Redox Reactions And Electrochemistry Question 518
Question: The e.m.f. of a Daniell cell at 298 K is $ E_1. $
When the concentration of $ ZnSO_4 $ is 1.0 M and that of $ CuSO_4 $ is 0.01 M, the e.m.f. changed to $ E_2 $ . What is the relationship between $ E_1 $ and $ E_2 $ -
Options:
A) $ E_2=0\ne E_1 $
B) $ E_1 \gt E_2 $
C) $ E_1 \lt E_2 $
D) $ E_1=E_2 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Cell reaction is, $ Zn+C{u^{2+}}\to Z{n^{2+}}+Cu $ $ E_{cell}=E_{_{Cell}}^{{}^\circ }-\frac{RT}{nF}\ell n\frac{[Z{n^{2+}}]}{[C{u^{2+}}]} $ Greater the factor $ [ \frac{(Z{n^{2+}})}{(C{u^{2+}})} ] $ , less is the EMF Hence $ E_1>E_2 $
BETA
