#Curves in the Ellingham diagrams for the formation of metallic oxides are basically straight lines with a positive slope. The slope is proportional to ΔS, which is approximately constant with temperature.
#The lower the position of a metal's line in the Ellingham diagram, the greater is the stability of its oxide. For example, the line for Al (oxidation of aluminium) is found to be below that for Fe (formation of ) meaning that aluminium oxide is more stable than iron(III) oxide.
#Stability of metallic oxides decreases with increase in temperature. Highly unstable oxides like  and HgO easily undergo thermal decomposition.
#The formation free energy of carbon dioxide () is almost independent of temperature, while that of carbon monoxide (CO) has negative slope and crosses the  line near 700 °C. According to the Boudouard reaction, carbon monoxide is the dominant oxide of carbon at higher temperatures (above about 700 °C), and the higher the temperature (above 700 °C) the more effective a reductant (reducing agent) carbon is.
#If the curves for two metals at a given temperature are compared, the metal with the lower Gibbs free energy of oxidation on the diagram will reduce the oxide with the higher Gibbs free energy of formation. For example, metallic aluminium can reduce iron oxide to metallic iron, the aluminium itself being oxidized to aluminium oxide. (This reaction is employed in thermite.)
#The greater the gap between any two lines, the greater the effectiveness of the reducing agent corresponding to the lower line.
#The intersection of two lines implies an oxidation-reduction equilibrium. Reduction using a given reductant is possible at temperatures above the intersection point where the ΔG line of that reductant is lower on the diagram than that of the metallic oxide to be reduced. At the point of intersection the free energy change for the reaction is zero, below this temperature it is positive and the metallic oxide is stable in the presence of the reductant, while above the point of intersection the Gibbs energy is negative and the oxide can be reduced.