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`DeltaG` for the reaction `(4)/(3)Al+O_(2)to(2)/(3)Al_(2)O_(3)` is -772kJ `mol^(-1)` of `O_(2)`. Calculate the minimum EMF in volts requird to carry out and electrolysis of `Al_(2)O_(3)` |
Answer» `AltoAl^(3+)+3e^(-)` `(4)/(3)mol" of "Al=(4)/(3)xx3mol" "e^(-)=4mol" "e^(-)` n=4 `DeltaG=-nFE` `-772xx1000J=-4xx96500xxE" "thereforeE=(772xx1000)/(4xx96500)=2.0V` |
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