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.
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Answer
We can study the choice of a reducing agent in a particular case using Ellingham diagram. It is evident from the diagram that metals for which the standard free energy of formation oftheir 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 which lie above it in the Ellingham diagram. This is because the standard free energy change $(\Delta_{\text{r}}\text{G}^{\circ})$ of the combined redox reaction will be negative by an amount equal to the difference in $\Delta_{\text{f}}\text{G}^{\circ}$ of the two metal oxides. Thus both Al and $Zn$ can reduce $FeO$ to Fe but Fe cannot reduce $Al_2O_3$ to $A1$ and $ZnO$ to $Zn$. In the same way, G can reduce $ZnO$ to Zn but not $CO.$
Note: Only that reagent will be preferred as reducing agent which will lead to decrease in free energy value $(\Delta\text{G}^{\circ})$ at a certain specific temperature.
Note: Only that reagent will be preferred as reducing agent which will lead to decrease in free energy value $(\Delta\text{G}^{\circ})$ at a certain specific temperature.
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