15 Facts on HI + K2Cr2O7: What, How To Balance & FAQs

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K2Cr2O7 is a common inorganic chemical reagent with the chemical formula K2Cr2O7, and Hydrogen Iodide is a diatomic molecule. Let us study detailed facts about these elements.

K2Cr2O7 is a bright red crystalline compound, K2Cr2O7 is a hexavalent chromium compound, and K2Cr2O7 is chronically not good to health. In contrast, HI is a diatomic, non-flammable, colorless, corrosive gas and strong inorganic acid. HI in an aqueous solution is called hydroiodic acid. 

This article will discuss details about K2CrO3 and HI, like reaction type, net ionic equation, balance equation, buffer solution, etc.

What is the product of HI and K2Cr2O7

KI (Potassium iodide), Cr(III)I3 (chromium(III) iodide) and I2 (solid Iodide) are the products of K2Cr2(VI)O7 (Potassium dichromate) and HI (Hydroiodic acid). The complete reaction equation is

K2Cr2(VI)O7 + 14HI ——–> 2KI + 2Cr(III)I3 + 3I2 + 7H2O

What type of reaction is HI + K2Cr2O7

K2CrO3 and HI is a Oxidation-Reduction(redox) reaction.

How to balance HI + K2Cr2O7

HI and K2CrO3 can be balanced by following the steps.

  • K2(+1)Cr2(+6)O7(-2) +H(+1)I(-1) ——–> 2K(+1)I(-1)+ 2Cr(+3)I(-1)3 + 3I(0)2 + 7H2O
  • The oxidation number of Cr is decreased; on the other hand, the oxidation number of Iodine is increased.
  • K2Cr2(+6)O7 —–>2Cr(+3)I3 ———-(1)
  • HI(-1) ——–>I(0) ——-(2)
  • The decrease in the oxidation number of Cr is 6 electrons per molecule in K2CrO3, while the increase in the oxidation number of I is 1 electron per molecule in HI.
  • So equation 2 is multiplied by 6
  • K2Cr2O7 + 6HI ——-> 2CrI3
  • To balance the Iodine and Potassium, 14 moles of HI required
  • K2Cr2(VI)O7 + 14HI ——–> 2KI + 2CrI3 + 3I2
  • To balance the hydrogen and oxygens on the reactant side, 7H2O must be added to the product side. So the balanced equation is
  • K2Cr2(VI)O7 + 14HI ——–> 2KI + 2Cr(III)I3 + 3I2 + 7H2O

HI + K2Cr2O7 titration

We cannot titrate K2Cr2O7 and HI. Although K2Cr2O7 is used for volumetric titrations, , and also, K2Cr2O7 and HI titration was not reported yet, because the titration of K2Cr2O7 must need to dilute H2SO4. In contrast, the titration of H2SO4 may form H2S by reacting with hydrogen in HI.

HI + K2Cr2O7 net ionic equation

The net ionic equation for the HI + K2Cr2O7 can be written by following steps.

  • K2Cr2(VI)O7 + 14HI ——–> 2KI + 2Cr(III)I3 + 3I2 + 7H2O
  • Disassociate all the molecules into ions; covalent compounds cannot be disassociated.
  • 2K(+1)(aq)+ Cr2O7(-2)(aq)+ H(+1)(aq) + I(-1)(aq) —> K(+1)(aq) + I(-1)(aq) + Cr(+3)(aq)+ 3I(-1)(aq)+ I2(s)+ H2O(l)
  • Remove the spectator ion, which means the ion has the same oxidation state in the reactants and the product.
  • Cr2O7(-2)(aq)+ I(-1)(aq) ——-> Cr(+3)(aq)+ I2(s)
  • Write half-reaction for oxidation and reduction.
  • I(-1)(aq)———-> I2(s) ——- Oxidation
  • Cr2O7(-2)(aq) ———> Cr(+3)(aq) ——- Reduction
  • Balance the ionic form obtained
  • 2I(aq)———-> I2(s)
  • Cr2O7(-2)(aq) ———> 2Cr(+3)(aq)
  • Add spectator ion again to obtain a molecular form
  • K(+1)+ Cr2O7(-2)+ 14H(+1) + 6I(-1) ——-> K(+1)+ I(-1)+ 2Cr(+3)+ I(-1)+ 3I2+ 7H2O
  • Finally, ensure all the ions are balanced. If not, balance by adding ions on the reactants and product side. So the balanced redox reaction is 
  • K2Cr2(VI)O7 + 14HI ——–> 2KI + 2Cr(III)I3 + 3I2 + 7H2O

HI + K2Cr2O7 conjugate pairs

HI + K2Cr2O7 has the following conjugate pairs

  • During the reaction, HI can donate a proton( act as an acid) and I– (act as a base), so the HI is a conjugate acid-base pair.
  • K2Cr2O7 does not contain any protons; hence it is not conjugate pair.

HI and K2Cr2O7 intermolecular forces

HI and K2Cr2O7 has the following intermolecular forces.

  • Because of HI acidity and polarity, HI has dipole-dipole and London dispersion forces are present
  • K2Cr2O7 disassociates to Cr2O7– and K+ ions. K2Cr2O7 shows weak Wander Waals forces

HI + K2Cr2O7 reaction enthalpy

The reaction enthalpy of HI + K2Cr2O7 is -3950.1173 which is negative.

  • The enthalpy of reactant K2Cr2O7 enthalpy is -2035
  • The enthalpy of reactant HI enthalpy is 26.48472
  • The enthalpy of product KI enthalpy is 0
  • The enthalpy of product Cr2I3 enthalpy is 2X26.103916 = 59.2078
  • The enthalpy of product I2 enthalpy is 0The enthalpy of product H2O enthalpy is 7X[-285.82996]= 2000.80972
  • [(K2Cr2O7 + 14HI) – (2KI + 2CrI3 + 3I2 + 7H)] =[(-2008.51588)- (-1941.60142)] = -3950.1173

Is HI + K2Cr2O7 a buffer solution

HI + K2Cr2O7 is not a buffer solution since the HI is a strong acid, and it disassociates to H+ and I– in an aqueous medium, and they don’t form a buffer solution. 

Is HI + K2Cr2O7 a complete reaction

HI + K2Cr2O7 is a complete reaction once it reaches equilibrium.

Is HI + K2Cr2O7 an exothermic or endothermic reaction

Since the HI + K2Cr2O7 is a redox reaction, it is exothermic; during the reaction, the amount of the energy released is greater than the amount of energy absorbed as the reaction enthalpy is negative.

Is HI + K2Cr2O7 a redox reaction

HI + K2Cr2O7 reaction is a redox(oxidation-reduction reaction) reaction since the 6I(-1) oxidized to 6I(0) and 2Cr(+4) reduced to 2Cr(+3).

Is HI + K2Cr2O7 a precipitation reaction

HI + K2Cr2O7 is not a precipitation reaction because HI is a strong acid and easily soluble in aqueous, and K2Cr2O7 is in salt form; hence it doesn’t form any precipitates. 

Is HI + K2Cr2O7 reversible or irreversible reaction

HI + K2Cr2O7 is an irreversible reaction because redox reactions are spontaneous. They transfer electrons in only one way.

Is HI + K2Cr2O7 displacement reaction

HI +K2Cr2O7 is not a displacement reaction, as no displacements of molecules or ions from the reactants were observed.

Conclusion

Besides the above facts,K2Cr2O7 is widely used in organic synthesis as a strong oxidizing and common reagent in analytical chemistry. On the other hand, HI is used as a strong acid, reducing reagent and analytical reagent. HI is supplied as compressed gas.