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Difficulty: MediumOxidizing and Reducing Agents and Tests

Fill in the blanks to complete the chemical observation and oxidation state change during the test for sulfur(IV) oxide gas. What are the correct terms to fill in the blanks?

Answer:When SO2SO_2 gas is passed through an acidified solution of potassium heptaoxodichromate(VI), K2Cr2O7K_2Cr_2O_7, the solution turns from orange to 【green】 because the dichromate ion is reduced, changing the oxidation state of chromium from +6+6 to 【+3】.

Answer

The solution turns green because chromium is reduced from an oxidation state of +6 to +3.
Sulfur(IV) oxide (SO2SO_2) is a strong reducing agent. When passed into an acidified solution of potassium heptaoxodichromate(VI), it reduces the orange dichromate ion (Cr2O72−Cr_2O_7^{2-}, where CrCr is in the +6+6 oxidation state) to the green chromium(III) ion (Cr3+Cr^{3+}, where CrCr is in the +3+3 oxidation state). Thus, the color turns green and the final oxidation state is +3+3.

Step-by-Step Solution

1
Identify the role of SO2SO_2 and K2Cr2O7K_2Cr_2O_7 in the redox reaction.
SO2SO_2 acts as a reducing agent and is oxidized to SO42−SO_4^{2-}, while K2Cr2O7K_2Cr_2O_7 acts as an oxidizing agent.
Reducing agents cause the reduction of other species while being oxidized themselves.
2
Determine the color change of the acidified K2Cr2O7K_2Cr_2O_7 solution.
The orange Cr2O72−Cr_2O_7^{2-} ion is reduced to the green Cr3+Cr^{3+} ion.
The formation of hydrated chromium(III) ions in solution imparts a distinct green color.
3
Determine the initial and final oxidation states of chromium.
In Cr2O72−Cr_2O_7^{2-}, chromium has an oxidation state of +6+6. Upon reduction to Cr3+Cr^{3+}, its oxidation state becomes +3+3.
The half-reaction is Cr2O72−+14H++6e−↛2Cr3++7H2OCr_2O_7^{2-} + 14H^+ + 6e^- \nrightarrow 2Cr^{3+} + 7H_2O.

Key Concept

Laboratory identification test for reducing agents using acidified potassium heptaoxodichromate(VI)
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Topics

Oxidizing and Reducing Agents and Tests

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