_3^{2-} + 14 \text{H}^+ \rightarrow 2 \text{Cr}^{3+} + 3 \text{S}_4\text{O}_6^{2-} + 7 \text{H}_2\text{O} \]
1. ⃝ Cr\(_2\)O\(_7^{2-}\)
2. ⃝ S\(_2\)O\(_3^{2-}\)
3. ⃝ S\(_4\)O\(_6^{2-}\)
4. ⃝ H\(^+\)
**Explanation:**
In the given reaction:
- Cr\(_2\)O\(_7^{2-}\) (dichromate ion)
- S\(_2\)O\(_3^{2-}\) (thiosulfate ion)
- H\(^+\) (proton)
The products are:
- Cr\(^{3+}\) (chromium(III) ion)
- S\(_4\)O\(_6^{2-}\) (tetrathionate ion)
- H\(_2\)O (water)
To identify the oxidizing agent, we need to determine which species is being reduced. The oxidizing agent is the species that gains electrons.
1. Cr\(_2\)O\(_7^{2-}\) is reduced to Cr\(^{3+}\), since chromium goes from an oxidation state of +6 in Cr\(_2\)O\(_7^{2-}\) to +3 in Cr\(^{3+}\).
2. S\(_2\)O\(_3^{2-}\) is oxidized to S\(_4\)O\(_6^{2-}\).
3. H\(^+\) remains as H\(^+\) in this context.
Therefore, the oxidizing agent in this reaction is Cr\(_2\)O\(_7^{2-}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F985fbbf0-f8fa-4c87-a1a5-f926559935c9%2Fa6ffbb2d-8928-4b61-ad06-38a236ba75da%2F1zul6t8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 25**
*Which is the oxidizing agent in the reaction below?*
\[ \text{Cr}_2\text{O}_7^{2-} + 6 \text{S}_2\text{O}_3^{2-} + 14 \text{H}^+ \rightarrow 2 \text{Cr}^{3+} + 3 \text{S}_4\text{O}_6^{2-} + 7 \text{H}_2\text{O} \]
1. ⃝ Cr\(_2\)O\(_7^{2-}\)
2. ⃝ S\(_2\)O\(_3^{2-}\)
3. ⃝ S\(_4\)O\(_6^{2-}\)
4. ⃝ H\(^+\)
**Explanation:**
In the given reaction:
- Cr\(_2\)O\(_7^{2-}\) (dichromate ion)
- S\(_2\)O\(_3^{2-}\) (thiosulfate ion)
- H\(^+\) (proton)
The products are:
- Cr\(^{3+}\) (chromium(III) ion)
- S\(_4\)O\(_6^{2-}\) (tetrathionate ion)
- H\(_2\)O (water)
To identify the oxidizing agent, we need to determine which species is being reduced. The oxidizing agent is the species that gains electrons.
1. Cr\(_2\)O\(_7^{2-}\) is reduced to Cr\(^{3+}\), since chromium goes from an oxidation state of +6 in Cr\(_2\)O\(_7^{2-}\) to +3 in Cr\(^{3+}\).
2. S\(_2\)O\(_3^{2-}\) is oxidized to S\(_4\)O\(_6^{2-}\).
3. H\(^+\) remains as H\(^+\) in this context.
Therefore, the oxidizing agent in this reaction is Cr\(_2\)O\(_7^{2-}\).
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