Is the reaction between thiosulfate and hydronium an acid-base reaction?

Is the following reaction an acid-base reaction:


$$\ce{2 H3O+(aq) + S2O3^2-(aq) -> 3H2O(l) + SO2(g) + 1/8 S8(s)}$$


And if so, which is the acid and which is the base?




Two of the three acid-base theories (Brønsted and Lewis) also define an acid-base reaction while the third theory (Arrhenius) does not — it merely defines what acids and bases *are.*


In the Lewis theory, an acid-base reaction occurs when a lone pair donor interacts with a lone pair acceptor to form a new bond *without* a simultaneous redox reaction happening. The less complicated Brønsted-Lowry theory describes an acid-base reaction as a reaction in which a proton $\ce{H+}$ is transferred from an acid to a base, giving us a conjugate base and a conjugate acid, respectively.


As has been pointed out, your reaction is not a one-step reaction. Rather, we have to break it down into its constituent steps. The *first* step, a reaction of hydronium $\ce{H3O+}$ with thiosulfate $\ce{S2O3^2-}$ *has* to be an acid-base reaction because that is practically the only reaction hydronium undergoes.


$$\ce{2H3O+ + S2O3^2- -> H2S2O3 + 2 H2O}\tag{1}$$


The acid is the species that *donates* the transferred proton, the base is the species that *accepts* it. Therefore, we can assign the *acid* and *base* roles as shown in equation $(1')$.


$$\ce{2 \underset{acid}{H3O+} + \underset{base}{S2O3^2-} -> 2 \underset{conjugate base}{H2O} + \underset{conjugate acid}{H2S2O3}}\tag{$1'$}$$


The second reaction is a redox reaction. Thiosulfate (and therefore also its conjugate acid) has two sulfur atoms in the oxidation states of $\mathrm{+V}$ and $\mathrm{-I}$, respectively. These undergo a redox reaction leading to compounds of the oxidation state $\mathrm{+IV}$ (sulfurous acid $\ce{H2SO3}$) and $\pm0$ (elemental sulfur). We can use the full redox reaction determination algorithm, but I’m skipping that since it is a simple one-electron transfer.


$$\ce{H2S2O3 -> H2SO3 + 1/8 S8}\tag{2}$$


And finally, we have another decomposition namely that of sulfurous acid to sulfur dioxide $\ce{SO2}$ and water. This can be seen as the back-reaction of a Lewis acid-base reaction wherein water is the liberated Lewis base and sulfur dioxide the liberated Lewis acid. Indeed, the forward Lewis acid-base reaction (the reverse direction of this reaction) occurs, too. This reaction is in equilibrium, driven to the product side because of the volatility of $\ce{SO2}$.


$$\ce{H2SO3 (aq) <=>> H2O (l) + SO2 (g) ^}\tag{3}$$


Therefore, the overall reaction is the sum of one Brønsted acid-base reaction, one redox reaction, and one Lewis acid-base reverse reaction.


$$\ce{2 H3O+ (aq) + S2O3^2- (aq) -> 3 H2O (l) + 1/8 S8 (s) + SO2 (g) ^}\tag{4}$$


 3