# In the following reaction is C reduced? 2C + O_2 -> 2CO

May 30, 2017

Carbon is formally $\text{oxidized}$.

#### Explanation:

And oxygen is formally $\text{reduced}$.

And we could assign oxidation numbers..........

$\stackrel{0}{C} + \frac{1}{2} {\stackrel{0}{O}}_{2} \left(g\right) \rightarrow : \stackrel{+ I I}{C} \equiv \stackrel{- I I}{O} \left(g\right)$

Carbon can also undergo complete oxidation to $\stackrel{+ I V}{C}$:

$\stackrel{0}{C} + \stackrel{0}{O} = \stackrel{0}{O} \left(g\right) \rightarrow \stackrel{- I I}{O} = \stackrel{+ I V}{C} = \stackrel{- I I}{O} \left(g\right)$

As always, the oxidation number is the charge left on the central atom, when all its bonds are broken, with the charge assigned to the most electronegative atom. Capisce?

As is usual, elements are assigned an oxidation number of $0$, given that they are conceived to have neither accepted nor donated electrons. The assignment of oxidation numbers is a CONCEPTUAL process; oxidation numbers are useful for balancing equations; they do not have fundamental significance. See here and links.