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Q. The. free energy change for the decomposition reaction, $ \frac{2}{3}A{{l}_{2}}{{O}_{3}}\xrightarrow[{}]{{}}\frac{4}{3}Al+{{O}_{2}} $ is $ \Delta G=+960\,kJ $ ( $ F=96500\text{ }C\,mo{{l}^{-1}} $ ) The minimum potential difference required to reduce $ A{{l}_{2}}{{O}_{3}} $ at $ 500{}^\circ C $ will be

JamiaJamia 2013

Solution:

$ A{{l}_{2}}{{O}_{3}}(2A{{l}^{3+}}+3{{O}^{2-}})\xrightarrow[{}]{{}}2Al+\frac{3}{2}{{O}_{2}}; $ $ n=6{{e}^{-}} $ $ \frac{2}{3}A{{l}_{2}}{{O}_{3}}\xrightarrow[{}]{{}}\frac{4}{3}Al+{{O}_{2}}; $ $ n=4{{e}^{-}} $ $ \Delta G=+960\,kJ=+960000J $ $ \Delta G=-nFE $ $ 960000=-4\times 96500\times E $ $ E=-2.487\,V $ Minimum potential difference required to reduce $ A{{l}_{2}}{{O}_{3}} $ is 2.487 V.