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On the bassis of the information available from the reaction ` 4/3 Al + O_2 rarr 2/3 Al_2 O_3 . Delta G =- 82 7 k J "mol"^(-1)` of `O_2` the minimum emf required to carry out an electorlysis of `Al_2 O_3` is `(F= 96500 C "mol"^(-1))`

A

2.14 V

B

4.28 V

C

6.42 V

D

8.56 V

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The correct Answer is:
A
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On the basis of information available from the reaction (4)/(3)Al+O_(2) rarr (2)/(3)Al_(2)O_(3),DeltaG = -827kJ mol^(-1) of O_(2) , the minimum emf required to carry out of the electrolysis of Al_(2)O_(3) is (F=96,500 C mol^(-1))

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)

The Gibbs energy for the decomposition of Al_2 O_3 at 2/3 Al_2O_3 rarr 4/3 Al = O_2 Delta_r G = + 966 k J "mol"^(-1) The potential differnce needed for electrolytic reduction of Al_2 O_3 at 500^@C is at least.

27 gm of Al will react completely with: 4Al + 3O_(2) rarr 2Al_(2)O_(3)

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