Redox Reactions And Electrochemistry Question 758
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 $ .
BETA
