The Ellingham Diagram: Which Metal Oxide Reduces Which

What the diagram plots — and the one convention that makes it work

An Ellingham diagram is a plot of the standard Gibbs energy of formation of a metal oxide, ΔG°f, against absolute temperature. The single non-negotiable convention is that every reaction is written to consume exactly one mole of O₂. So we write 2Mg + O₂ → 2MgO and ⁴⁄₃Al + O₂ → ⅔Al₂O₃, not the per-mole-of-metal forms. Normalizing to one mole of O₂ is what lets you compare a divalent metal like Mg directly against a trivalent one like Al on a single vertical scale: the y-axis is energy per mole of oxygen reacted, a common currency.

The lines run from the top-right (weakly negative, near 0 kJ) down to the bottom-left (strongly negative, below −1100 kJ near room temperature for CaO and MgO, though these low-temperature lines rise substantially with heating — the MgO line reaches about −960 kJ by 1000 °C). More negative means the oxide is more thermodynamically stable — the metal has a stronger appetite for oxygen. Reactive metals such as Ca, Mg, Al and Ti anchor the bottom; noble metals such as Ag and the Ag₂O/Au lines sit near the top, barely below zero, which is precisely why silver tarnishes reluctantly and gold does not oxidize at all under normal conditions.

Because ΔG° = ΔH° − TΔS° and both ΔH° and ΔS° are roughly constant over a wide temperature window, each curve is close to a straight line with slope −ΔS° and a y-intercept of ΔH° when extrapolated back to T = 0 K (at the origin of the plotted axis, typically 298 K, ΔG° ≠ ΔH°). That linearity is the whole reason the diagram is legible: it turns a mess of temperature-dependent equilibria into a ruler-and-eye exercise. The lines are not perfectly straight — ΔH° and ΔS° drift slightly with T — but the deviation is small enough to ignore for the qualitative reasoning the chart is used for.

Why the lines slope the way they do — the entropy argument

Take the differential of the Gibbs relation: from ΔG° = ΔH° − TΔS°, the slope of an Ellingham line is d(ΔG°)/dT = −ΔS°. Everything about the diagram's shape follows from the sign and size of ΔS° for the specific reaction written.

For a typical oxidation, a solid (or liquid) metal plus gaseous O₂ becomes a solid oxide: 2Zn(s) + O₂(g) → 2ZnO(s). One mole of gas disappears and nothing gaseous is produced, so ΔS° is strongly negative (roughly −190 to −220 J K⁻¹ mol⁻¹, dominated by the standard entropy of O₂, S°(O₂) ≈ 205 J K⁻¹ mol⁻¹). A negative ΔS° gives a positive slope, so these lines rise as temperature increases. Physically, heat makes the oxide progressively less stable relative to metal-plus-oxygen, because the entropy penalty of locking up gaseous oxygen costs more free energy (the TΔS° term) the hotter you go.

The lines also have kinks. Wherever a species undergoes a phase change, its entropy jumps, so ΔS° of the reaction — and hence the slope — changes abruptly. When the metal melts, ΔS° of the reaction becomes more negative (the reactant gained entropy on melting), so the line bends to a steeper positive slope; the effect is larger, and more visible, when the metal boils. Classic examples are the sharp upward break in the Mg line at its boiling point (~1090 °C) and the kink in the zinc line at zinc's boiling point (~907 °C). Reading these breaks correctly is essential: they can move a crossover temperature by hundreds of degrees.

The reading rule: lower line reduces higher oxide

Here is the operational heart of the diagram. Suppose you want to know whether element A can reduce the oxide of element B. Write the two formation reactions, both per mole of O₂, and subtract to get the metallothermic reaction. The overall ΔG° equals ΔG°(A-oxide) − ΔG°(B-oxide). If A's line lies below B's line at your temperature, that difference is negative and the reduction is spontaneous: A pulls oxygen off B's oxide.

• Aluminothermic (thermite) reduction: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe. The Al line sits far below the Fe line at all accessible temperatures, so ΔG° is strongly negative (around −840 kJ for the overall reaction, which corresponds to ≈ −560 kJ per mole of O₂ transferred). This is the thermite reaction used to weld rail track — it reaches ~2500 °C and casts molten iron in situ.
• Kroll process: 2Mg + TiCl₄ → 2MgCl₂ + Ti runs on the chloride analogue of the same logic; the oxide diagram's cousin, the Ellingham chloride diagram, shows Mg below Ti. Mg is chosen because its line lies below Ti's.
• Silver won't reduce iron: the Ag₂O line is far above the FeO line, so silver metal cannot pull oxygen from iron oxide — the sign flips.

A crucial subtlety: the diagram tells you only whether a reaction is thermodynamically allowed (ΔG° < 0), not whether it is fast. Kinetics, mass transport, and the need to reach ignition or diffusion temperatures are invisible on the chart. A reaction with ΔG° = −500 kJ can still sit inert at room temperature for centuries if the activation barrier is high or the reactants are separated by a passivating oxide film — exactly why aluminium metal survives in air despite its enormous oxygen affinity.

Carbon: the reductant whose line goes the wrong way

Carbon is special, and its behavior is the single most important feature of the diagram for industrial metallurgy. Three carbon reactions matter: C + O₂ → CO₂ (nearly horizontal, ΔS° ≈ 0 because gas moles are conserved, 1 → 1), 2C + O₂ → 2CO (steep downward slope, because 1 mole of gas becomes 2, ΔS° ≈ +180 J K⁻¹ mol⁻¹), and 2CO + O₂ → 2CO₂ (upward slope like a metal, 3 gas moles → 2).

The 2C + O₂ → 2CO line is the only one on a standard oxide diagram that falls with temperature. Because it slopes down while every metal-oxide line slopes up, the carbon-to-CO line will eventually cross below every metal oxide line if you go hot enough. At the crossover temperature, C becomes able to reduce that oxide via C(s) + MₓO(s) → xM + CO(g). Below the crossover carbon cannot do it; above, it can. Typical crossovers: carbon reduces ZnO above ~950 °C, FeO/Fe₂O₃ above ~700 °C (the working principle of the blast furnace), MgO only above ~1900 °C, and Al₂O₃ only above ~2000 °C — which is why aluminium is never carbon-smelted but electrolyzed (Hall–Héroult).

The three carbon lines intersect near ~700 °C. Below that the CO₂-forming route (C + O₂ → CO₂) is lower and CO₂ is the stable product; above it, the CO-forming route dominates and CO is the effective reductant. This crossover is intimately tied to the Boudouard equilibrium, C + CO₂ ⇌ 2CO, which is endothermic and pushed toward CO at high temperature (Le Chatelier: 2 gas moles favored by heat and low pressure). In a blast furnace, hot coke regenerates CO from CO₂, and it is CO gas — not solid carbon — that does most of the actual reduction of iron ore in the upper stack.

A worked example: reading equilibrium pressures and crossovers

The diagram is quantitative, not just qualitative. Two auxiliary scales are usually printed around the border and read by drawing a straight line from a labelled origin point through a point on a curve. The p(O₂) scale gives the oxygen partial pressure at which metal and oxide coexist, because ΔG° = RT ln p(O₂) for the reaction 2M + O₂ → 2MO (with unit-activity solids). The p(CO)/p(CO₂) and p(H₂)/p(H₂O) scales give the reducing gas ratio needed to reduce a given oxide with CO or H₂.

Concretely, take the reduction of hematite by CO at 1000 K (~727 °C). For 2CO + O₂ → 2CO₂ we look up ΔG° ≈ −450 kJ, and for ⁴⁄₃Fe + O₂ → ⅔Fe₂O₃, ΔG° ≈ −370 kJ (values per mole O₂). Subtracting, the net ΔG° for Fe₂O₃ + 3CO → 2Fe + 3CO₂ is negative, so the reduction proceeds — consistent with blast-furnace practice. Because it is close to the crossover, the required gas is CO-rich; the equilibrium p(CO)/p(CO₂) ratio computed from ΔG° = −RT ln K is substantial, which — together with finite residence time and counter-current gas–solid contacting — is why furnaces run CO-rich and never reach full conversion in a single pass.

Another number worth internalizing: to decompose an oxide without a reductant (2MO → 2M + O₂), you need ΔG° > 0, i.e. temperatures above where the metal's line crosses ΔG° = 0. Silver oxide (Ag₂O) crosses zero near ~190 °C, so gentle heating decomposes it to silver and oxygen — the reason Ag₂O is an unstable curiosity while Al₂O₃ (which would need temperatures well over 3000 °C) is a refractory ceramic. Mercuric oxide (HgO) decomposes similarly on heating, and it was the thermal decomposition of HgO — focusing sunlight on it with a burning lens — that Joseph Priestley used to discover oxygen in 1774, the classic lecture demonstration.

History, limits, and where the diagram breaks down

Harold Johnson Thomas Ellingham published the original chart in 1944 in the Journal of the Society of Chemical Industry, plotting oxide and chloride formation energies for extractive metallurgists. Francis D. Richardson and Jeffes extended it in 1948 with the nomographic p(O₂), CO/CO₂ and H₂/H₂O scales, so the fully annotated version is sometimes called a Richardson–Ellingham diagram. The same construction is now drawn for sulfides, chlorides, fluorides and nitrides, each with its own reference gas (S₂, Cl₂, etc.).

The diagram's limits are as important as its power. It is purely thermodynamic: it never predicts rate, ignition temperature, or whether a protective oxide film will passivate the surface. It assumes unit activity for pure condensed phases; real ores are solutions and slags where activities differ from one, shifting effective lines. The straight-line idealization fails near phase transitions and over very wide temperature ranges, where ΔHₚ and ΔSₚ drift. And it silently assumes the stated stoichiometry of the oxide — many metals form several oxides (FeO, Fe₃O₄, Fe₂O₃; Cu₂O vs CuO), each with its own line, and the stable one changes with temperature and p(O₂).

Finally, the diagram cannot capture reductions that beat thermodynamics by removing product from the system. Electrolytic reduction (Hall–Héroult for Al, Downs cell for Na) does work that ΔG° alone forbids by supplying electrical energy, and vacuum or inert-sweep processes push otherwise-unfavorable reactions by dropping p(O₂) or p(CO) far below standard state — which is exactly what the border scales quantify. Understood with these caveats, the Ellingham diagram remains the most compact and durable predictive tool in extractive metallurgy: one chart that explains why iron is smelted with coke, aluminium with electricity, and titanium with magnesium.

Why oxide lines slope up but the carbon-to-CO line slopes down

Feature	Typical metal oxide, e.g. 2Zn + O₂ → 2ZnO	Carbon to CO, 2C + O₂ → 2CO
Gas moles: reactant → product	1 mol O₂ → 0 mol gas (net −1)	1 mol O₂ → 2 mol CO (net +1)
Sign of ΔS° for reaction	Negative (gas consumed, disorder falls)	Positive (gas produced, disorder rises)
Slope of line (= −ΔS°)	Positive: line rises with T	Negative: line falls with T
Consequence	Oxide gets LESS stable as T rises	CO gets MORE favorable as T rises
Practical upshot	Reduction easier at high T	Carbon becomes a universal reductant above its crossover with each metal