1.

If decomposition of Al_(2)O_(3) at 500 ""^(@)C temperature gives following reaction and gives Gibb's free energy : (2)/(3)Al_(2)O_(3) to (4)/(3) Al+O_(2). . . DeltaG=966kJ" "mol^(-1) then what is the difference of minimum required electrical energy for the reduction of Al_(2)O_(3) by electrolysis ?

Answer»

5.0V
4.5V
2.5V
3.0V

Solution :`(2)/(3)Al_(2)O_(3)""THEREFORE DeltaG=966kJ`
`therefore Al_(2)O_(3)""DeltaG=(966xx3)/(2)=1449kJ`
`2Al^(3+)+6e^(-) to 2Al`
`DeltaG=nFE_(CELL)`, `therefore E_(cell)=(DeltaG)/(nF)|"Where, "n=6," "DeltaG=1449xx10^(3)J," "E=96500`
`=(1449xx1000)/(6xx96500)`
`=2.5V`


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