A) 8.56 V
B) 2.14 V
C) 4.28 V
D) 6.42 V
Correct Answer: B
Solution :
[b] \[\Delta G=-nEF\] For 1 mole of Al, n = 3 \[\therefore \] for \[\frac{4}{3}\] mole of Al, \[n=3\times \frac{4}{3}=4\] According to question, \[-827\times 1000=-4\times E\times 96500\] \[E=\frac{827\times 1000}{4\times 96500}=2.14V\] \[\therefore \] minimum e.m.f required = 2.14 VYou need to login to perform this action.
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