Question

The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

Answer 1

The above figure is a plot of Gibbs energy (ΔG°) vs. T for the formation of some oxides.

It can be observed from the above graph that a metal can reduce the oxide of other metals if the standard free energy of formation (Δ_fG°) of the oxide of the former is more negative than the latter. For example, since \[\ce{\Delta_f G^{\circ}_{Al, Al_2, O_3}}\] is more negative than \[\ce{\Delta_f G^{\circ}_{Cu, Cu_2. O}}\], Al can reduce Cu₂O to Cu, but Cu cannot reduce Al₂O₃. Similarly, Mg can reduce ZnO to Zn, but Zn cannot reduce MgO because \[\ce{\Delta_f G^{\circ}_{Mg, MgO}}\] is more negative than \[\ce{\Delta_f G^{\circ}_{Zn, ZnO}}\]

Answer 2

We can study the choice of a reducing agent in a particular case using an Ellingham diagram.

It is evident from the diagram that metals for which the standard free energy of formation of their oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative. It means that any metal will reduce the oxides of other metals that lie above it in the Ellingham diagram. This is because the standard free energy change (Δ_rG°) of the combined redox reaction will be negative by an amount equal to the difference in Δ_f G° of the two metal oxides. Thus, both Al and Zn can reduce FeO to Fe, but Fe cannot reduce Al₂O₃ to Al and ZnO to Zn. In the same way, G can reduce ZnO to Zn but not CO.

Only that reagent will be preferred as reducing agent, which will lead to decrease in free energy value (ΔG°) at a certain specific temperature