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4/3Al + O2 gives 2/3 Al2O3#G of oxygen is _827.The minimum Emf required to carry out the electrolysis of Al2Cl3 ?

4/3Al + O2 gives 2/3 Al2O3#G of oxygen is _827.The minimum Emf required to carry out the electrolysis of Al2Cl3 ?

Grade:12

1 Answers

Juhi
15 Points
5 years ago
∆G for O2 = -nFE°
 
So, E°= ∆G/-nF
 
E°= -827000/2×96500C
 
E°= 4.28
 
For Al2O3 minimum emf required=
 
4.28/2= 2.14 Answer

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