K2Cr2O7 + H2S + H2SO4 = S + K2SO4 + Cr2(SO4)3 + H2O

The Potassium Dichromate, Hydrogen Sulfide and Sulfuric acid (K2Cr2O7 H2S H2SO4) are very common chemicals. They react to form sulfur, potassium sulfate, Chromium Sulfate as well as water. Actually K2Cr2O7 and H2S react in acidic environment. On the other hand the reaction is an Oxidation-reduction reaction.

About K2Cr2O7 H2S H2SO4

Potassium dichromate (K2Cr2O7) is a common inorganic chemical reagent, most commonly used as an oxidizing agent in various laboratory. It is a crystalline ionic solid with a very bright, red-orange color. The Potassium Dichromate is very popular among the chemists to determine the unknown concentration of the solution of secondary standard substances. That’s means the Potassium Dichromate is a primary standard substance.

Hydrogen sulfide is the chemical compound with the chemical formula H2S. It is a colorless gas with the characteristic foul odor of rotten eggs. In this reaction it is reducing agent. it can act as oxidizing as well as the reducing agent. Hydrogen sulfide is very poisonous as well as corrosive.

Sulfuric acid is a mineral acid with molecular formula H₂SO₄. It is a colorless, odorless, and syrupy liquid that is soluble in water. In this reaction it acts as an acidic medium only.

Hydrogen sulphide reacts with potassium dichromate in presence of sulfuric acid

The Potassium Dichromate and Hydrogen Sulfide in presence of Sulfuric acid (K2Cr2O7 H2S H2SO4) react. The Potassium Dichromate reduced to Chromium Sulfate Cr2(SO4)3. On the other hand Hydrogen Sulfide (H2S) oxidized to solid Sulfur. The skeleton reaction equation is-

K2Cr2O7 + H2S + H₂SO₄ = S + K2SO4 + Cr2(SO4)3 + H2O

The above reaction is a Redox reaction. Therefore we can balance the reaction equation by Ion-electron method. By using this method the balanced equation is-

K2Cr2O7 + 3H2S + 4H₂SO₄ = 3S + K2SO4 + Cr2(SO4)3 + 7H2O

Proof by the Ion-electron method

The skeleton reaction for the redox reaction is-

K2Cr2O7 + H2S + H₂SO₄ = S + K2SO4 + Cr2(SO4)3 + H2O

Here,

The Oxidizing agent: K2Cr2O7 or Cr2O72-
The Reducing agent: H2S or, S-2

Reduction Half Reaction: 

Cr2O72- + 6e + 14H+ = 2Cr3+ + 7H2O … … … (1)

Oxidation Half Reaction:

S2- – 2e = S … … … … (2)

Now, equation (1)+(2)x3,

Cr2O72- + 6e + 14H+ = 2Cr3+ + 7H2O

3S2- – 6e = 3S


Cr2O72-+ 14H+ + 3S2- = 2Cr3+ + 7H2O + 3S

Cr2O72-+ 8H+ + (6H++ 3S2-)= 2Cr3+ + 7H2O + 3S

⇒ adding necessary ions and radicals we get,

K2Cr2O+ 4H2SO4 3H2S K2SO4 Cr2(SO4)3 + 7H2O + 3S

Click on the reaction to learn balancing redox reaction easily by ion electron method

 

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