Redox Reactions And Electrochemistry Question 42
Question: The e.m.f. of the cell $ Zn|Z{n^{2+}}(0.01M)||F{e^{2+}}|Fe(0.001M) $ at 298 K is 0.2905 then the value of equilibrium for the cell reaction is [IIT-JEE Screening 2004]
Options:
A) $ \frac{0.32}{{e^{0.0295}}} $
B) $ \frac{0.32}{{10^{0.0295}}} $
C) $ \frac{0.26}{{10^{0.0295}}} $
D) $ \frac{0.32}{{10^{0.0591}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
For this cell, reaction is: $ Zn+F{e^{2+}}\to Z{n^{2+}}+Fe $ $ E=E^{0}-\frac{0.0591}{n}\log \frac{c_1}{c_2} $ ;
$ E^{0}=E+\frac{0.0591}{n}\log \frac{c_1}{c_2} $ $ =0.2905+\frac{0.0591}{2}\log \frac{{10^{-2}}}{{10^{-3}}}=0.32\ V $ .
$ E^{0}=\frac{0.0591}{2}\log K_{c} $ ; $ \log K_{c}=\frac{0.32\times 2}{0.0591}=\frac{0.32}{0.0295} $
$ \therefore \ \ K_{c}=\frac{0.32}{10^{0295}} $ .
BETA
