During the qualitative analysis of SO3^2- using dilute H2 SO4, SO2 gas is evolved which turns K2 Cr2 O7 solution (acidified with dilute H2 SO4):

Given: SO3^2- is treated with dilute H2 SO4 and the evolved SO2 gas is passed through acidified K2 Cr2 O7 solution. Find: The colour produced in the acidified dichromate solution. Apply the principle: SO2 acts as a reducing agent and reduces dichromate ion to Cr^3+. The reaction described in the solution is: Cr2 O7^2- + SO3^2- + H^+ Cr^3+ + SO4^2- Dichromate ion is orange, but on reduction it forms Cr^3+ ions, which are green in colour. Therefore, the solution turns Green, so the correct option is C.

During the qualitative analysis of SO3^2- using dilute H2 SO4, SO… — Solution | JEE Chemistry

MCQEasyJEE 2023Salt Analysis (Cations & Anions)

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

A
Black
B
Red
C
Green
D
Blue

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $\text{SO}_3^{2-}$ is treated with dilute $\text{H}_2\text{SO}_4$ and the evolved $\text{SO}_2$ gas is passed through acidified $\text{K}_2\text{Cr}_2\text{O}_7$ solution.

Find: The colour produced in the acidified dichromate solution.

Apply the principle: $\text{SO}_2$ acts as a reducing agent and reduces dichromate ion to $\text{Cr}^{3+}$ .

The reaction described in the solution is:

\text{Cr}_2\text{O}_7^{2-} + \text{SO}_3^{2-} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{SO}_4^{2-}

Dichromate ion is orange, but on reduction it forms $\text{Cr}^{3+}$ ions, which are green in colour.

Therefore, the solution turns Green, so the correct option is C.

Common mistakes

Confusing the original orange colour of dichromate with the final observed colour. The question asks for the colour after reaction, not before reaction. Track the product formed after reduction.
Assuming $\text{SO}_2$ is only an acidic gas and not a reducing agent. Here $\text{SO}_2$ reduces dichromate to $\text{Cr}^{3+}$ , and that reduction causes the green colour.

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