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

K2Cr2O7, FeSO4, and HCl are the reactants of the above reaction. Mainly, K₂Cr₂O₇ and 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, since the reactants exchange elections to become product. In this electron exchanging reaction, the HCl plays a vital role. It creates an acidic environment.

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

If we notice the above redox reaction carefully then we can understand the oxidation number of the Cr in K2Cr2O7 become +3 in the product from +6. It clearly indicates that the Cr⁺⁶ in K₂Cr₂O₇ accepts a total of six electrons, three electrons each. That means it is the oxidizing agent for this reaction. 

On the other hand, FeSO4 releases one electron, and the oxidation number become +3 from +2. Therefore, Fe⁺² in FeSO4 is the reducing agent in the above reaction.

Now the name of the reactant chemicals as follows-

K₂Cr₂O₇ = Potassium dichromate which is the Oxidizing agent

FeSO₄ = Iron(II) sulfate which is the Reducing agent

HCl = Hydrochloric acid (creates acidic environment)

Balancing the reaction among K2Cr2O7, FeSO4, and HCl

From the above discussion,  it is clear like the water that the reaction is an oxidation-reduction reaction. Therefore, we can easily use the ion-electron method to balance this particular reaction. So further not do any delay, let’s get started-

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

Here,

The Oxidizing agent: K₂Cr₂O₇ or Cr₂O₇^2-

The Reducing agent: FeSO₄ or, Fe²⁺

Oxidation Half Reaction

Here, Fe atom in the reductant FeSO4 releases one electron and its oxidation number increased to +3 from +2. As we all know that the reaction in which a chemical species released electron(s) is known as the oxidation half reaction. The oxidation half of the concerning reaction is-

⇒ Fe²⁺ – e^– = Fe³⁺ … … … (1)

Reduction Half Reaction

As the oxidation-reduction reaction is a symultanious process, the reduction half reaction also takes place along with oxidation half reaction. The each Cr atom in the oxidizing agent K₂Cr₂O₇ or Cr₂O₇^2- takes three electrons released by the Fe, so oxidation number of Cr atom becomes +3 from +6.

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

Now, one mole of oxidizing agent needs 6 mole of electrons. On the other side, reducing agent iron releases only one electron which have to full fill the demand of the electrons for the reduction half reaction. Therefore, one mole of oxidizing agent needs six times of mole of reducing agent. So we should multiply the respective reaction equation with corresponding number. Then we add them to get a full redox reaction.

Here, equation (1)x6 + equation (2),

6Fe²⁺ – 6e^– = 6Fe³⁺

Cr₂O₇^2-  + 14H⁺ + 6e^– = 7H₂O + 2Cr³⁺

---

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

Cr₂O₇^2- + 6Fe²⁺ + 14H⁺ = 2Cr³⁺ + (2Fe³⁺ + 4Fe³⁺) + 7H₂O 

Here, adding necessary ions and radicals we get,

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

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