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Question

You are given an Ellingham diagram (deliberately not shown). It is given that for almost all practically achievable temperatures, the FeFe2O3 line lies above the AlAl2O3 line and ΔfG for both Al2O3 and Fe2O3 are negative. Which of the following statements is/are true?


A

Under the given conditions, neither metal can reduce the other oxide

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B

Fe can reduce Al2O3 into Al

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C

Al can reduce Fe2O3 into Fe

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D

Both b and c

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Solution

The correct option is C

Al can reduce Fe2O3 into Fe


A simple metric is that - provided at least one of the lines has a negative ΔfG at a particular Temperature, the oxide that lies “above” will be reduced by the metal that lies below in the Ellingham diagram. Note that at least the ΔfG for the formation of the oxide of the reducing agent/metal must be negative.

Given that ΔfG Fe Fe2O3 is negative and this line lies above the AlAl2O3 line in the Ellingham diagram for oxides. This means that

ΔfG Fe Fe2O3 is <0 and

ΔfG Al Al2O3 is<0

Such that ΔfG Fe Fe2O3 (less negative) >ΔfG Al Al2O3 (more negative)

Now both the free energy changes are negative. Also ΔfG Fe2O3 Fe is >0 and

ΔfG Al Al2O3 (more negative) - ΔfG Fe Fe2O3 (less negative) <0 and hence Al reducing Fe2O3 is a spontaneous process.

Note: - ΔfG Fe Fe2O3 = ΔfG Fe2O3 Fe and we are looking for thermodynamic viability of Fe2O3+Al Al2O3+Fe or
Al2O3 + FeFe2O3+Al

Please refer to the video on Ellingham diagram


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