Ellingham Diagrams

Introduction

To decide whether carbon, CO or another metal can remove oxygen from an oxide, compare the free energies of competing oxidation reactions. An Ellingham diagram places standard Gibbs energy of oxide formation on a temperature axis. It turns many tables of thermodynamic data into a visual guide for extraction, provided every plotted reaction uses consistent oxygen normalisation.

Core explanation

An Ellingham line commonly represents formation of an oxide from an element and O₂, normalised per mole of O₂ consumed. For example, 2Mg + O₂ → 2MgO can be plotted with its standard ΔG° at each temperature. A more negative formation free energy means the oxide is thermodynamically more favoured from those elements under standard conditions. When comparing two metals at the same temperature and same O₂ normalisation, the metal whose oxide line lies lower can often reduce the oxide of a metal whose line lies higher, because forming the lower-line oxide releases more Gibbs energy than decomposing the upper-line oxide costs.

The slope of a line is approximately −ΔS° when ΔH° and ΔS° vary slowly with temperature, because ΔG° = ΔH° − TΔS°. For forming a condensed metal oxide from solid metal and gaseous O₂, gas is consumed, so ΔS° is often negative and the line tends to slope upward. Carbon oxidation to CO, 2C + O₂ → 2CO, increases gas moles from one O₂ to two CO, so its ΔS° can be positive and its line may slope downward. This different slope helps explain why carbon monoxide-forming reduction becomes more favourable at some higher temperatures. Exact line shapes can have kinks at melting, boiling or phase transformations where entropy and enthalpy change.

An intersection marks a temperature where two standard oxide-formation free energies are equal on the chosen normalisation. On one side, one oxidation route is thermodynamically favoured; on the other, the order reverses. This does not mean an industrial furnace instantaneously changes behaviour at the crossing. Real gas partial pressures change ΔG from ΔG°, reactants may not make contact, and kinetic barriers may be large. An Ellingham diagram establishes thermodynamic possibility under its assumptions, not actual reaction speed or final process yield.

For a reductant R competing with metal M for oxygen, subtract the metal-oxide formation reaction from the reductant-oxide formation reaction to obtain a net reduction. If ΔG° for R oxidation is more negative than for M oxidation at the same temperature and oxygen amount, the net M oxide reduction by R has negative ΔG°. A line far below does not by itself guarantee a clean product: carbon can form carbides with some metals, and metal volatility or slag chemistry can complicate extraction.

Oxygen pressure matters through the reaction quotient. For 2M + O₂ → 2MO, ΔG = ΔG° + RT ln Q, with Q involving oxygen activity or partial pressure and condensed-phase activities. Lower oxygen pressure changes the equilibrium. A diagram based on standard O₂ pressure is therefore a reference map; practical furnaces use specified gas compositions.

Step-by-step reasoning

1. Verify axes and that all reactions use the same amount of O₂. 2. At the chosen temperature, compare ΔG° positions for metal and reductant oxidation lines. 3. A reductant-oxide line below the target metal-oxide line suggests a negative standard net reduction ΔG°. 4. Use slope ≈ −ΔS° to explain broad trends and note phase-change kinks. 5. Qualify with oxygen pressure, kinetics, side products and physical separation needs.

Visual explanation

Plot temperature horizontally and ΔG° per mole O₂ vertically. Draw an upward-sloping metal-oxide line and a downward-sloping C-to-CO line crossing at one temperature. Shade the region where the carbon line is below as standard thermodynamic favourability for carbon to remove oxygen. Add a note “rate not shown” beside the graph.

Real-world analogy

Two competitors bid for the same oxygen. The more negative oxide-formation free energy is the lower bid and tends to win under the reference conditions. The analogy expresses thermodynamic preference, but the competitors still need a path to exchange oxygen and their actual concentrations can change the result.

Real-world example

A metallurgist considering CO reduction of iron oxide can compare Fe-oxide formation and CO-to-CO₂ formation free energies at furnace temperature. A favourable standard comparison helps justify the process, while furnace gas ratios and reaction kinetics determine how far reduction proceeds in each zone.

Why?

Why do many metal-oxide lines slope upward? Formation consumes O₂ gas and usually lowers entropy. Since slope is approximately −ΔS°, a negative formation entropy gives a positive slope in a ΔG° versus T plot.

Common misconception

“A lower line means the oxide is easier to reduce” reverses the basic reading: a lower formation line usually means a more thermodynamically stable oxide. Another mistake is interpreting a crossing as an exact reaction-onset temperature; nonstandard gas pressures and kinetics can shift observed behaviour.

Worked example

At a hypothetical temperature, let ΔG° for forming one O₂-equivalent of MO be −300 kJ and for forming the reductant oxide RO be −450 kJ. Reversing MO formation costs +300 kJ, while RO formation releases −450 kJ. Adding gives net reduction ΔG° = −150 kJ for transfer of that oxygen amount from M to R. The lower RO line therefore indicates a thermodynamically favourable standard reduction. The calculation says nothing about reaction rate.

Quick check

1. If a reductant's oxide-formation line lies above the target metal-oxide line at a chosen temperature, is standard reduction generally favoured by that reductant? Answer: No. The reductant's oxide formation is less negative, so oxygen transfer from the target oxide to the reductant would generally have positive standard ΔG° on a matched oxygen basis.

Exam focus

Check reaction normalisation and read both lines at the same temperature. State “thermodynamically favourable under standard conditions,” not “automatically occurs rapidly.” Explain slope from entropy and line kinks from phase transitions. If numerical ΔG° values are given, reverse the target oxide formation and add the reductant oxidation with correct signs.

Advanced insight

An Ellingham plot can be related to equilibrium oxygen partial pressure: a sufficiently low pO₂ can destabilise an oxide even when its standard formation ΔG° is negative. Gas mixtures such as CO/CO₂ set an effective oxygen chemical potential. The diagram is therefore a graphical thermodynamic map of redox conditions, not merely a ranking of metals.

Summary

Ellingham diagrams plot standard oxide-formation Gibbs energy versus temperature, usually per common O₂ amount. A reductant whose oxidation line lies lower than a metal-oxide line can often reduce that oxide thermodynamically at that temperature. Slopes reflect entropy, crossings reflect standard free-energy equality, and practical extraction still depends on gas conditions, kinetics and side chemistry.

Practice questions

1. Why should two oxide formation reactions be normalised to the same O₂ amount before comparing line heights? Answer: Gibbs energies scale with reaction stoichiometry. Comparing different oxygen amounts would mix different reaction extents and could give a false conclusion about which oxygen-transfer route has lower free energy.

2. What does an upward slope suggest about ΔS° for oxide formation? Answer: Since slope ≈ −ΔS°, an upward slope indicates negative ΔS°, often because gaseous O₂ is consumed to make condensed oxide.

3. A metal-oxide line and a carbon-to-CO line cross at T . Does the diagram guarantee rapid reduction above T ? Answer: No. A crossing changes the standard thermodynamic ordering, but reaction kinetics, gas partial pressures, contact, phase behaviour and side reactions control actual reduction.