Q. The emf of a Daniell cell at 298 K is $E_{1},\ Zn|{\ ZnSO}_{4}\ ||\ {CuSO}_{4}\ |\ Cu.$ When the concentration of (0.01 M) (1.0 M) ${ZnSO}_{4}$ is 1.0 M and that of ${\ CuSO}_{4}$ is 0.01 M, the emf changed to $E_{2}.$ What is the relationship between $E_{1}\ and\ E_{2}?$
A) $E_{1} = \ E_{2}$
B) $E_{1} > E_{2}$
C) $E_{1} < \ E_{2}$
D) $E_{2} = 0 \neq \ E_{1}$
Explanation
(b)
$E_{1} = E_{0} - \frac{0.0591}{2}\log\frac{0.01}{1} = E_{0} + \frac{0.0591}{2} \times 2$
$E_{2} = E_{0} - \frac{0.0591}{2}\log\frac{1}{0.01} = E_{0} - \frac{0.0591}{2} \times 2$
$\therefore E_{1} > E_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.