Concept 06 / 06General Principles and Processes of Isolation of Elements
Thermodynamic Principles of Metallurgy: Ellingham Diagrams
Gibbs energy criterion for reduction of metal oxides, coupling of reactions, Ellingham diagrams (Delta G versus T for oxide formation), slopes of the lines and the special behaviour of the C to CO line, reading the diagram to choose reducing agents and temperatures (iron, zinc, aluminium), and limitations.
Reduction of a metal oxide, , is feasible only if its is negative. Since , temperature matters. Reduction can be viewed as coupling two reactions: the decomposition of the metal oxide (unfavourable, ) and the oxidation of the reducing agent (favourable, ); the overall is the sum.
Ellingham diagram (see figure). A plot of against for the formation of oxides, all written per mole of (such as ).
Most metal oxide lines slope upwards: the reaction consumes gaseous oxygen, so is negative and becomes more positive as rises. A sudden change of slope occurs at a phase change of the metal (melting, or boiling as for Zn and Mg), where becomes more negative.
The line () slopes downwards, because one mole of gas gives two (). The line is almost horizontal (). The line slopes upwards (three moles of gas give two).
Reading the diagram. A metal (or reducing agent) whose line lies lower can reduce the oxide of any metal whose line lies higher at that temperature; the vertical gap gives of the overall reduction. Where the carbon line crosses a metal oxide line, carbon becomes able to reduce that oxide above the crossing temperature.
Iron: the line crosses the FeO line near , so above this temperature coke reduces FeO; below about , the line lies lower than the FeO line, so CO is the effective reductant in the cooler upper part of the blast furnace. Both roles appear in the blast furnace.
Zinc: carbon reduces ZnO only at high temperature (about in practice), and zinc distils off as vapour.
Aluminium and magnesium: their oxide lines lie so low that carbon could reduce them only above about to , which is uneconomic (and Al forms carbides); these metals are obtained by electrolysis. Conversely, Al lies below and , so it reduces them (thermite process).
Copper: lies high, so it is easily reduced (even by in self-reduction).
Limitations. The diagram gives only thermodynamic feasibility, not rates (kinetics); it assumes reactants and products are in equilibrium, which is not always true in practice.
Key Takeaways:
- Plot per mol against ; lower line reduces higher oxide.
- Metal oxide lines slope up (); slopes down ().
- C reduces FeO above about ; CO is the better reductant below about .
- Al and Mg oxides need electrolysis, not carbon.
- Pitfall: a feasible reaction by Ellingham may still be too slow; the diagram says nothing about rate.
Brain-Teaser 1: Using the approximate lines and (kJ), find the minimum temperature at which carbon reduces FeO.
Solution: For , . This is negative when , about .
Brain-Teaser 2: Why does the slope of the Mg to MgO line increase sharply above about ?
Solution: Magnesium boils near this temperature. Above it, the reaction is , consuming three moles of gas instead of one, so becomes much more negative and the slope () becomes much larger.
Brain-Teaser 3: Can aluminium reduce MgO? Can magnesium reduce ? Answer with reference to the Ellingham diagram.
Solution: Below the crossing temperature (roughly to ) the MgO line lies below the line, so Mg can reduce but Al cannot reduce MgO. Above the crossing point (where the lines interchange, due to the boiling of Mg), aluminium can reduce MgO.
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