Identify the oxidizing agent: #2"S"_2"O"_3 ^(2-) + "I"_2 -> "S"_4"O"_6^(2-) + 2"I"^(-)#?
1 Answer
The oxidizing agent is
Explanation:
A quick technique to use here would be to look at the fact that you're going from iodine,
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,
Consequently, you can say that iodine,
#"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
Similarly, the oxidation number of sulfur goes from
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.