Question

You are given an Ellingham diagram (deliberately not shown). It is given that for almost all practically achievable temperatures, the $Fe\to F{e}_2{O}_3$ line lies above the $Al\to A{l}_2{O}_3$ line and ${\Delta }_f{G}^{\circ }$ for both $A{l}_2{O}_3$ and $F{e}_2{O}_3$ are negative. Which of the following statements is/are true?

• A. Under the given conditions, neither metal can reduce the other oxide
• B. Fe can reduce $A{l}_2{O}_3$ into Al
• C. Al can reduce $F{e}_2{O}_3$ into Fe
• D. Both b and c

Answer 1

The correct option is C

Al can reduce $F{e}_2{O}_3$ into Fe

A simple metric is that - provided at least one of the lines has a negative ${\Delta }_f{G}^{\circ }$ 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 ${\Delta }_f{G}^{\circ }$ for the formation of the oxide of the reducing agent/metal must be negative.

Given that ${\Delta }_f{G}^{\circ }$ Fe $\to$ $F{e}_2{O}_3$ is negative and this line lies above the $Al\to A{l}_2{O}_3$ line in the Ellingham diagram for oxides. This means that

${\Delta }_f{G}^{\circ }$ Fe $\to$ $F{e}_2{O}_3$ is <0 and

${\Delta }_f{G}^{\circ }$ Al $\to$ $A{l}_2{O}_3$ is<0

Such that ${\Delta }_f{G}^{\circ }$ Fe $\to$ $F{e}_2{O}_3$ (less negative) $>{\Delta }_f{G}^{\circ }$ Al $\to$ $A{l}_2{O}_3$ (more negative)

Now both the free energy changes are negative. Also ${\Delta }_f{G}^{\circ }$ $F{e}_2{O}_3$ $\to$ Fe is >0 and

${\Delta }_f{G}^{\circ }$ Al $\to$ $A{l}_2{O}_3$ (more negative) - ${\Delta }_f{G}^{\circ }$ Fe $\to$ $F{e}_2{O}_3$ (less negative) <0 and hence Al reducing $F{e}_2{O}_3$ is a spontaneous process.

Note: - ${\Delta }_f{G}^{\circ }$ Fe $\to$ $F{e}_2{O}_3$ = ${\Delta }_f{G}^{\circ }$ $F{e}_2{O}_3$ $\to$ Fe and we are looking for thermodynamic viability of $F{e}_2{O}_3+Al$ $\to$ $A{l}_2{O}_3+Fe$ or
$A{l}_2{O}_3$ + $Fe\to F{e}_2{O}_3+Al$

Please refer to the video on Ellingham diagram