Redox Reactions And Electrochemistry Ques 175

Question: The emf of a Daniel cell at 298K is $ E_1 $ $ Zn|\underset{(0.01M)}{\mathop{ZnSO_4}},||\underset{(1.0M)}{\mathop{CuSO_4}},|Cu $ when the concentration of $ ZnSO_4 $ is 1.0 M and that of $ CuSO_4 $ is 0.01 M, the emf changed to $ E_2 $ . What is the relationship between $ E_1 $ and $ E_2 $ [CBSE PMT 2003]

Options:

A) $ E_2=0\ne E_1 $

B) $ E_1>E_2 $

C) $ E_1<E_2 $

D) $ E_1=E_2 $

Show Answer

Answer:

Correct Answer: B

Solution:

$ E_1=E_{o}-\frac{0.0591}{2}\log \frac{0.01}{1}=E_{o}+\frac{0.0591}{2}\times 2 $ $ E_2=E_{o}-\frac{0.0591}{2}\log \frac{100}{0.01}=E_{o}-\frac{0.0591}{2}\times 4 $
$ \therefore E_1>E_2 $ .



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