Question

When \(\text{SO}_2\) is passed through acidified \(\text{K}_2\text{Cr}_2\text{O}_7\), the process that takes place is

• A. the solution turns blue
• B. the solution is decolourised
• C. \(\text{SO}_2\) is reduced
• D. green \(\text{Cr}_2(\text{SO}_4)_3\) is formed

Hint:

Reduction of \(\text{Cr}^{6+}\) to \(\text{Cr}^{3+}\) is indicated by orange to green colour change.

Answer 1

The Correct Option is D

Step 1: Identify the nature of reactants.
Acidified potassium dichromate \((\text{K}_2\text{Cr}_2\text{O}_7)\) is a strong oxidising agent, while sulfur dioxide \((\text{SO}_2)\) acts as a reducing agent.
Step 2: Oxidation–reduction process.
In acidic medium, \(\text{Cr}^{6+}\) ions are reduced to \(\text{Cr}^{3+}\), and \(\text{SO}_2\) is oxidised to sulfate ions.
Step 3: Colour change and product formation.
The orange colour of dichromate changes to green due to formation of chromium(III) sulfate: \[ \text{K}_2\text{Cr}_2\text{O}_7 + 3\text{SO}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{Cr}_2(\text{SO}_4)_3 + \cdots \] Step 4: Conclusion.
Hence, green \(\text{Cr}_2(\text{SO}_4)_3\) is formed.