# Consider the following reaction. What is being oxidized?

## Sep 12, 2017

Dihydrogen is being oxidized.....

#### Explanation:

Oxidation entails a formal LOSS of electrons, and an increase in oxidation number corresponding to the NUMBER of electrons lost...

We gots zerovalent dihydrogen gas, ${\stackrel{0}{H}}_{2}$, that is oxidized to water, i.e. $\stackrel{+ I}{H}$:

$\frac{1}{2} {H}_{2} \left(g\right) \rightarrow {H}^{+} + 1 {e}^{-}$ $\left(i\right)$

And since for every oxidation, every electron loss, there is a corresponding reduction, an electron gain....

$\stackrel{+ I V}{C} {O}_{2} \left(g\right) + 2 {H}^{+} + 2 {e}^{-} \rightarrow \stackrel{+ I I}{C} O \left(g\right) + {H}_{2} O$ $\left(i i\right)$

And so we adds, $2 \times \left(i\right) + \left(i i\right)$.......to retire the electrons.....

$C {O}_{2} \left(g\right) + \cancel{2 {H}^{+}} + {H}_{2} \left(g\right) \rightarrow C O \left(g\right) + {H}_{2} O + \cancel{2 {H}^{+}}$

To give....

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

And thus dihydrogen is oxidized and carbon dioxide is reduced......