Graphical Representation Ellingham Diagrams

The most compact display of free energies of formation is a plot of AGo Vs T. According to Eq (9.53), if AHo and ASo are independent of temperature, this plot is a straight line with a slope equal to -ASo and an intercept (at T = 0 K) of AHo. Such a line is called an Ellingham line, and is shown in Fig. 9.6b for formation of a metal oxide from the elemental constituents. A family of these lines for a number of metals with a common nonmetal (e.g., oxygen, chlorine) is called an Ellingham diagram. By far the most common is the oxide Ellingham diagram.

In order to permit comparison of different metal/metal oxide couples in the diagram, the formation reactions are written on a per-mole-O2 basis, rather than for one mole of the oxide. The general formula of a metal oxide is MmOn, where the integers m and n depend on the oxidation state (i.e., the valence) of the metal ion in the oxide. For all oxides, the valence of the oxygen ion is -2. The values of m and n must satisfy electrical neutrality of the crystal. Starting from the lowest-valence metal, the oxide formation reactions included in the Ellingham disgram are:

Monovalent (M = Ag, Cs,...) 4M + O2 = 2M2O Divalent (M = Mg, Ba,...) 2M + O2 = 2MO Trivalent (Al, La,.) 4/3M + O2 = 2/3 M2O3

Pentavalent (Nb, V, ...) 4/5M + O2 = 2/5M2O5 Hexavalent (Mo, U, .) 2/3M + O2 = 2/3MO3

The tetravalent metal reaction was treated in detail in Sect. 9.9. The general reaction is:

Since the metal and its oxide are pure in formation reactions, the analysis of Sect. 9.6 applies; the free energy of formation of the oxide is identical to the oxygen potential of the M/MaOb couple:

Figure 9.8 shows the Ellingham diagram for oxides. The diagram provides a direct visual assessment of the relative stabilities of the oxides of the metals. The metal/oxide couple with the lowest oxygen pressure is the most stable. According to Eq (9.56), this condition is equivalent to assigning oxide stability to the metal whose oxide has the most negative AGo.

Example: mixing calcium metal with uranium dioxide at any temperature results in complete reduction of UO2 by the reaction:

The arrow separating reactants and products in place of an equal sign indicates that the reaction is not an equilibrium reaction. With three components and four phases, the phase rule allows this system no degrees of freedom. The only condition that would permit equilibrium of the above reaction is

AGU /UO, _ AGCa/CaO . ^WCT^ Fig. 9.8 ShoWS that the Ca/CaO line doeS not intersect the U/UO2 line. Ca and UO2 react until one of the two is consumed.

Problems 9.12 and 9.21 deal with the conditions for equilibrium in other systems containing two different metals and their oxides.

The slopes of the lines in Fig. 9.8 are approximately equal, indicating that all reactions have roughly the same value of ASo. The reason for this uniformity of slope is the reaction entropy rule-of-thumb, illustrated by Eq (9.14) for dioxides. Applying this rule to the general oxidation reaction of Eq (9.55) suggests that the slopes of the lines in Fig. 9.8 should be ~ 200 J/mole-K, which is the molar entropy of one mole of O2(g). The actual slopes in Fig. 9.8 range from slightly more than 200 J/mole-K for simple oxides such as NiO and CaO to ~ 170 J/mole-K for more complex oxides such as UO2 and ZrO2.

The relative vertical positions of the lines in the diagram are a direct consequence of their enthalpies of formation, AHo, which are the intercepts at 0 K. The most exothermic reaction is 2Ca + O2 = 2CaO and the least exothermic is 2Hg + O2 = 2HgO.

Several metals exhibit multiple valence states; Fig. 9.8 contains lines for TiO and TiO2, Cu2O and CuO, and three oxides of uranium, UO2, U3O8 and UO3.Uranium ions in UO2 and UO3 have valences (or oxidation states) of 4+ and 6+, respectively. The intermediate oxide U3O8 has a mean valence of 5.33. Since ionic charges must be integers, this oxide contains 2/3 U5+ and 1/3 U6+. This combination is more stable than the pure 5+ oxide, U2O5, which does not exist. The higher the oxide (i.e., the larger the oxygen-to-metal ratio), the larger (less negative) is the free energy of formation. That is, the line for Ti/TiO2 lies above that for Ti/TiO.

Example Over what range of oxygen pressures is Cu2O the stable oxide of copper?

There are two Ellingham lines for copper oxides. The lowest oxide is represented by the equilibrium reaction:

for which Fig. 9.8 gives AGa/Ql 0 = -55 kcal/mole (-230 kJ/mole)at 400oC (673 K). At this temperature, the oxygen pressure of the Cu/Cu2O couple is:

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