Redox Reactions And Electrochemistry Question 47
Question: $ Zn(s)+Cl_2(1\ \text{atm)}\to Z{n^{2+}}+2C{l^{-}} $ . $ E^0_{cell} $ of the cell is 2.12 V. To increase E [BVP 2004]
Options:
A) $ [Z{n^{2+}}] $ should be increased
B) $ [Z{n^{2+}}] $ should be decreased
C) $ [C{l^{-}}] $ should be decreased
D) $ {P_{Cl_2}} $ should be decreased
Show Answer
Answer:
Correct Answer: B
Solution:
According to nernst’s equation $ E_{cell}^{{}}=E_{cell}^{0}-\frac{nRT}{F}\log \frac{c_1}{c_2} $
For $ Z{n_{(s)}}+C{l_{2(1\ atm)}}\to Z{n^{2+}}+2C{l^{-}} $ $ c_1=[Z{n^{2+}}] $ and $ c_2=[C{l^{-}}] $
Hence to increase E, $ c_1 $ should be decreased and $ c_2 $ should be increased is $ [Z{n^{2+}}] $ should be decreased and Cl should be increased.
BETA
