The Gas That Removes the Color

Oct 13, 2025 by sorusgentrification.com 130 views

**What is the Gas Which Removes the Color of KMnO₄ Solution Acidified with Sulphuric Acid?**

Understanding the Chemistry Behind KMnO₄

KMnO₄, or potassium permanganate, is a strong oxidizing agent commonly used in various chemical reactions. When acidified with sulphuric acid (H₂SO₄), it forms a purple-colored solution. However, this color can be removed by a specific gas. In this article, we will explore the chemistry behind this phenomenon and identify the gas responsible for removing the color of the KMnO₄ solution.

The Role of KMnO₄ in Acidified Sulphuric Acid

KMnO₄ is a powerful oxidizing agent that can donate oxygen atoms to other substances. When it is acidified with sulphuric acid, it forms a purple-colored solution due to the presence of the permanganate ion (MnO₄⁻). This ion is responsible for the characteristic color of the solution.

The Gas That Removes the Color

The gas that removes the color of the KMnO₄ solution acidified with sulphuric acid is a reducing agent that can donate electrons to the permanganate ion, thereby reducing it to a colorless form. Among the options provided, the correct answer is:

a) SO₂ (Sulphur Dioxide)

Sulphur dioxide (SO₂) is a reducing agent that can donate electrons to the permanganate ion, reducing it to a colorless form. When SO₂ is bubbled through the KMnO₄ solution, it reacts with the permanganate ion to form manganese(II) sulfate and sulfuric acid.

The Reaction

The reaction between SO₂ and KMnO₄ can be represented by the following equation:

MnO₄⁻ + 2H⁺ + SO₂ → Mn²⁺ + SO₄²⁻ + H₂O

In this reaction, the permanganate ion (MnO₄⁻) is reduced to manganese(II) ion (Mn²⁺), and the color of the solution is removed.

Other Options

The other options provided, NO₂ (nitrogen dioxide), P₂O₅ (phosphorus pentoxide), and CO₂ (carbon dioxide), are not the correct answers. While these gases can react with KMnO₄, they do not remove the color of the solution.

NO₂ (Nitrogen Dioxide)

Nitrogen dioxide (NO₂) is a strong oxidizing agent that can react with KMnO₄ to form nitrogen oxide and manganese(II) ion. However, this reaction does not remove the color of the solution.

P₂O₅ (Phosphorus Pentoxide)

Phosphorus pentoxide (P₂O₅) is a strong dehydrating agent that can react with KMnO₄ to form phosphoric acid and manganese(II) ion. However, this reaction does not remove the color of the solution.

CO₂ (Carbon Dioxide)

Carbon dioxide (CO₂) is a neutral gas that does not react with KMnO₄ to remove the color of the solution.

Conclusion

In conclusion, the gas that removes the color of KMnO₄ solution acidified with sulphuric acid is SO₂ (sulphur dioxide). This gas is a reducing agent that can donate electrons to the permanganate ion, reducing it to a colorless form. The reaction between SO₂ and KMnO₄ can be represented by the equation: MnO₄⁻ + 2H⁺ + SO₂ → Mn²⁺ + SO₄²⁻ + H₂O.

Frequently Asked Questions

Q: What is the role of KMnO₄ in acidified sulphuric acid? A: KMnO₄ is a powerful oxidizing agent that can donate oxygen atoms to other substances. When it is acidified with sulphuric acid, it forms a purple-colored solution due to the presence of the permanganate ion (MnO₄⁻).

Q: What is the gas that removes the color of the KMnO₄ solution? A: The gas that removes the color of the KMnO₄ solution is SO₂ (sulphur dioxide).

Q: What is the reaction between SO₂ and KMnO₄? A: The reaction between SO₂ and KMnO₄ can be represented by the equation: MnO₄⁻ + 2H⁺ + SO₂ → Mn²⁺ + SO₄²⁻ + H₂O.

Q: Why does SO₂ remove the color of the KMnO₄ solution? A: SO₂ removes the color of the KMnO₄ solution because it is a reducing agent that can donate electrons to the permanganate ion, reducing it to a colorless form.

Q: What are the other options provided, and why are they not correct? A: The other options provided are NO₂ (nitrogen dioxide), P₂O₅ (phosphorus pentoxide), and CO₂ (carbon dioxide). These gases do not remove the color of the KMnO₄ solution because they do not react with the permanganate ion in the same way that SO₂ does.