Redox Reactions And Electrochemistry Ques 606
Question: On the basis of the information available from the reaction $ \frac{4}{3}Al+O_2\to \frac{2}{3}Al_2O_3,\Delta G $ $ =-827kJmo{l^{-1}} $ of $ O_2 $ the minimum e.m.f required to carry out an electrolysis of $ Al_2O_3 $ is $ ( F=96500Cmo{l^{-1}} ) $
Options:
A) 8.56 V
B) 2.14 V
C) 4.28 V
D) 6.42 V
Show Answer
Answer:
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 V
