MCQ Detailed View

This reaction is an important example of redox chemistry in inorganic chemistry. Sulfur dioxide (SO₂) is a well-known reducing agent, while potassium dichromate (K₂Cr₂O₇) is a strong oxidizing agent.

Potassium dichromate normally gives an orange solution due to the presence of Cr⁶⁺ ions. When SO₂ is passed into this solution, a redox reaction occurs:

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  SO₂ gets oxidized to sulfuric acid (H₂SO₄).

  
  


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  Cr⁶⁺ ions in dichromate get reduced to Cr³⁺ ions.

  
  

The Cr³⁺ ions impart a green color to the solution.

The reaction can be represented as:

K2Cr2O7+3SO2+H2SO4    ⟶    Cr2(SO4)3+K2SO4+H2OK_2Cr_2O_7 + 3SO_2 + H_2SO_4 \;\; \longrightarrow \;\; Cr_2(SO_4)_3 + K_2SO_4 + H_2OK2Cr2O7+3SO2+H2SO4⟶Cr2(SO4)3+K2SO4+H2O

Here:

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  The orange color of dichromate disappears.

  
  


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  A green solution of chromium(III) sulfate is produced.

  
  

This test is often used in laboratories as a qualitative reaction to detect sulfur dioxide gas or sulfite ions (which release SO₂ in acidic solution). The color change is distinctive and easily observed, making it a reliable detection method.

It is also a classic demonstration of how the oxidizing ability of dichromate and the reducing ability of SO₂ complement each other in a redox process.

Key Facts:

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  SO₂ → reducing agent.

  
  


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  K₂Cr₂O₇ → oxidizing agent.

  
  


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  Color change: Orange → Green.

  
  


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  Final product: Chromium(III) sulfate solution.

  
  

Thus, when SO₂ solution is passed, the substance that turns green is Potassium dichromate (K₂Cr₂O₇).