# What is Fe_3O_4, and how do we represent its formation from iron, and water?

Aug 28, 2017

This is a mixed valence oxide of $F e O$ and $F {e}_{2} {O}_{3}$........

#### Explanation:

In the original redox reaction, elemental iron is OXIDIZED to $F e O + F {e}_{2} {O}_{3}$, i.e. $F e O \cdot F {e}_{2} {O}_{3} \equiv F {e}_{3} {O}_{4}$.

And so we write 2 separate oxidation rxns....

$F e + {H}_{2} O \rightarrow F e O + 2 {e}^{-} + 2 {H}^{+}$, and....

$2 F e + 3 {H}_{2} O \rightarrow F {e}_{2} {O}_{3} + 6 {H}^{+} + 6 {e}^{-}$

We adds these together and get.....

$3 F e + 4 {H}_{2} O \rightarrow {\underbrace{F {e}_{2} {O}_{3} + F e O}}_{F {e}_{3} {O}_{4}} + 8 {H}^{+} + 8 {e}^{-}$ $\left(i\right)$

And for every oxidation, electron loss, there is a corresponding reduction, electron gain, and the most likely oxidant is dioxygen gas.....

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

And so we take $4 \times \left(i i\right) + \left(i\right)$ to eliminate the electrons from our final redox equation.......

$3 F e + 2 {O}_{2} + \cancel{4 {H}_{2} O + 8 {H}^{+} + 8 {e}^{-}} \rightarrow {\underbrace{F {e}_{2} {O}_{3} + F e O}}_{F {e}_{3} {O}_{4}} + \cancel{8 {H}^{+} + 8 {e}^{-} + 4 {H}_{2} O}$

To give finally.......

$3 F e + 2 {O}_{2} \rightarrow {\underbrace{F {e}_{2} {O}_{3} + F e O}}_{F {e}_{3} {O}_{4}}$

$F {e}_{3} {O}_{4}$ is a mixed valence iron oxide.....$F {e}_{2} {O}_{3} \cdot F e O$