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DeltaG for the reaction : (4)/(3) Al+...

`DeltaG` for the reaction `:`
`(4)/(3) Al+O_(2)rarr (2)/(3)Al_(2)O_(3)`
is `-772 kJ mol^(-1)` of `O_(2)`.
Calculate the minimum `EMF` in volts required to carry out an electrolysis of `Al_(2)O_(3)`

Text Solution

Verified by Experts

`Al to Al^(3+)+underset(=4" mol electrons") underset(4//3xx3" mol")(3e^(-))`
`DeltaG=-nFE_(cell)`
`E_(cell)=-(DeltaG)/(nF)`
`DeltaG=772" kJ "mol^(-1)=772xx1000" J "mol^(-1)`
`=772xx1000" CV "mol^(-1)`
`n=4,=96500" C "mol^(-1)`
`E_(cell)=((772xx1000" CV "mol^(-1)))/((4)xx(96500" C "mol^(-1)))=2V`
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