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

Potassium dichromate iron(II) sulfuric acid (K2Cr2O7 FeSO4 H2SO4) is accessible in nearly every chemical laboratory worldwide. In the H2SO4 medium, K2Cr2O7 and FeSO4 respond mainly. To balance the redox reaction we use the ion-electron technique.

Potassium dichromate reacts in sulfuric acid with iron(II) sulfate

The skeleton reaction of Potassium dichromate and iron(II) sulfate in 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

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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-  + 14H⁺ + 6e^– = 7H₂O + 2Cr³⁺

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

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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FeSO4 + KMnO4 + H2SO4 = Fe2(SO4)3 + MnSO4 + H2O + K2SO4

Al + H2SO4 = Al2(SO4)3 + SO2 + H2O

Al + H2SO4 = Al2(SO4)3 +H2