NEET Solved Paper 2017 Question 45
Question: In the electrochemical cell $ Zn|ZnSO _4(0.01)||CuSO _4(1.0,M)Cu, $ the emf of this Daniel cell is $ E _1. $ When the concentration of $ ZnSO _4 $ is changed to 1.0 M and that of $ CuSO _4 $ changed to 0.01 M, emf changes to $ E _2. $ From the following, which one is the relationship between $ E _1 $ and $ E _2 $ ? (Given, $ \frac{RT}{F}=0.059 $ )
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: D
Solution:
- $ Zn|ZnSO _4(0.01,M)||CuSO _4(1.0,M)|Cu $
$ \therefore $ $ E _1=E_cell^{0}-\frac{2.303RT}{2\times F}\times \log \frac{(0.01)}{1} $ When concentrations are changed
$ \therefore $ $ E _2=E_cell^{o}-\frac{2.303RT}{2F}\times \log \frac{1}{0.01} $ i.e., $ E _1>E _2 $
