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`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)` |
Answer» `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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