Question

Using the reaction in the picture, XeO3 + 4NaOH + O3 ---> 4Na^+ + XeO6^4- + O2 + 2H2O, write a reasonable mechanism for your balanced redox reaction. Use only nucleophilic/electrophilic steps, even though there may actually be some single electron transfer steps involved.

Now we're ready to balance the redox reaction. To the equation below, first add the product that comes from O3, and then add the correct coefficients in front of the NaOH and the H2O species: XeO3 +19 NaOH(aq) +03 → 4Nat + Xe064 +1 0.1 + 2 H2O Now comes a harder part. On the next page, write a reasonable mechanism for your balanced redox reaction. Use only nucleophilic/electrophilic steps, even though there may actually be some single electron transfer steps involved. Hint: The following two mechanistic steps from alkene ozonolysis may give you some guidance. Xe Oz + 4 Naah +O3 → Wat + XeO" + O2 + H H C CH3 HC CHE 3 Nay Xella 20- O 0-0 Lood S-CH3 HC Now comes the hardest part. What additional chemical reaction needs to occur in order to reach Rogers' Equation 19.20? (*That equation is correct, but it does not reveal all that is going on.) 4- Il

Answer 1

The mechanism is very simple.

In the given reaction, XeO3 is the reducing agent, i.e. it reduces other species by itself undergoing oxidation.

The oxidation state of Xe in XeO3 = +6, which undergoes oxidation by losing 2 electrons, and forms a compound in which Xe is in an oxidation state of +8 (XeO3²⁺).

In the given reaction, O3 is the oxidizing agent, i.e. it oxidizes other species by itself undergoing reduction.

The oxidation state of O in O3 = 0, which undergoes reduction by gaining the 2 electrons lost by XeO3, and forms O2 and O^2-.

So, the Xe-species produced is XeO3²⁺ + O^2- = XeO4.

This XeO4 further takes 2 more O's from 4 moles of NaOH (4 Na⁺ and 4 OH^-) and forms XeO₆^4-, the remaining 2OH^- and 2 H⁺ form two water (H₂O) molecules.

Now, your equation is ready.

XeO3 + O3 + 4 NaOH ----> XeO6^4- + O2 + 4 Na⁺ + 2 H2O