Question

During the qualitative analysis of $SO _3^2$ using dilute $H _2 SO _4, SO _2$ gas is evolved which turns $K _2 Cr _2 O _7$ solution (acidified with dilute $H _2 SO _4$ ) :

• A. red
• B. blue
• C. green
• D. black

Answer 1

The Correct Option is C

To solve this question, we need to understand the chemical reaction taking place when sulfur dioxide \((SO_2)\) is evolved and comes in contact with an acidified solution of potassium dichromate \((K_2Cr_2O_7)\). The qualitative analysis described involves a redox reaction. Let's break it down step by step:

1. Chemical Reaction Involved: When \(SO_3^{2-}\) ions react with dilute sulfuric acid \((H_2SO_4)\), sulfur dioxide gas \((SO_2)\) is evolved according to the following equation: \(SO_3^{2-} + 2H^+ \rightarrow SO_2 + H_2O\)
2. Interaction with Acidified Potassium Dichromate: The evolved \(SO_2\) is a reducing agent. When it is passed through an acidified solution of \((K_2Cr_2O_7)\), it reduces the dichromate ions \((Cr_2O_7^{2-})\) to chromium \((Cr^{3+})\) ions. The color change indicates this reaction. The balanced chemical equation is: \(Cr_2O_7^{2-} + 3SO_2 + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O\)
3. Color Change Observation: In the above reaction, the \(Cr_2O_7^{2-}\) ions, which are orange in the solution, get reduced to \(Cr^{3+}\) ions, which impart a green color. Hence, the solution turns green.

Conclusion: Among the given options, the correct color change observed when sulfur dioxide reduces potassium dichromate is green.