Ellingham diagrams and metal extraction

People have smelted iron with charcoal for more than three thousand years. Aluminium is the most abundant metal in the Earth’s crust (about 8 %), yet in the 1850s it cost more than gold — Napoleon III, the story goes, let only his most honoured guests eat with aluminium cutlery. It became cheap only after electrolysis was invented in 1886. Both jobs mean pulling a metal out of its oxide, so why does a charcoal fire work for iron and fail for aluminium? In 1944 the British chemist Harold Ellingham drew the answer as a single chart.

To compare oxides fairly, every reaction is written per mole of O₂ consumed, for example $\ce{2Zn} + \ce{O2} \mathrel{\ce{->}} \ce{2ZnO}$ or $\ce{4/3Al + O2} \longrightarrow \ce{2/3Al2O3}$. Then, with the last lessons’ approximation $\Delta G^\circ(T) = \Delta H^\circ - T\Delta S^\circ$, ΔG° is drawn as a straight line against temperature:

• the intercept (extrapolated to 0 K) is ΔH°, and the slope is −ΔS°;
• forming a metal oxide consumes one mole of gaseous O₂, so ΔS° is about −180 to −220 J/K; the lines are nearly parallel and all rise to the right — oxides become less stable as it gets hotter;
• the lower the line, the more stable the oxide. At a given temperature a metal can reduce every oxide whose line lies above its own, and ΔG° per mole of O₂ transferred is the vertical gap between the lines;
• once a line crosses ΔG° = 0 the oxide decomposes by itself in 1 bar of O₂: Ag₂O at about 470 K, HgO at about 740–840 K. In 1774 Priestley made oxygen exactly this way, heating HgO with sunlight focused by a lens.

One line on the chart is different: $\ce{2C} + \ce{O2} \mathrel{\ce{->}} \ce{2CO}$. It turns one mole of gas into two, ΔS° = +179 J/K, so it slopes down to the right — the hotter it is, the better carbon is at grabbing oxygen. It crosses one metal line after another: FeO at about 1020 K, ZnO at about 1260 K (about 1220 K once zinc’s boiling at 1180 K is included), Cr₂O₃ at about 1490 K, SiO₂ at about 1910 K, and Al₂O₃ only near 2300 K.

The two other carbon lines: $\ce{C} + \ce{O2} \mathrel{\ce{->}} \ce{CO2}$ is almost flat (no change in moles of gas, ΔS° ≈ +3 J/K); $\ce{2CO} + \ce{O2} \mathrel{\ce{->}} \ce{2CO2}$ rises. The three meet at about 980 K: below that temperature carbon burns mainly to CO₂, above it CO is favoured (the Boudouard equilibrium $\ce{CO2} + \ce{C} \mathrel{\ce{<=>}} \ce{2CO}$). That is why the working reducing agent at high temperature is really CO.

A blast furnace is charged from the top with iron ore, coke and limestone while hot air is blown in at the bottom. Near the tuyeres the coke burns and the temperature exceeds 2000 K; the CO₂ formed is immediately reduced to CO by the glowing coke. The CO flows upward and reduces the iron oxides step by step: $\ce{Fe2O3} \longrightarrow \ce{Fe3O4} \longrightarrow \ce{FeO} \longrightarrow \ce{Fe}$.

• $\ce{Fe2O3 + 3CO} \longrightarrow \ce{2Fe + 3CO2}$ has ΔG° < 0 at every temperature — easy;
• $\ce{FeO} + \ce{CO} \mathrel{\ce{<=>}} \ce{Fe} + \ce{CO2}$ has ΔG° = +6.4 kJ at 1000 K, K ≈ 0.47: this step runs to the right only when the gas is more than about 68 % CO. Excess coke keeps topping up the CO through the Boudouard reaction, meeting exactly this requirement.

The limestone decomposes to CaO, which combines with the SiO₂ in the ore to form molten slag floating on the iron. Zinc has a trickier story: carbon reduces ZnO only above about 1200 K, but zinc boils at 1180 K, so what forms is zinc vapour that must be condensed quickly before CO₂ oxidises it back — which is why zinc was isolated long after copper, tin and iron.