Q. In the electrochemical cell : $Zn|ZnSO_4$ (0.01 M)$||$ $CuSO_4$ (1.0 M)$|Cu$, the emf of this Daniel cell is $E_1$. When the concentration of $ZnSO_4$ is changed to 1.0 M and that of $CuSO_4$ changed to 0.01 M, the emf changes to $E_2$. From the followings, which one is the relationship between $E_1$ and $E_2$ ? (Given, $\frac{RT}{F}$ = 0.059)
A) $E_1 = E_2$
B) $E_1 < E_2$
C) $E_1 > E_2$
D) $E_2 = 0 \neq E_1$
Explanation
[Solution] $E = E^\circ - \frac{0.059}{2}\log\frac{[Zn^{2+}]}{[Cu^{2+}]}$: $E_1 = E^\circ + 0.059$, $E_2 = E^\circ - 0.059$, so $E_1 > E_2$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.