Ellingham diagram — how to predict which reducing agent to use

Question

What is an Ellingham diagram? How do we use it to predict the feasibility of a metal oxide reduction? Why is carbon such an effective reducing agent at high temperatures? Why can aluminium reduce iron oxide but iron cannot reduce aluminium oxide?

(JEE Main + CBSE Board — graphical interpretation + reasoning)

Solution — Step by Step

The Ellingham diagram plots standard Gibbs free energy of formation ( $\Delta_f G^\circ$ ) of metal oxides against temperature. Each line represents the reaction:

2\text{M}(s) + \text{O}_2(g) \rightarrow 2\text{MO}(s)

Lines that are lower on the diagram have more negative $\Delta G^\circ$ — meaning the oxide is more stable and the metal has a stronger affinity for oxygen.

A metal can reduce the oxide of another metal only if its line lies below the oxide’s line on the Ellingham diagram at that temperature.

Why? If metal A has a more negative $\Delta G^\circ$ for oxide formation than metal B, then A has a stronger tendency to combine with oxygen. So A will “steal” oxygen from B’s oxide, reducing B to its metallic form.

Example: The Al₂O₃ line is below the Fe₂O₃ line at all temperatures. So aluminium can reduce iron oxide (this is the thermite reaction):

2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}

But iron CANNOT reduce aluminium oxide — its line is above Al₂O₃.

The carbon-oxygen line ( $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ and $2\text{C} + \text{O}_2 \rightarrow 2\text{CO}$ ) has a negative slope — it goes downward with increasing temperature (because CO is a gas, and $\Delta G$ becomes more negative as entropy increases at high T).

Most metal oxide lines have a positive slope (going upward).

This means: at some sufficiently high temperature, the carbon line crosses below the metal oxide line, making carbon a viable reducing agent. This is why carbon (as coke) can reduce ZnO, Fe₂O₃, and SnO₂ at high temperatures in blast furnaces.

However, carbon cannot reduce very stable oxides like Al₂O₃, MgO, or CaO — their lines are always below the carbon line.

Locate the metal oxide you want to reduce
Find a reducing agent whose line is BELOW the oxide line at your working temperature
The vertical gap between the two lines = the driving force ( $\Delta G$ for the overall reaction)
Larger gap = more feasible reaction

graph TD
    A["Ellingham Diagram Reading"] --> B["Find oxide to reduce"]
    B --> C["Find agent with line BELOW"]
    C --> D{"Gap between lines?"}
    D -->|Large| E["Reaction highly feasible"]
    D -->|Small| F["Reaction barely feasible"]
    D -->|"Agent line above"| G["Reduction NOT possible"]
    A --> H["Carbon line slopes DOWN"]
    H --> I["Crosses many oxide lines at high T"]
    I --> J["Carbon reduces ZnO, Fe2O3, SnO2"]
    style A fill:#fbbf24,stroke:#000,stroke-width:2px
    style E fill:#86efac,stroke:#000
    style G fill:#fca5a5,stroke:#000

Why This Works

The Ellingham diagram is a thermodynamic tool. The Gibbs free energy determines spontaneity — a reaction is spontaneous when $\Delta G < 0$ . If metal A forms a more stable oxide than metal B ( $\Delta G$ more negative), then the reaction “A reduces B-oxide” has $\Delta G < 0$ and is thermodynamically favourable.

The power of the diagram is visual prediction — you do not need to calculate $\Delta G$ values. Just compare line positions at the temperature of interest.

Common Mistake

The classic error: students write “the metal with a more negative $\Delta G$ can be reduced by carbon.” This is backwards. A more negative $\Delta G$ means the oxide is MORE stable and HARDER to reduce. Carbon can reduce oxides whose lines are ABOVE the carbon line, not below it. The reducing agent’s line must be below the oxide’s line.

For JEE: the reason electrolysis is needed for aluminium (Hall-Heroult process) is visible on the Ellingham diagram — the Al₂O₃ line is below the carbon line at all practical temperatures. Carbon simply cannot reduce aluminium oxide thermally. This connects metallurgy with electrochemistry — a cross-topic connection JEE loves.

Question

Solution — Step by Step

Why This Works

Common Mistake

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