Question

What will be the minmum potential difference needed to reduce $A{l}_2{O}_3$ at ${500}^{o}C$?
Given $\Delta G=960kJ$

Answer 1

$2A{l}_2{O}_3⟶\frac{4}{3}Al+O_{2}is\Delta G=+960kJwehave,\Delta G^0=-nF{E}^{0}960\times {10}^{3}J=-4\times E^0\times 96500=\frac{-960\times {10}^3}{4\times 96500}=E_{Cell}^0{E}_{Cell}^0=\frac{-96500}{3860}=-2.487VSince,\Delta G=+960kJ=+960\times {10}^{3}J{E}^0=?F=96500C{mol}^{-1}\frac{4}{3}A{l}^{3+}+4e^{-}\to \frac{4}{3}Aln=4$