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

1 Answer
Jan 26, 2016

Answer:

This is a redox reaction.

Explanation:

You're dealing with a redox reaction in which free iodine, #"I"_2#, oxidizes the thiosulfate anions, #"S"_2"O"_3^(2-)#, to thetrathionate anions, #"S"_4"O"_6^(2-)#.

At the same time, iodine is reduced to iodide anions, #"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(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 #color(blue)(+2)#.

On the products' side, you have a total of four sulfur atoms, each with an average oxidation state of #color(blue)(+2.5)#.

If you go by individual atoms, four sulfur atoms change oxidation state going from #color(blue)(+2)# to #color(blue)(+2.5)#, with a total of two electrons being lost in the process (#+8 -> +10#).

So, iodine picks up electrons and the sulfur loses them #-># 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.