Understanding the Reaction Between H₂S and Acidic K₂Cr₂O₇

When hydrogen sulfide (H₂S) reacts with acidified potassium dichromate (K₂Cr₂O₇), it is a redox reaction where H₂S acts as a reducing agent and K₂Cr₂O₇ acts as an oxidizing agent. The dichromate ion (Cr₂O₇²⁻) is reduced, while sulfide (S²⁻) is oxidized.

Step 1: Write the Half-Reactions

Oxidation Half-Reaction (H₂S oxidized to S):

Reduction Half-Reaction (Cr₂O₇²⁻ reduced to Cr³⁺ in acidic medium):

Step 2: Balance the Electrons and Combine the Reactions

To balance electrons, multiply the oxidation half-reaction by 3:

Now, add this to the reduction half-reaction:

Step 3: Write the Complete Molecular Equation

Since the reaction occurs in acidic medium with K₂Cr₂O₇, the complete balanced equation is:

(Note: H₂SO₄ provides the acidic medium and SO₄²⁻ ions)

Step 4: Analyze the Products Formed

From the balanced equation, the products are:

• Potassium sulfate (K₂SO₄)
• Chromium(III) sulfate (Cr₂(SO₄)₃)
• Elemental sulfur (S)
• Water (H₂O)

Final Answer

CrSO₄ is not formed in this reaction. Chromium is reduced to the +3 oxidation state, forming Cr₂(SO₄)₃, not CrSO₄ (which would imply chromium in the +2 oxidation state).

Related Topics

• Redox Reactions
• Oxidizing and Reducing Agents
• Properties of Dichromates
• Reactions of Hydrogen Sulfide

Important Formulae and Theory

Key Theory: In acidic medium, dichromate (Cr₂O₇²⁻) is a strong oxidizing agent. It is reduced to green Cr³⁺ ions. Sulfide ions (S²⁻) are oxidized to elemental sulfur (S). The reaction is often used as a test for sulfide ions, characterized by the formation of a green solution and a yellow precipitate of sulfur.

Common Oxidizing Agents: K₂Cr₂O₇ (in acid), KMnO₄, HNO₃.

Common Reducing Agents: H₂S, SO₂, Fe²⁺, I⁻.