K2Cr2O7 + FeSO4 + HCl = KCl + CrCl3 + FeCl3 + Fe2(SO4)3 + H2O

K2Cr2O7, FeSO4, and HCl are the reactants of the above reaction. Mainly, K2Cr2Oand FeSO4 react with each other, while they exchange the electrons. Yes, K2Cr2O7 + FeSO4 + HCl = KCl + CrCl3 + FeCl3 + Fe2(SO4)3 + H2O is a redox (Oxidation-Reduction) reaction. In this electron exchanging reaction, the HCl plays a vital role. It creates an acidic environment. Now the name of the reactant chemicals as follows-

K2Cr2O7 = Potassium dichromate which is Oxidizing agent

FeSO4 = Iron(II) sulfate which is Reducing agent

HCl = Hydrochloric acid (creates acidic environment)

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

The skeleton reaction of Potassium dichromate and iron(II) sulfate in hydrochloric acid (K2Cr2O7 HCl H2SO4) is–

K2Cr2O7 + FeSO4 + HCl = KCl + CrCl3 + FeCl3 + Fe2(SO4)3 + H2O

Read more:

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

K2Cr2O7 + FeSO4 + HCl = KCl + CrCl3 + FeCl3 + Fe2(SO4)3 + H2

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+


Cr2O72- + 6Fe2+ + 14H= 2Cr3+ + 7H2O + 6Fe3+

Cr2O72- + 6Fe2+ + 14H= 2Cr3+ + (2Fe3+ + 4Fe3+) + 7H2

Here,

adding necessary ions and radicals we get,

K2Cr2O7 +14HCl + 6FeSO4 = 2KCl + 2FeSO4 + 2Fe2(SO4)3 + 2CrCl3 + 7H2O

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