Ellingham DiagramJEE Advanced

Concept

An Ellingham diagram plots ΔG° of oxide formation against temperature. Metal lines slope up (gas O₂ consumed, ΔS < 0). The C→CO line slopes down (gas count doubles). Wherever carbon's line dips below a metal's line, carbon can strip that oxide — the thermodynamic logic of every blast furnace.

Key formula

Derivation

Couple the reactions: MO + C → M + CO has ΔG = ΔG(C→CO) − ΔG(M→MO). Negative exactly when the carbon line is lower.

Slopes are −ΔS: metal oxidation loses one O₂ (slope +0.2 kJ/K-ish); 2C + O₂ → 2CO gains net one gas mole (slope −0.18). Lines must cross — above the crossover, coke wins. Fe: ~1000 K (easy); Al: >2300 K (hence electrolysis); Mg: hotter still.

Scenarios to explore

• Ellingham Diagram — ΔG° vs T — when carbon can steal a metal's oxygen.

Real-world applications

• Blast-furnace iron (coke reduces Fe₂O₃ above ~1000 K).
• Why aluminium needs Hall–Héroult electrolysis, not smelting.
• Pidgeon process: Si reduces MgO only under vacuum at 1400 °C.

JEE exam tips

• The C→CO line is the magic one — its downward slope guarantees carbon beats everything eventually.
• Below ~980 K CO is the better reducer; above it, C itself — crossover of the CO/CO₂ lines.
• A metal can reduce any oxide whose line lies ABOVE its own (Al reduces Fe₂O₃ — thermite).

Common mistakes

• Reading the lowest line as least stable — LOWER = MORE stable oxide.
• Ignoring the kink at melting/boiling points of the metal (slope changes).
• Thinking Ellingham says anything about RATE — pure thermodynamics.

Exam traps to avoid

• ΔS ≈ 0 for C + O₂ → CO₂ (1 gas → 1 gas): that line is nearly horizontal.
• Ellingham uses ΔG°, so predictions assume standard pressures — real furnaces shift slightly.