K2Cr2O7, FeCl2, and HCl are very common chemicals in all most every laboratory in the world. Besides, all the science students, as well as the teachers, know that K2Cr2O7 is a very good oxidizing agent and FeCl2 is a reducing agent. The remaining reactant HCl is an acid.
About the reaction among K2Cr2O7, FeCl2, and HCl
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In the above reaction, although there are three reactants HCl only creates an acidic environment. Mainly the K2Cr2O7 (Potassium dichromate) and FeCl2(Iron(II)chloride) take part in the reaction. If we observed sincerely then we will notice that in this reaction there is an exchange of electrons between the reactants of this particular reaction.
K2Cr2(+6)O7 + Fe(+2)Cl2 + HCl = KCl + Cr(+3)Cl3 + Fe(+3)Cl3 + H2O
Since there is an exchange of electron, i.e. exchange of oxidation number of the ions or atoms, the above reaction id an oxidation-reduction (redox) reaction.
When we look at the reaction equation we can see that K2Cr2O7 (Potassium dichromate) released in total 6 electrons (3 electrons each atom) on the other hand FeCl2 (Iron(II)chloride) accepts only one electron. That’s means-
K2Cr2O7 = Potassium dichromate which is Oxidizing agent ( or Cr2O72-)
FeCl2 = Iron(II)chloride which is Reducing agent (or Fe2+)
HCl = Hydrochloric acid (creates acidic environment)
Balancing the reaction equation
As we learn that the reaction is a redox (oxidation-reduction) reaction, we can apply the popular ion-electron method and try vĩ đại balance the chemical reaction equation. Let us use the ion-electron method and balance the chemical reaction equation.
To bởi so sánh, we need the skeleton chemical reaction equation. Therefore, the skeleton chemical reaction equation of K2Cr2O7, FeCl1 in the presence of HCl is-
K2Cr2O7 + FeSO4 + HCl = KCl + CrCl3 + FeCl3 + H2O
The above reaction is a full redox reaction. We must calculate it by the half-reaction; i.e. oxidation half-reaction as well as reduction-half reaction. Let us bởi it-
Oxidation Half Reaction
The reducing agent, Fe2+ donates an electron and becomes Fe3+. The oxidizing agent accepts this one electron.
⇒ Fe2+ – e– = Fe3+ … … … (1)
Reduction Half Reaction
As we discuss above oxidizing agent Cr2O72- accepts six electrons donated by the six reducing agents and produces two Cr3+ ions. The reduction half reaction-
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⇒ Cr2O72- + 6e– + 14H+ = 2Cr3+ + 7H2O … … … (2)
As we notice that oxidizing agent K2Cr2O7 (Potassium dichromate) or Cr2O72- released 6 electrons, on the other hand, reducing agent FeCl2 (Iron(II) chloride) or Fe2+ received only one electron each. So, in this case, one molecule of oxidizing agent Potassium dichromate needs 6 times reducing agent Iron(II) chloride. For this reason, we multiply equation (2) 6 times and add equation (1) vĩ đại get the full oxidation-reduction reaction.
6Fe2+ – 6e– = 6Fe3+
Cr2O72- + 14H+ + 6e– = 7H2O + 2Cr3+
Cr2O72- + 6Fe2+ + 14H+ = 2Cr3+ + 7H2O + 6Fe3+
Do you remember that we excluded the K+ and Cl– ions for our calculation benefit? At this point, we must add the ions vĩ đại fulfill the balanced redox reaction.
Adding necessary ions and radicals
K2Cr2O7 + 6FeCl2 + 14HCl = 2KCl + 2CrCl3 + 7H2O + 6FeCl3
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KMnO4 + HCl + FeCl2 = FeCl3 + MnCl2 + KCl + H2O
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