Question

\(K_2Cr_2O_7\) paper acidified with dil. \(H_2SO_4\) turns green when exposed to :

• A. Sulphur dioxide
• B. Sulphur trioxide
• C. Carbon dioxide
• D. Hydrogen sulphide

Hint:

The reduction of \(K_2Cr_2O_7\) by SO2 produces Cr³⁺ ions, which impart a green colour to the solution.

Answer 1

The Correct Option is A

To determine why K_2Cr_2O_7 paper acidified with dilute H_2SO_4 turns green when exposed to sulphur dioxide, we need to understand the basic chemistry involved in the redox reactions.

Step 1: Understanding the reagents

K_2Cr_2O_7 (Potassium dichromate) is an oxidizing agent. In an acidic solution, it can be reduced to chromic ion, Cr^{3+}, which has a green color.

Step 2: Reaction with Sulphur Dioxide

When sulphur dioxide (SO_2) comes in contact with K_2Cr_2O_7 in acidic medium, it gets oxidized to sulphuric acid (H_2SO_4) while the dichromate is reduced to the chromic ion:

The balanced chemical equation for this redox reaction is:

K_2Cr_2O_7 + 3SO_2 + 4H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 4H_2O

Here, Cr_2(SO_4)_3 contains Cr^{3+} ions which are green in color. This is why the paper turns green.

Step 3: Evaluating Other Options

• Sulphur trioxide (SO_3) does not have a reducing effect on Cr_2O_7^{2-} ions, hence it does not cause the color change.
• Carbon dioxide (CO_2) is not a reducing agent and does not affect the chromate ions in this context.
• Hydrogen sulphide (H_2S) can reduce chromate, but it typically results in a different color reaction due to the production of sulfur or other compounds.

Therefore, the correct answer is sulphur dioxide, which effectively reduces K_2Cr_2O_7 to the green chromic ion. This reflects the option "Sulphur dioxide" as the answer.