Applications of the Ellingham Diagram

The Ellingham diagram's central practical use is to help select a suitable reducing agent and an appropriate temperature range for a given reduction. Conceptually, reducing a metal oxide to its metal is a COMPETITION for oxygen between the metal already in the oxide and the element being used as the reducing agent: if the metal's own oxide is the more thermodynamically stable of the two, oxygen stays bound to the metal and no reduction occurs; but if the OXIDE OF THE REDUCING ELEMENT is the more stable of the two, then oxygen transfers away from the metal oxide onto the reducing element, and reduction proceeds. Because the diagram plots ΔG of oxide FORMATION, and a more negative ΔG means a more stable (lower-energy) oxide, this comparison reduces to a simple visual rule: at any chosen temperature, whichever oxide-formation line sits LOWER on the diagram represents the more stable oxide, and the element of the LOWER line can reduce the oxide of any element whose line lies ABOVE it at that temperature.

Three worked applications follow from this rule.

1. Oxides that decompose without any reducing agent. The formation lines for Ag2O and HgO sit at the UPPER part of the diagram, with decomposition temperatures of only about 600 K and 700 K respectively. Because these lines are so high (their oxides so weakly stable), Ag2O and HgO decompose on heating alone, even in the complete absence of any reducing agent.

2. One metal reducing another metal's oxide. Any metal can reduce the oxide of another metal whose formation line lies ABOVE its own. For example, in the Ellingham diagram the chromium-oxide line (formation of Cr2O3) lies above the aluminium-oxide line (formation of Al2O3), meaning Al2O3 is the more stable oxide of the two -- so aluminium CAN be used to reduce chromic oxide (this is exactly the aluminothermic process of Section 1.3.2: Cr2O3 + 2Al → 2Cr + Al2O3). The same aluminium line, however, sits ABOVE the magnesium-oxide and calcium-oxide lines, so aluminium CANNOT reduce the oxides of magnesium or calcium, whose oxides are thermodynamically more stable still.

3. Carbon as a near-universal reducing agent, but only above a threshold temperature. Because the carbon (2C+O2=2CO) line has a NEGATIVE slope while almost every metal-oxide line has a positive slope, the carbon line cuts diagonally across nearly every metal-oxide line somewhere on the diagram -- meaning carbon (coke) can reduce essentially any of these oxides, PROVIDED the temperature is raised high enough to reach that crossing point. This is illustrated quantitatively for iron: the formation lines for FeO and CO intersect at around 1000 K. BELOW 1000 K, the carbon line lies ABOVE the iron line, meaning FeO is the more stable oxide at that temperature, so reduction of FeO by carbon is NOT thermodynamically feasible there. ABOVE 1000 K, the carbon line drops below the iron line, so coke CAN be used as the reducing agent above this temperature. This is confirmed by an explicit free-energy calculation at 1500 K: for 2Fe(s) + O2(g) → 2FeO(s), ΔG1 = -350 kJ mol⁻¹ ...... (1); for 2C(s) + O2(g) → 2CO(g), ΔG2 = -480 kJ mol⁻¹ ...... (2). Reversing reaction (1) gives 2FeO(s) → 2Fe(s) + O2(g), -ΔG1 = +350 kJ mol⁻¹ ...... (3); coupling reactions (2) and (3) gives 2FeO(s) + 2C → 2Fe(l,s) + 2CO(g), ΔG3 = ΔG2 + (-ΔG1) = -480 + 350 = -130 kJ mol⁻¹ ...... (4) -- a negative overall ΔG, confirming the reduction is thermodynamically favourable at 1500 K, and giving a standard free energy change per mole of FeO reduced of ΔG3/2 = -65 kJ mol⁻¹. …