K2Cr2O7 + FeSO4 + H2SO4

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

K2Cr2O7 FeSO4 H2SO4

Here,

The Oxidizing agent: K2Cr2Oor Cr2O72-

The Reducing agent: FeSO4 or, Fe2+

Oxidation Half Reaction:

Fe2+ – e = Fe3+ … … … (1)

Reduction Half Reaction:

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

Now,

equation (1)x6 + (2),

6Fe2+ – 6e = 6Fe3+

Cr2O72- + 6e + 14H+ = 2Cr3+ + 7H2O


Cr2O72- + 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

There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.

-Albert Einstein

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