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...
Read More
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:
SO₂ gets oxidized to sulfuric acid (H₂SO₄).
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:
The orange color of dichromate disappears.
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.
SO₂ → reducing agent.
K₂Cr₂O₇ → oxidizing agent.
Color change: Orange → Green.
Final product: Chromium(III) sulfate solution.
Thus, when SO₂ solution is passed, the substance that turns green is Potassium dichromate (K₂Cr₂O₇).
Discussion
Leave a Comment