Question

6.23. 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 1

Answer:

heya Your Answer:

Thermodynamic factor helps us in choosing a suitable reducing agent for the reduction of a particular metal oxide to the metallic state as discussed below.

From the diagram, it is evident that metals for which the standard free energy of formation (∆f G°)of their oxides is more negative can reduce those metal oxides for which the standard free energy of formation (∆f G°)of their respective oxides is less negative. In other words. Any metal will reduce the oxides of other metals which lie above it in Ellingham diagram because the standard free energy change (∆f G °) of the combined redox reaction will be negative by an amount equal to the difference in (∆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. Similarly, C can reduce ZnO to Zn but not CO.

Hope it's helpful to you

@ Dhruvshi23

Answer 2

◆ Answer with explanation -

I absolutely agree with the statement that choice of reducing agent depends on thermodynamic factors most importantly free energy change.

∆G° of the reducing agent must be more than the oxide it has to reduce. i.e. oxide with more negative ∆G° reduces oxides with less negative ∆G°.  

Hope this helps

Explanation:

itss_katy16  :p