How would you classify the reaction #I_2(aq)+2S_2O""_3^(2-) (aq) -> 2I^(-) (aq) + S_4O""_6^(2-)(aq)#?
1 Answer
This is a redox reaction.
Explanation:
You're dealing with a redox reaction in which free iodine,
At the same time, iodine is reduced to iodide anions,
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(color(blue)(0))("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(color(blue)(0))("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
On the products' side, you have a total of four sulfur atoms, each with an average oxidation state of
If you go by individual atoms, four sulfur atoms change oxidation state going from
So, iodine picks up electrons and the sulfur loses them
You can read more on sulfur's fractional oxidation number in the tetrathionate anion in this Socratic answer.
