The Ellingham Diagram: Thermodynamics Deciding an Industrial Process
A single plot that answers which reducing agent will work, and at what temperature, for any metal oxide.
What is plotted
The vertical axis is the standard free energy change for forming one mole of oxide from the metal and oxygen, normalised so every reaction consumes the same amount of oxygen. The horizontal axis is temperature. Normalising the oxygen is essential, or the lines could not be compared.
Why the lines slope as they do
Since ΔG° = ΔH° − TΔS°, the slope of each line is −ΔS°. So the sign of the entropy change determines the direction of the slope, and the entropy change is dominated by what happens to the gas.
| Reaction type | Gas moles | ΔS° | Slope |
|---|---|---|---|
| Solid metal to solid oxide | Consumes one mole of oxygen | Negative | Upward |
| Carbon to carbon monoxide | One mole in, two moles out | Positive | Downward |
| Carbon to carbon dioxide | One mole in, one out | Near zero | Nearly flat |
Reading the diagram
A metal can reduce the oxide of another if its own line lies below the other's at the temperature concerned. The vertical gap between them is the free energy change for the reduction, so a larger gap means a more strongly favourable reaction.
Where two lines cross, the reduction becomes feasible only above or below that temperature. Finding the crossing point is a standard task, and it identifies the minimum temperature at which a process will work.
Breaks in the lines
A sharp change of slope marks a phase change — melting or boiling of the metal or its oxide — because the entropy change of the reaction alters at that point. Boiling produces the larger break, since vaporisation involves a much bigger entropy change than melting.
Choosing a reduction method
- Locate the metal oxide line.
- See which potential reducing agent's line lies below it, and at what temperature.
- If carbon works at an accessible temperature, use it — it is cheapest.
- If not, consider a more reactive metal as reductant.
- If no chemical reductant works, electrolysis is required.
The most reactive metals sit at the very bottom of the diagram, below everything including carbon at all practical temperatures. That is precisely why they are obtained electrolytically, and stating the reason that way is a complete answer.
What the diagram does not tell you
- Nothing about rate. A thermodynamically feasible reduction may be impractically slow.
- Nothing about side reactions — carbide formation is a real problem for some metals and is invisible on the diagram.
- It assumes standard states, so real conditions may differ.
- It assumes pure phases, whereas real systems involve solutions and slags.
Carbide formation is the practical reason carbon is not used for certain metals despite the diagram permitting it, and mentioning it distinguishes a thorough answer from a purely thermodynamic one.
Frequently asked questions
Why must the oxygen be normalised?
Because the free energy change depends on how much reaction is written. Comparing lines requires each to refer to the same quantity of oxygen consumed.
Why does the carbon dioxide line stay nearly flat?
Because one mole of gas is consumed and one produced, so the entropy change is close to zero and the slope is nearly horizontal.
What does a line crossing mean?
That above the crossing temperature the lower reagent can reduce the other's oxide, and below it the reverse holds. It identifies the threshold temperature for a process.
Why is electrolysis needed for the most reactive metals?
Because their oxide lines lie below every available chemical reducing agent's line at all practical temperatures, so no chemical reduction is thermodynamically possible.
Preparing for a chemistry entrance exam?
ABC Chemistry runs focused IIT-JAM, CSIR-NET, GATE and CUET-PG Chemistry coaching at our centre and through live online classes for students across India.
Call / WhatsApp: +91 92121 42427