The reaction of potassium dichromate (K₂Cr₂O₇) with sulfur dioxide (SO₂) is a redox reaction where potassium dichromate acts as an oxidizing agent and sulfur dioxide acts as a reducing agent. Here's the balanced chemical equation:

K₂Cr₂O₇ + 3SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Explanation:

* Potassium dichromate (K₂Cr₂O₇) gets reduced, with Cr⁶⁺ ions in the dichromate ion (Cr₂O₇²⁻) being reduced to Cr³⁺ ions in chromium(III) sulfate (Cr₂(SO₄)₃).

* Sulfur dioxide (SO₂) gets oxidized, with sulfur in the +4 oxidation state being oxidized to sulfur in the +6 oxidation state in sulfate ions (SO₄²⁻).

Step-by-step breakdown:

1. Oxidation of SO₂:

* SO₂ → SO₄²⁻ + 2e⁻ (Each sulfur atom loses 2 electrons)

2. Reduction of K₂Cr₂O₇:

* Cr₂O₇²⁻ + 6e⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O (Each chromium atom gains 3 electrons)

3. Balancing the electrons:

* Multiply the oxidation half-reaction by 3 to balance the electrons: 3SO₂ → 3SO₄²⁻ + 6e⁻

* Now the electron gain and loss are equal.

4. Combining the half-reactions:

* K₂Cr₂O₇ + 3SO₂ + 14H⁺ → K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O

5. Adding H₂SO₄:

* Since the reaction requires H⁺ ions, we add H₂SO₄ to provide these ions:

* K₂Cr₂O₇ + 3SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Important Note:

* This reaction is often carried out in acidic conditions, which is why H₂SO₄ is added to the reaction.

* The reaction produces a green solution containing chromium(III) sulfate (Cr₂(SO₄)₃), which is a characteristic color change often used to indicate the presence of dichromate.