Half-equations for the combustion of sulfur dioxide

Question

Given the following redox equation:

$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$

$\ce{SO2}$ is being oxidized to $\ce{SO3}$ and $\ce{O2}$ is being reduced to $\ce{SO3}$.

I figured a way to write the half equations:

Oxidation half reaction: $$\ce{SO2 + H2O -> SO3 + 2H+ + 2e-}$$

Reduction half reaction: $$\ce{O2 + 4H+ + 4e- -> 2H2O}$$

After balancing and summing, I get: $$\ce{2SO2 + 2H2O + O2 + 4H+ -> 2SO3 + 2H2O + 4H+}$$

By cancelling $\ce{H2O}$ and $\ce{H+}$ we obtain the desired redox equation. The used half equations necessitate the presence of water and acidic medium at first, but in reality we need neither water nor an acidic medium.
Is there a better way of constructing half-equations that do not require any extra materials at the beginning? I mean the real half equations that express how combustion is made in reality. I thought that we don't need water and acid to perform it. In the "preparation" section here they didn't mention water or acid.
In addition, I guess oxygen must have sulfur trioxide as the corresponding reduced form instead of water.
Am I mistaken?

Answer 1

Orthocresol already sufficiently pointed out, that half-reactions are a book-keeping tool for electrons. When talking about electrolysis there might be coincidental overlap between half-reactions and the overall mechanism, but as the wording is supposed to indicate, this is not necessarily true.
Many reactions look simple at first glance, but on the second look they appear as highly complex systems of different equilibria. Actually the most simple looking reactions are often the ones that are painstakingly complex.

I actually mean the real half equations that express how combustion is made in reality. I thought that we don't need water and acid to perform it. Am I mistaking?

Since you have been asking about "the real half-reactions" I thought I give a little more insight into the mechanism.

I found a fairly old publication on the mechanisms of the formation of sulfur oxides during combustion.^[1] For the purpose of illustrating the complexity this shall be enough, but I am fairly certain there are more modern approaches and kinetic and mechanistic studies available today.

The reaction itself proceeds only at moderate temperatures (400-600 °C) and is slightly exothermic. $$\ce{SO2 + 1/2O2 <=> SO3 (g) + $99.0$~kJ}$$ Because the reaction proceeds quite slowly, catalysts are necessary. Here comes into play what you find in Orthocresol's answer: you need a species that provides you with oxygen(-2) ions. In praxis you use $\ce{V2O5}$ catalysts, e.g. in the synthesis of sulfuric acid.^[2]
Sulfurtrioxide is very hydroscopic, which indicates, that the reactions that you proposed cannot happen, since it would hydrolyse immediately in the presence of water. $$\ce{SO3 + H2O -> H2SO4(l)}$$

Sulfurtrioxide has many modifications, most commonly it exists as $\ce{(SO3)3}$ units in solid phase. Condensation is also one of the driving forces of this reaction, in the gas phase the equilibrium between $\ce{SO2}$, $\ce{O2}$, and $\ce{SO3}$ would not be entirely on the product side.
There is also a variety of different other sulfuroxides that are possibly involved in this complex equilibrium. This is due to the fact, that sulfurtrioxide is a powerful oxidiser.

Let's have a glance at a few reactions that are involved in the formation. The reaction itself most likely proceeds via a radical pathway in the gas phase, with $M$ being a catalyst of some sort (it can be the wall or some inert molecule): $$\begin{align} \ce{SO2 + O2 &<=> SO3 + O}\\ \ce{SO2 + O + $M$ &<=> SO3 + $M$} \end{align}$$

One important reaction that disturbs the equilibrium is $$\ce{SO_{$n$} + O <=> SO_{$n-1$} + O2}.$$

I also found a publication about the mechanism in the troposphere and things get even more complicated there.^[3] The reason for that is dilution, a reasonable amount of molecules in excited states, and various other species involved. This topic became interesting because of air pollution and acid rain.
I am only giving a few examples and leaving out a lot of details, as stated previously, this is for illustrative purposes. $$\begin{align} \ce{SO2 + O2 &<=> SO4}\\ \ce{SO4 + O2 &<=> SO3 + O3}\\ \ce{SO2 + SO2 &<=> SO3 + SO}\\ \ce{SO2 + O3 &<=> SO3 + O2}\\ &\text{etc.} \end{align}$$

As soon as you consider species that are usually available in the troposphere, e.g. $\ce{N2}$, $\ce{NO}$, $\ce{H2}$, $\ce{CO}$, $\ce{CO2}$, etc, things get really complicated.

Conclusion

You cannot write half reaction equations for this equilibrium. Half reactions in general do not represent the reaction mechanism; they are a just a tool to generate overall reaction equations.
The mechanism of the reaction is significantly more complex than only two half-reactions.
You are not mistaken, you don't need water to perform the combustion, however, you need some sort of catalyst, that is able to absorb the excess bonding energy.

---

References

1. Arthur Levy, Earl L. Merryman, William Thomas Reid, Environ. Sci. Technol., 1970, 4 (8), 653–662.

2. Arnold F. Holleman, Nils Wiberg, Egon Wiberg: Lehrbuch der Anorganischen Chemie. 102. Auflage. De Gruyter: 2008. (in German)
   English version: Nils Wiberg, A. F. Holleman, Egon Wiberg (Eds.): Holleman-Wiberg's Inorganic Chemistry. 1st Edition. Academic press: 2001. (Amazon, Google books)

3. Jack G. Calvert, Fu Su, Jan W. Bottenheim, Otto P. Strausz, Atmospheric Environ., 1978, 12 (1-3), 197-226.

Answer 2

That's a very interesting question and shows some good thinking on your part, so kudos on that first.

The simple answer is, no, there is no way you can construct half-equations that don't involve "extra species". Just consider the oxidation half-reaction. Since $\ce{SO2}$ and $\ce{SO3}$ have different numbers of oxygen, you need some way to balance the oxygen:

$$\ce{SO2 + $\text{oxygen(-2)-containing species}$ -> SO3 + 2e-}$$

The problem is that there are no oxygen(-2)-containing species in the reaction, apart from $\ce{SO2}$ and $\ce{SO3}$, of course! So the "closest" you could get to a pure set of half-reactions would be:

$$\begin{align} \ce{2SO2 + 2O^2-} &\ce{-> 2SO3 + 4e-} \\ \ce{O2 + 4e-} &\ce{-> 2O^2-} \end{align}$$

but even that involves some weird oxide ions that the combustion strictly doesn't need.

The actual problem is that they are only a formal way of representing redox reactions. The way you are taught to balance half-reactions, which involves adding $\ce{H2O}$ and $\ce{H+}$, only applies to the case of an electrochemical cell in aqueous solution.

Those half-reactions that you are trying to construct do not represent what is going on at all in the reaction. So, the issue is not that you cannot construct "pure" half-reactions. The issue is more of that you are trying to interpret the half-reactions wrongly. Half-reactions are merely a tool to 1) count electron transfer 2) construct a correct overall equation after cancelling out duplicate species - nothing more than that.