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



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