Identify the oxidizing agent: #2"S"_2"O"_3 ^(2-) + "I"_2 -> "S"_4"O"_6^(2-) + 2"I"^(-)#?

1 Answer
Jun 27, 2017

The oxidizing agent is #"I"_2#.


A quick technique to use here would be to look at the fact that you're going from iodine, #"I"_2#, on the reactants' side to the iodide anion, #"I"^(-)#, on the products' side.

In this case, you're going from a neutral molecule to a negatively charged ion, so right from the start, you know that iodine is being reduced, i.e. it is taking in electrons.

This can only mean that the thiosulfate anion, #"S"_2"O"_3^(2-)#, is acting as a reducing agent because it is reducing iodine to iodide anions.

Consequently, you can say that iodine, #"I"_2#, is acting as an oxidizng agent because it is oxidizing the thiosulfate anion to the tetrathionate anion, #"S"_4"O"_6^(2-)#.

#"S"_2"O"_3^(2-) -> "reduces I"_2 + "gets oxidzied to S"_4"O"_6^(2-)#

#"I"_ 2 -> "oxidizes S"_ 2"O"_ 3^(2-) + "gets reduced to I"^(-)#

You can see that this is the case by assigning oxidation numbers to the atoms that take part in the reaction--I won't add the states to keep the chemical equation simple

#2stackrel(color(blue)(+2))("S")_ 2stackrel(color(blue)(-2))("O")_ 3""^(2-) + stackrel(color(blue)(0))("I")_ 2 -> stackrel(color(blue)(+"5/2"))("S")_ 4 stackrel(color(blue)(-2))("O")_ 6 ""^(2-) + 2stackrel(color(blue)(-1))("I") ""^(-)#

Notice that the oxidation number of iodine goes from #color(blue)(0)# on the reactants' side to #color(blue)(-1)# on the products' side. This tells you that iodine is being reduced because its oxidation number is decreasing.

Similarly, the oxidation number of sulfur goes from #color(blue)(+2)# on the reactants' side to an average value of #color(blue)(+"5/2")# on the products' side. This tells you that sulfur is being oxidized because its oxidation number is increasing.

So once again, you can conclude that iodine is the oxidizing agent because it oxidizes the thiosulfate anions to the tetrathionate anions while being reduced to the iodide anions.

Similarly, the thiosulfate anion is the reducing agent because it reduces iodine to the iodide anions while being oxidized to the tetrathionate anions.