Questions: Ellingham Diagrams and the Extraction of Metals

Question 1

Why do most lines on an Ellingham diagram for metal oxides slope upward as temperature increases?

• A. Because ΔH° becomes more negative at high temperature
• B. Because the reactions consume gaseous O₂, so ΔS° is negative and the slope (−ΔS°) is positive
• C. Because metal oxides melt at high temperature
• D. Because the equilibrium constant increases with temperature

Answer: (B)

The slope of ΔG° against T is −ΔS°. Forming a solid oxide from a solid metal and O₂ gas removes gas, so ΔS° is negative and the line rises with temperature.

Question 2

The line for 2C(s) + O₂(g) → 2CO(g) slopes downward, so carbon becomes a more effective reducing agent at higher temperatures.

• T. True
• F. False

Answer: True

This reaction turns one mole of gas into two, so ΔS° is positive and ΔG° falls as T rises. At high enough temperature the CO line passes below many metal oxide lines, and carbon can then reduce those oxides.

Question 3

At a given temperature, metal A can reduce the oxide of metal B under standard conditions when:

• A. A's line lies above B's line
• B. A's line lies below B's line
• C. Both lines have the same slope
• D. B's line crosses ΔG° = 0

Answer: (B)

Subtracting the two formation reactions gives the displacement reaction. Its ΔG° is the lower line minus the upper line, which is negative, so the metal with the lower line takes the oxygen.

Question 4

Explain, using the Ellingham diagram, why aluminum is produced by electrolysis rather than by heating alumina with carbon.

Model answer: The Al₂O₃ line lies far below the carbon lines at practical furnace temperatures, so carbon reduction of alumina is not favorable there. Electrolysis of alumina dissolved in molten cryolite (the Hall–Héroult process) supplies the needed free energy electrically.

The diagram shows the free-energy gap directly. When no affordable chemical reductant lies below an oxide's line, an electrical driving force is used instead.