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NCERT Class XII Chemistry
Chapter - General Principle of Process of Isolation of Elements
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Question : 21 of 31
Marks: +1, -0
The value of ΔfG∘\Delta_{f}G^{\circ} for formation of Cr2O3\mathrm{Cr_2O_3} is – 540 kJ mol−1^{-1} and that of Al2O3\mathrm{Al_2O_3} is – 827 kJ mol−1^{-1}. Is the reduction of Cr2O3\mathrm{Cr_2O_3} possible with Al?
Solution:  
The two equations are :
43Al(s)+O2(g)⟶23Al2O3(s);ΔfGAl,A2O30=−827kJ mol−1\frac{4}{3} \mathrm{Al}_{(s)} + \mathrm{O}_{2(\mathrm{g})} \longrightarrow \frac{2}{3} \mathrm{Al}_2\mathrm{O}_{3(s)} ; \Delta_{\mathrm{f}} G_{\mathrm{Al},\mathrm{A}_2\mathrm{O}_3}^{0} = -827 \mathrm{kJ\,mol}^{-1}
43Cr(s)+O2(g)⟶23Cr2O3(s);ΔfG0=−540kJ mol−1\frac{4}{3} \mathrm{Cr}_{(s)} + \mathrm{O}_{2(\mathrm{g})} \longrightarrow \frac{2}{3} \mathrm{Cr}_2\mathrm{O}_{3(s)} ; \Delta_{\mathrm{f}} G^{0} = -540 \mathrm{kJ\,mol}^{-1}
Subtracting equation (ii) from equation (i), we get
43Al(s)+23Cr2O3(g)⟶23Al2O3(s)+43Cr(s);Δ,G0=−287kJ mol−1\frac{4}{3} \mathrm{Al}_{(\mathrm{s})} + \frac{2}{3} \mathrm{Cr}_2\mathrm{O}_{3(\mathrm{g})} \longrightarrow \frac{2}{3} \mathrm{Al}_2\mathrm{O}_{3(\mathrm{s})} + \frac{4}{3} \mathrm{Cr}_{(\mathrm{s})} ; \Delta, G^{0} = -287 \mathrm{kJ\,mol}^{-1}
Since ΔfG∘\Delta_{f}G^{\circ} of the combined redox reaction is negative, therefore, reductionof Cr2O3\mathrm{Cr_2O_3} by Al is possible.
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