Ellingham Diagram — Revision Notes

⚡ 30-Second Revision

Ellingham Diagram — Plot of $\Delta G^circ$ vs. T for oxide formation.
Equation — $\Delta G^circ = \Delta H^circ - T\Delta S^circ$.
Slope — Equal to $-\Delta S^circ$.

- Most metal oxides: Slope positive (upwards) because $\Delta S^circ < 0$ (gas consumed). - $2C(s) + O_2(g) \rightarrow 2CO(g)$: Slope negative (downwards) because $\Delta S^circ > 0$ (gas produced).

Intercept — $\Delta H^circ$ at $T=0$.
Stability — Lower line = more stable oxide = harder to reduce.
Reduction Feasibility — Reducing agent's oxidation line must be *below* the metal oxide's formation line.
Crossing Point — Temperature where $\Delta G^circ$ values are equal; indicates change in relative stability/reducing power.
Limitations — Predicts thermodynamic feasibility only, NOT reaction rate (kinetics).

2-Minute Revision

The Ellingham diagram is a crucial tool in metallurgy, graphically representing the standard Gibbs free energy change ($Delta G^circ$) for the formation of various metal oxides as a function of temperature.

It's based on $Delta G^circ = Delta H^circ - TDelta S^circ$. The y-axis shows $Delta G^circ$, and the x-axis shows temperature. Most metal oxide lines slope upwards because their formation consumes gaseous oxygen, leading to a decrease in entropy ($Delta S^circ < 0$), making the slope ($-Delta S^circ$) positive.

A lower line on the diagram signifies a more stable oxide, meaning it's harder to reduce.

For a substance to act as a reducing agent for a metal oxide, its own oxidation line must lie *below* the metal oxide's formation line at the desired temperature. This ensures that the overall $Delta G^circ$ for the coupled reduction reaction is negative.

The carbon line for the formation of CO ($2C + O_2 \rightarrow 2CO$) is unique as it slopes downwards ($Delta S^circ > 0$), making carbon an increasingly effective reducing agent at higher temperatures.

Crossing points between lines indicate temperatures where the relative stability of oxides changes, or where a reducing agent becomes thermodynamically effective. Remember, the Ellingham diagram only predicts thermodynamic feasibility, not the rate of reaction.

5-Minute Revision

The Ellingham diagram is a powerful graphical representation used in extractive metallurgy to predict the thermodynamic feasibility of reducing metal oxides. It plots the standard Gibbs free energy change ($Delta G^circ$) for the formation of various metal oxides against temperature. The fundamental equation governing these plots is the Gibbs-Helmholtz equation: $\Delta G^circ = \Delta H^circ - T\Delta S^circ$.

Key Interpretations:

1
Slope — The slope of each line is equal to $-Delta S^circ$. For most metal oxidation reactions (e.g., $2M(s) + O_2(g) \rightarrow 2MO(s)$), gaseous oxygen is consumed, leading to a decrease in entropy ($Delta S^circ < 0$). This results in a positive slope ($-Delta S^circ > 0$), meaning $Delta G^circ$ becomes less negative (or more positive) at higher temperatures, indicating that metal oxides generally become less stable at elevated temperatures.
2
Carbon Lines — The oxidation of carbon is crucial. The reaction $2C(s) + O_2(g) \rightarrow 2CO(g)$ involves an increase in gas moles ($Delta S^circ > 0$), giving it a negative slope. This makes carbon an increasingly powerful reducing agent at higher temperatures. The reaction $C(s) + O_2(g) \rightarrow CO_2(g)$ has $Delta S^circ \approx 0$, resulting in a nearly horizontal line.
3
Oxide Stability — A lower position on the diagram (more negative $Delta G^circ$) indicates a more stable oxide, which is harder to reduce.
4
Reduction Feasibility — For a reducing agent 'R' to reduce a metal oxide $M_xO_y$, the $Delta G^circ$ for the overall coupled reaction must be negative. Graphically, this means the line for the oxidation of the reducing agent ($R + O_2 \rightarrow RO_z$) must lie *below* the line for the formation of the metal oxide ($xM + O_2 \rightarrow M_xO_y$) at the temperature of interest. The $Delta G^circ$ for the reduction reaction is calculated as $Delta G^circ_{oxidation(R)} - Delta G^circ_{formation(MO)}$.
5
Crossing Points — The intersection of two lines signifies the temperature at which their $Delta G^circ$ values are equal. Above this temperature, the substance whose line was initially higher becomes thermodynamically easier to reduce by the substance whose line is now lower.

Example: In a blast furnace, CO reduces iron oxides at lower temperatures ($500-800^circ C$) because the $2CO + O_2 \rightarrow 2CO_2$ line is below the iron oxide lines. At higher temperatures (above $800^circ C$), the $2C + O_2 \rightarrow 2CO$ line drops below the iron oxide lines, making carbon itself the dominant reducing agent.

Limitations: The Ellingham diagram is a thermodynamic tool. It predicts whether a reaction *can* occur spontaneously, but it provides no information about the reaction rate (kinetics). A thermodynamically feasible reaction might still be very slow in practice.

Prelims Revision Notes

1
Definition — Ellingham diagram plots $\Delta G^circ$ (y-axis) vs. T (x-axis) for metal oxide formation reactions.
2
Equation — Based on $\Delta G^circ = \Delta H^circ - T\Delta S^circ$.
3
Slope Interpretation

* Slope = $-\Delta S^circ$. * Most metal oxidations ($M(s) + O_2(g) \rightarrow MO(s)$): $\Delta n_{gas} < 0$, so $\Delta S^circ < 0$. Thus, slope is positive (lines go upwards). Oxides become less stable at higher T.

* Carbon to Carbon Monoxide ($2C(s) + O_2(g) \rightarrow 2CO(g)$): $\Delta n_{gas} > 0$, so $\Delta S^circ > 0$. Thus, slope is negative (line goes downwards). Carbon becomes a stronger reducing agent at higher T.

* Carbon to Carbon Dioxide ($C(s) + O_2(g) \rightarrow CO_2(g)$): $\Delta n_{gas} = 0$, so $\Delta S^circ \approx 0$. Line is nearly horizontal.

1
Intercept — Represents $\Delta H^circ$ at $T=0$ K.
2
Phase Transitions — Sudden change in slope indicates melting or boiling point of metal or oxide, as $\Delta S^circ$ changes significantly.
3
Oxide Stability — A lower line (more negative $\Delta G^circ$) indicates a more stable oxide, harder to reduce.
4
Reduction Principle — For a metal oxide $M_xO_y$ to be reduced by a reducing agent $R$, the $\Delta G^circ$ for the overall reaction must be negative. Graphically, the line for $R$'s oxidation must be *below* the line for $M_xO_y$'s formation.
5
Crossing Points — The temperature at which two lines intersect. Above this temperature, the metal corresponding to the lower line can reduce the oxide corresponding to the upper line.
6
Example: Iron Extraction — CO reduces iron oxides at $500-800^circ C$. Above $800^circ C$, carbon (C) directly reduces iron oxides.
7
Limitations — The diagram only predicts thermodynamic feasibility (spontaneity), not the rate of reaction (kinetics). It assumes reactants/products are in standard states.

Vyyuha Quick Recall

Every Line Looks Interesting, Not Giving How Any Metal Reduces At Moment's Time.

Ellingham Lines: $\Delta G^circ$ vs.

*Self-correction during mnemonic creation: The mnemonic 'Lower line = Less stable' is incorrect. It should be 'Lower line = MORE stable'. This highlights a common misconception that the mnemonic should help avoid. Let's refine it.*

Revised Mnemonic:

Every Line Looks Interesting, Not Giving How Any Metal Reduces At Moment's Time.

Ellingham: $\Delta G^circ$ vs.