What is the oxidizing agent in 1/8"S"_8("orthorhombic") + "O"_2(g) -> "SO"_2(g)18S8(orthorhombic)+O2(g)SO2(g), the formation reaction for sulfur dioxide gas?

1 Answer
Nov 13, 2016

Let's put it this way: what do you think of assigning oxygen the role of oxidizing agent? Does that make sense?

If not, then we can do it another way. The oxidation state of sulfur in its natural state at 25^@ "C"25C and "1 atm"1 atm is 00, but the natural state of sulfur is "S"_8("orthorhombic")S8(orthorhombic), so let's rewrite this reaction as:

1/8"S"_8("orthorhombic") + "O"_2(g) -> "SO"_2(g)18S8(orthorhombic)+O2(g)SO2(g)

That aside, "O"_2(g)O2(g) is also in its natural state at 25^@ "C"25C and "1 atm"1 atm, so both of their oxidation states are 00.

But, both of them combine to form a compound in which they are both ions. Therefore, there must be oxidation-reduction going on.

In fact, they've changed oxidation state as follows:

stackrel(0)("O"_2) -> stackrel(-2)("O"_2)0O22O2
1/8stackrel(0)("S"_8) -> stackrel(+4)("S")180S8+4S

where the oxidation states are for each individual atom, NOT the entire set of "O"O or "S"S atoms. "O"O often has the oxidation state of -22 as an ion, with some rare exceptions of -11 in peroxides.

Since the oxidation state of "S"S increased to become more positive, it means that "S"_8S8 was oxidized by the only other reactant, "O"_2O2.

Therefore, "O"_2O2 is the oxidizing agent.