K2Cr2O7 + FeSO4 + H2SO4 = Cr2(SO4)3 + Fe2(SO4)3 + K2SO4 + H2O

Potassium dichromate iron(II) sulfate sulfuric acid (K2Cr2O7 FeSO4 H2SO4) are available in almost all chemical lab across the World. Mainly K2Cr2O7 and FeSO4  react in the H2SO4 medium. We use the ion-electron method to balance the redox reaction.

Potassium dichromate react with iron(II) sulfate in sulfuric acid medium

The skeleton reaction for the redox reaction of Potassium dichromate iron(II) sulfate sulfuric acid (K2Cr2O7 FeSO4 H2SO4) is-

K2Cr2O7 + FeSO4 + H2SO4 = Cr2(SO4)3 + Fe2(SO4)3 + K2SO4 + H2O

Or,

K₂Cr₂O₇ + FeSO₄ + H₂SO₄ = Cr₂(SO₄)₃ + Fe₂(SO₄)₃ + K₂SO₄ + H₂O

Read more:

How to balance the redox reaction FeSO4 + KMnO4 + H2SO4 = Fe2(SO4)3 + MnSO4 + H2O + K2SO4

Here,

The Oxidizing agent: K₂Cr₂O₇ or Cr₂O₇^2-

The Reducing agent: FeSO₄ or, Fe²⁺

Oxidation Half Reaction:

Fe²⁺ – e = Fe³⁺ … … … (1)

Reduction Half Reaction:

Cr₂O₇^2- + 6e + 14H⁺ = 2Cr³⁺ + 7H₂O … … … (2)

Now,

equation (1)x6 + (2),

6Fe²⁺ – 6e = 6Fe³⁺

Cr₂O₇^2- + 6e + 14H⁺ = 2Cr³⁺ + 7H₂O

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Cr₂O₇^2- + 6Fe²⁺ + 14H⁺ = 2Cr³⁺ + 6Fe³⁺ + 7H₂O

Now

adding necessary ions and radicals we get,

K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ = Cr₂(SO₄)₃ + 3Fe₂(SO₄)₃ + K₂SO₄ + 7H₂O

“Answer”

K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ = Cr₂(SO₄)₃ + 3Fe₂(SO₄)₃ + K₂SO₄ + 7H₂O

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-Albert Einstein