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

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 ${25}^{\circ} \text{C}$ and $\text{1 atm}$ is $0$, but the natural state of sulfur is "S"_8("orthorhombic"), so let's rewrite this reaction as:

$\frac{1}{8} {\text{S"_8("orthorhombic") + "O"_2(g) -> "SO}}_{2} \left(g\right)$

That aside, ${\text{O}}_{2} \left(g\right)$ is also in its natural state at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, so both of their oxidation states are $0$.

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:

$\stackrel{0}{{\text{O"_2) -> stackrel(-2)("O}}_{2}}$
$\frac{1}{8} \stackrel{0}{\text{S"_8) -> stackrel(+4)("S}}$

where the oxidation states are for each individual atom, NOT the entire set of $\text{O}$ or $\text{S}$ atoms. $\text{O}$ often has the oxidation state of $- 2$ as an ion, with some rare exceptions of $- 1$ in peroxides.

Since the oxidation state of $\text{S}$ increased to become more positive, it means that ${\text{S}}_{8}$ was oxidized by the only other reactant, ${\text{O}}_{2}$.

Therefore, ${\text{O}}_{2}$ is the oxidizing agent.