Question

During the qualitative analysis of \( SO_{3}^{-2}\) using dilute H₂SO₄, SO₂ gas evolved which turns K₂Cr₂O₇ solution (acidified H₂SO₄)

• A. Green
• B. Black
• C. Blue
• D. Red

Hint:

In qualitative analysis, dichromate ions (Cr₂O₇^2–) are commonly used as oxidizing agents. Reduction of these ions often results in a color change, which can help identify the reducing species.

Answer 1

The Correct Option is A

The given question involves the qualitative analysis of sulfite ion (\( SO_{3}^{2-} \)) with dilute sulfuric acid (\( H_{2}SO_{4} \)), resulting in the evolution of sulfur dioxide (\( SO_{2} \)) gas. The task is to identify the color change observed when this \( SO_{2} \) gas interacts with acidified potassium dichromate (\( K_{2}Cr_{2}O_{7} \)) solution.

Let's go through the step-by-step reasoning for this chemical reaction:

1. When sulfite ions (\( SO_{3}^{2-} \)) react with dilute sulfuric acid, sulfur dioxide (\( SO_{2} \)) gas is produced.
2. Sulfur dioxide (\( SO_{2} \)) is a reducing agent. In an acidic medium, it reduces orange potassium dichromate (\( K_{2}Cr_{2}O_{7} \)) to green chromium sulfate (\( Cr_{2}(SO_{4})_{3} \)).
3. The dichromate ion (\( Cr_{2}O_{7}^{2-} \)) is reduced to chromium ion (\( Cr^{3+} \)), and the solution changes color from orange to green.
4. The balanced chemical equation for the reduction of dichromate by sulfur dioxide in acidic solution is:

K_{2}Cr_{2}O_{7} + 3SO_{2} + H_{2}SO_{4} \rightarrow Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + H_{2}O

Thus, the correct answer is that the solution turns green.

Let's examine the options to rule out others:

• Green: Correct, as explained by the reduction process of the dichromate ion.
• Black: Incorrect, as there is no reaction leading to the formation of a black substance in this context.
• Blue: Incorrect, since no blue compound is formed during this reaction.
• Red: Incorrect, because the reaction does not produce or change into any red-colored species.

Therefore, the correct answer is "Green".