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Which is the oxidizing agent in the reaction below? Cr₂O72- + 6 S₂O32- + 14 H+ O Cr₂O7²- O S203²- 2- O S406² O Ht - 2 Cr³+ + 3 S406² + 7 H₂O

Which is the oxidizing agent in the reaction below? Cr₂O72- + 6 S₂O32- + 14 H+ O Cr₂O7²- O S203²- 2- O S406² O Ht - 2 Cr³+ + 3 S406² + 7 H₂O

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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--- ## Question 25 ### Which is the oxidizing agent in the reaction below? \[ \text{Cr}_2\text{O}_7^{2-} + 6 \text{S}_2\text{O}_3^{2-} + 14 \text{H}^+ \rightarrow 2 \text{Cr}^{3+} + 3 \text{S}_4\text{O}_6^{2-} + 7 \text{H}_2\text{O} \] - \(\bigcirc\) Cr₂O₇²⁻ - \(\bigcirc\) S₂O₃²⁻ - \(\bigcirc\) S₄O₆²⁻ - \(\bigcirc\) H⁺ --- ### Explanation: In a redox reaction, the oxidizing agent is the substance that gains electrons (is reduced). Here, to identify the oxidizing agent, we examine the changes in oxidation states of the elements in the reactants and products. ### Answer Options and Justifications: - **Cr₂O₇²⁻**: Dichromate ions, \(\text{Cr}_2\text{O}_7^{2-}\), are commonly known oxidizing agents. In this reaction, chromium (Cr) is reduced from +6 oxidation state in \(\text{Cr}_2\text{O}_7^{2-}\) to +3 oxidation state in \(\text{Cr}^{3+}\). - **S₂O₃²⁻**: Thiosulfate ions, \(\text{S}_2\text{O}_3^{2-}\), generally act as reducing agents. In this reaction, sulfur is oxidized from +2 oxidation state in \(\text{S}_2\text{O}_3^{2-}\) to +2.5 in \(\text{S}_4\text{O}_6^{2-}\). - **S₄O₆²⁻**: Tetrathionate ions, \(\text{S}_4\text{O}_6^{2-}\), are products formed in this reaction, hence not relevant as an oxidizing agent. - **H⁺**: Hydrogen ions, \(\text{H}^+\), are neither oxidized nor reduced in this equation;
Transcribed Image Text:--- ## Question 25 ### Which is the oxidizing agent in the reaction below? \[ \text{Cr}_2\text{O}_7^{2-} + 6 \text{S}_2\text{O}_3^{2-} + 14 \text{H}^+ \rightarrow 2 \text{Cr}^{3+} + 3 \text{S}_4\text{O}_6^{2-} + 7 \text{H}_2\text{O} \] - \(\bigcirc\) Cr₂O₇²⁻ - \(\bigcirc\) S₂O₃²⁻ - \(\bigcirc\) S₄O₆²⁻ - \(\bigcirc\) H⁺ --- ### Explanation: In a redox reaction, the oxidizing agent is the substance that gains electrons (is reduced). Here, to identify the oxidizing agent, we examine the changes in oxidation states of the elements in the reactants and products. ### Answer Options and Justifications: - **Cr₂O₇²⁻**: Dichromate ions, \(\text{Cr}_2\text{O}_7^{2-}\), are commonly known oxidizing agents. In this reaction, chromium (Cr) is reduced from +6 oxidation state in \(\text{Cr}_2\text{O}_7^{2-}\) to +3 oxidation state in \(\text{Cr}^{3+}\). - **S₂O₃²⁻**: Thiosulfate ions, \(\text{S}_2\text{O}_3^{2-}\), generally act as reducing agents. In this reaction, sulfur is oxidized from +2 oxidation state in \(\text{S}_2\text{O}_3^{2-}\) to +2.5 in \(\text{S}_4\text{O}_6^{2-}\). - **S₄O₆²⁻**: Tetrathionate ions, \(\text{S}_4\text{O}_6^{2-}\), are products formed in this reaction, hence not relevant as an oxidizing agent. - **H⁺**: Hydrogen ions, \(\text{H}^+\), are neither oxidized nor reduced in this equation;
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