# How is "ferric oxide" reduced by carbon to give steel?

Mar 5, 2016

$\text{Reduction:}$ $F {e}_{2} {O}_{3} \left(s\right) + 6 {H}^{+} + 6 {e}^{-} \rightarrow 2 F e \left(s\right) + 3 {H}_{2} O \left(l\right)$ $\left(i\right)$

$\text{Oxidation:}$ $C \left(s\right) + 2 {H}_{2} O \left(l\right) \rightarrow C {O}_{2} \left(s\right) + 4 {H}^{+} + 4 {e}^{-}$ $\left(i i\right)$

#### Explanation:

Overall: $2 \times \left(i\right) + 3 \times \left(i i\right)$

$2 F {e}_{2} {O}_{3} + 3 C \left(s\right) \rightarrow 4 F e \left(s\right) + 3 C {O}_{2} \left(g\right)$

These are balanced with respect to (i) mass, and (ii) charge, as in fact it must be!

Clearly I have been able to cancel the waters and ${H}^{+}$. Charge is balanced; mass is balanced. This represents a reasonable reaction for iron smelting. Industrially, coke is heated to give (red-hot!) carbon monoxide, which is injected (blasted!) into the iron oxide slag, where it performs the redox reaction. Note that I do not utilize dioxygen gas in these equations, because I want to represent the redox behaviour of iron and carbon only, so I need separate redox couples.