K2Cr2O7 + FeSO4 + H2SO4

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

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-  + 14H+ + 6e= 7H2O + 2Cr3+


14H+ + 6Fe2+ + Cr2O72- = 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

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