# How do we represent the formation of Fe_3O_4 by the oxidation of iron metal in basic conditions?

Jul 11, 2017

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

#### Explanation:

If you got $F {e}_{3} {O}_{4}$, this is mixed valence oxide of iron ......it occurs as the mineral magnetite, which we could formulate as $F e O \cdot F {e}_{2} {O}_{3} \equiv F {e}_{3} {O}_{4}$.......

And so metallic iron is oxidized to $F {e}^{2 +} + 2 \times F {e}^{3 +}$............

$3 F e \left(s\right) \rightarrow F {e}^{2 +} + 2 F {e}^{3 +} + 8 {e}^{-}$ $\left(i\right)$

But something must be correspondingly reduced.....and here it is the protium ion in water........

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

And so I take $\left(i\right) + 8 \times \left(i i\right)$ to gives...........

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

Charge and mass are balanced as is required...........