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.

Question

Owen Ambrose · Accepted Answer

#SO_2# is neither an oxidising or reducing agent.

This is not a redox reaction because, as you can see, none of the elements involved have changed in the oxidation numbers that I have assigned to them. #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. This is untrue, though, as you are actually transferring an oxide ion to create the sulfate(IV) ion: #SO_2+O^(2-)rarrSO_(3)^(2-)# Sulfur is still +4. To expand the concept of redox to systems without oxygen or hydrogen, the addition/subtraction of electrons and the use of oxidation numbers were introduced.

Luna Chen · Answer

undefined SO2 is the reducing agent in this reaction because it loses electrons and undergoes oxidation, causing another substance (in this case, CaCO3) to gain electrons and undergo reduction.