Which is the oxidizing agent in the reaction below? → Cr₂O72- + 6 S2₂032 + 14 H+ 2 Cr3+ + 3 S4062 +7 H₂O O Cr₂O7²- O S₂O3²- O S406²- OHt

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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\(_7^{2-}\).
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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