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
The Oxidizing agent: K2Cr2O7 or Cr2O72-
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 Cr2O72- takes three electrons released by the Fe, so oxidation number of Cr atom becomes +3 from +6.
⇒ Cr2O72- + 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+
Cr2O72- + 14H+ + 6e– = 7H2O + 2Cr3+
Cr2O72- + 6Fe2+ + 14H+ = 2Cr3+ + 7H2O + 6Fe3+
Cr2O72- + 6Fe2+ + 14H+ = 2Cr3+ + (2Fe3+ + 4Fe3+) + 7H2O
Here, adding necessary ions and radicals we get,
K2Cr2O7 +14HCl + 6FeSO4 = 2KCl + 2FeSO4 + 2Fe2(SO4)3 + 2CrCl3 + 7H2O