How would you classify the reaction I_2(aq)+2S_2O""_3^(2-) (aq) -> 2I^(-) (aq) + S_4O""_6^(2-)(aq)?

Jan 26, 2016

This is a redox reaction.

Explanation:

You're dealing with a redox reaction in which free iodine, ${\text{I}}_{2}$, oxidizes the thiosulfate anions, ${\text{S"_2"O}}_{3}^{2 -}$, to thetrathionate anions, ${\text{S"_4"O}}_{6}^{2 -}$.

At the same time, iodine is reduced to iodide anions, ${\text{I}}^{-}$.

You can confirm that this is what's going on by assigning oxidation numbers to the elements that are taking part in the reaction - since the reaction takes place in aqueous solution, I won't add state symbols.

$\stackrel{\textcolor{b l u e}{0}}{{\text{I"_2) + 2stackrel(color(blue)(+2))("S"_2)stackrel(color(blue)(-2))("O"_3^(2-)) -> 2stackrel(color(blue)(-1))("I"^(-)) + stackrel(color(blue)(+2.5))("S"_4)stackrel(color(blue)(-2))("O}}_{6}^{2 -}}$

Don't be confused by the fact that sulfur has a fractional oxidation number in the tetrathionate anion, that just means that not all the sulfur atoms that form the anion have the same oxidation state.

The reduction half-reaction has iodine reduced to iodide anions

$\stackrel{\textcolor{b l u e}{0}}{{\text{I"_2) + 2"e"^(-) -> 2stackrel(color(blue)(-1))("I}}^{-}}$

The iodine is reduced because it gains two electrons.

The oxidation half-reaction has the thiosulfate anion oxidized to tetrathionate anion

2stackrel(color(blue)(+2))("S"_2)stackrel(color(blue)(-2))("O"_6^(2-)) -> stackrel(color(blue)(+2.5))("S"_4)stackrel(color(blue)(-2))("O"_6^(2-)) + 2"e"^(-)

Notice what happens here. Each thiosulfate anion loses one electron. On the reactants' side, you have a total of four sulfur atoms, each with an oxidation state of $\textcolor{b l u e}{+ 2}$.

On the products' side, you have a total of four sulfur atoms, each with an average oxidation state of $\textcolor{b l u e}{+ 2.5}$.

If you go by individual atoms, four sulfur atoms change oxidation state going from $\textcolor{b l u e}{+ 2}$ to $\textcolor{b l u e}{+ 2.5}$, with a total of two electrons being lost in the process ($+ 8 \to + 10$).

So, iodine picks up electrons and the sulfur loses them $\to$ you're indeed dealing with a redox reaction.

You can read more on sulfur's fractional oxidation number in the tetrathionate anion in this Socratic answer.