Consider the following reaction: #CaCO_3+SO_2 to CaSO_3+CO_2#. Here #SO_2# is oxidized to #CaSO_3#, so is #SO_2# a reducing agent or oxidizing agent? Kindly give your reason.

1 Answer
Feb 18, 2016

Answer:

#SO_2# is neither an oxidising or reducing agent.

Explanation:

I have assigned oxidation numbers to all the elements involved and you can see that none are changed, so this is not a redox reaction.

#stackrelcolor(blue)(+2)("Ca")stackrelcolor(blue)(+4)("C")stackrelcolor(blue)(-2)("O")_3+stackrelcolor(blue)(+4)("S")stackrelcolor(blue)(-2)"O"_2rarrstackrelcolor(blue)(+2)("Ca")stackrelcolor(blue)(+4)("S")stackrelcolor(blue)(-2)("O")_3+stackrelcolor(blue)(+4)("C")stackrelcolor(blue)(-2)"O"_2#

As #"SO"_2# forms #"CaSO"_3# you might think sulfur is being oxidised as the original, historical definition of oxidation referred to the addition of oxygen and/or removal of hydrogen.

However, this is not the case since you are, in effect, transferring an oxide ion to make the sulfate(IV) ion:

#SO_2+O^(2-)rarrSO_(3)^(2-)#

Sulfur remains +4.

The idea of adding/removing electrons and the use of oxidation numbers was introduced so that the concept of redox could be extended to other systems that did not involve oxygen or hydrogen.