Redox Reactions And Electrochemistry Ques 609
Question: The electrode potential $ {E_{(Z{n^{2+}}/Zn)}} $ of a zinc electrode at $ 25{}^\circ C $ with an aqueous solution of $ 0.1MZnSO_4 $ is $ [E{{{}^\circ }_{(Z{n^{2+}}/Zn)}}~=-0.76V $ . Assume $ \frac{2.303RT}{F}=0.06at298K]. $
Options:
A) +0.73
B) $ -0.79 $
C) $ -0.82 $
D) $ -0.70 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] For $ Z{n^{2+}}\xrightarrow{{}}Zn $ $ {E _{Z{n^{2+}}/Zn}}=E{{{}^\circ } _{Z{n^{2+}}/Zn}}-\frac{2.303RT}{nF}\log \frac{[Zn]}{[Z{n^{2+}}]} $
$ =-0.76-\frac{0.06}{2}log\frac{1}{[0.1]}=-0.76-0.03 $ $ {E_{Z{n^{2+}}/Zn}}=-0.79V $
