# When hydrogen burns, what is oxygen's role?

Feb 11, 2016

#### Answer:

Oxygen acts as the $\text{oxidant}$, the acceptor of electrons.

#### Explanation:

The above definition is a formalism, but it does help us to rationalize reactivity. Oxygen lies on the right of the Periodic Table; its high nuclear charge tends to accept electrons, i.e. becoming reduced, and oxidizing another species:

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(g\right)$

In the reaction above, the reactants are elemental, and have an oxidation state of $0$. In the products, hydrogen has a formal oxidation state of $+ I$, and the oxygen has a formal oxidation state of $- I I$. Because of its tendency to accept electrons, $O$ commonly has a formal oxidation state of $- I I$ in its compounds. When we assign oxidation states, we remember that these are formalisms; nevertheless, they give us an idea of electron transfer in the reaction.