1.

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»

Solution :`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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