Ellingham Diagram-Predicting the Feasibility, Examples, Applications, Limitations, Practice problems and FAQs

Have you participated in a race?

For sure you have! We all have participated in the race in some event or the other in our school.

So whoever crosses the finish line wins the race.

Let us take two metal oxides. What happens when we take two metal oxides who are competing to get reduced first. The one with a lower affinity toward oxygen will be reduced by another metal that has higher affinity toward oxygen. But how can we compare the affinities of metal towards oxygen?

Ellingham diagram is the answer.

Ellingham diagram helps us to compare the affinities of metal towards oxygen. Such that when the plot of one metal oxide is below the others then the former metal is capable of reducing the later metal oxide into its corresponding metal.

Let's study the Ellingham diagram in detail!

Table of contents:

Predicting the Feasibility of Thermal Reduction of an Ore
Ellingham Diagram
Example of Reduction
Applications of Ellingham diagram
Limitations
Practice problems
Frequently Asked Questions-FAQs

Predicting the Feasibility of Thermal Reduction of an Ore

Let us consider various cases for change in gibbs free energy.

If ΔG = -ve , reaction is spontaneous

If ΔG = +ve , reaction is non-spontaneous

If ΔG = O , reaction is in equilibrium

For the process to be spontaneous, change in gibbs free energy must be negative.

Lets us understand a mathematical relation between H, G and S graphically. In the above given reaction entropy decreases and hence with increasing temperature magnitude of TS decreases. The magnitude of H also decreases. But rate of decrease of H is less as compared to the S. and at some point both become equal and G become zero.

Ellingham Diagram:

It is a graph that provides the basis for considering the choice of reducing agent in reduction of oxides. 1. Generally, it consists of the plots of Δ_G_f₀ vs T for the formation of oxides.

2. The slope of the graph for formation of oxides(xM MxO) is observed to be positive. ΔG further increases with increases in temperature.

3. Generally, free energy follows a straight line. Except when the element melts or evaporates.

Example:

Hg HgO line changes its slope at 3650C because at this temperature Hg boils.

Similarly,

Mg MgO line changes its slope at 11200C because at this temperature Mg boils.

4. There is a point in the curve of the graph above which metal oxide (MxO) decomposes on its own and below that point ΔG = -ve, which means metal oxide is stable below that point. This point is the point of intersection of two lines. Here one metal is used to reduce the oxide of another metal at a particular temperature. So any metal can reduce the oxide of another metal which lies above it in the Ellingham diagram. This is because the ΔG becomes more negative by an amount equal to the difference between the two graphs at a particular temperature.

At the point of intersection of the Al2O3 and MgO curves (marked as “A” in the diagram), the ΔGr becomes zero for the two reactions. The temperature at point A is 1500°C. So, below 1500°C temperature, the graph of MgO lies below the graph of Al2O3, so below this temperature Al cannot reduce MgO , but above this temperature graph of Al2O3 shifts below the graph of MgO so above 1500°C, Al can reduce MgO .

Also Al reduces the oxides of Zn, Fe, Ti as it lies below all these metals.

The reducing agent forms an oxide when a metal oxide is reduced.

The oxide of the metal decomposes as

MxO(s) xM(s or l) + ½ O2(g)………(i)

Carbon as reducing agent:

Oxidation of C is given as:

C(s) + ½ O2(g) CO(g)………(ii)

Adding eq(i) and (ii)

MxO(s) +C(s) xM(s) + CO(g)

Here C (coke) acts as a reducing agent.

All the graphs are increasing and on increasing temperature, G of formation of metal oxide become less negative, while on increasing temperature, G of C CO reaction becomes more negative. So C can be used to reduce any metal from their corresponding metal oxide.

Carbon monoxide as a reducing agent:

Oxidation of CO is given as:

CO (s)+ ½O2 (g) CO2 (g………(iii)

Adding eq (i) and (iii)

MxO(s) + CO(s) xM(s or l) + CO2(g)………(iv)

Here, CO will act as a reducing agent.

Example of Reduction:

Cr2O3 can be reduced by Al at 12000C.

$\frac{2}{3} C r_{2} O_{3} \to \frac{4}{3} C r + O_{2}; Δ G^{0} = + 475 k J . . . . . . . . (1)$

$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}; Δ G^{0} = - 800 k J . . . . . . . . (2)$

On adding eq (1) and (2)

$\frac{2}{3} C r_{2} O_{3} + \frac{4}{3} A l \to \frac{4}{3} C r + \frac{2}{3} A l_{2} O_{3}; Δ G^{0} = - 300 k J$

As ΔG0 is negative, the reduction of Cr2O3 by Aluminium is possible.

Limitations:

1. It cannot tell us about the kinetics of the reaction; it tells us whether the reaction is viable or not. There is no explanation for the reaction's rate.

2. It is assumed that the reactants and products of a reaction are always in equilibrium, although this may not always be the case. It's possible for the reactant and products to also exist in the solid-liquid state.

3. The study is thermodynamic, the process of studying the reactions depicted in the diagram might be slow.

Applications of Ellingham diagram:

1. It is used to see how easily metal oxides and sulfides can be reduced.

2. It is used in metallurgy to forecast the temperature at which elements and their oxides reach equilibrium.

3. Using the Ellingham diagram, we can forecast the conditions(temperature) under which a mineral can be converted to its metal.

4. It is used to figure out which reducing agent is optimal for reducing a particular metallic oxides.

5. It can be used in the extraction of various metals:

Example:

1. Extraction of Fe from its oxide 2. Extraction of Cu from Cu2O 3. Extraction of Zn from ZnO

Practice Problems:

1. Ellingham diagram is a graphical representation of:

A. ΔH vs T
B. ΔG vs T
C. ΔH vs P
D. (ΔG – TΔS) vs T

Answer: B

Solution: Ellingham diagram depicts the free energy of reaction with respect to temperature.

2. With respect to an ore, Ellingham diagram helps to predict feasibility of its ………………… .

A. Electrolysis
B. Zone refining
C. Thermal reduction
D. Vapour phase refining

Answer: C

Solution: Ellingham diagram is the graph between ΔG and T of various oxides. These help in predicting the feasibility of thermal reduction of ores.

3. The correct statement regarding the given Ellingham diagram is:

A. At 8000C, Cu can be used for the extraction of Zn from ZnO
B. At 14000C, Al can be used for the extraction of Zn from ZnO
C. At 5000C, coke can be used for the extraction of Zn from ZnO
D. Coke cannot be used for extraction of Cu from Cu2O

Answer : B

Solution: According to the given diagram, Al can reduce ZnO.

$3 Z n O + 2 A l ⟶ 3 Z n + A l_{2} O_{3}$

At 1400°C, ΔG for the formation of Al₂O₃ is a more negative compound than 'zero'.

4. According to the following diagram, A reduces BO₂ when the temperature is:

A. < 14000C
B. > 14000C
C. > 12000C but < 14000C
D. < 12000C

Answer : B

Solution: In the Ellingham diagram, the line of the element that lies below can reduce the oxide of the element which lies above it. Therefore, when T > 14000C, the gibbs free energy for the formation of AO2 will be more negative. For T > 14000C, △Gr<0 for the formation of AO2 So option B is correct.

Frequently Asked Questions-FAQs

1. What is the reason behind the change in the slope of xM MxO lines in ellingham diagram?
Answer : This is because there is large change in entropy associated with change in state which causes change in slope of formation of metal oxides from their corresponding metal.

2. What happens after the point of intersection of two curves in the ellingham diagram?
Answer : After the point of intersection, △rG0 becomes more negative for the redox reaction combined including the MxO reduction.

3. What does the difference in two rG values signifies?
Answer : It determines whether the reduction of an oxide of the upper line is feasible by an element present in the lower line. Larger the difference between the upper and lower lines, the easier the reduction.

4. What does the positive value of G slopes signifies?
Answer : At high temperature G formation of metal oxide becomes positive, indicating that metal oxide has become unstable and can decompose into metal & oxygen.

Related topics:

Extraction of Crude Metal from concentrated Ore	Extraction of iron
Occurrence of Metals	Concentration of ores
Some important Terms used in Metallurgy	Refining of metals from concentrated ores