In the electrochemical (Daniell) cell $\mathrm{Zn}|\mathrm{ZnSO}{}_4(0.01\mathrm{M})||\mathrm{CuSO}{}_4(1.0\mathrm{M})|\mathrm{Cu}$, the emf is $E_1$. When the concentration of $\mathrm{ZnSO}{}_4$ is changed to $1.0\text{M}$ and that of $\mathrm{CuSO}{}_4$ to $0.01\text{M}$, the emf changes to $E_2$. Which one of the following is the correct relationship between $E_1$ and $E_2$? (Given $\frac{RT}{F}=0.059$)

Solution & Explanation

\textbf{Answer:} (C) $E_1>E_2$ \textbf{Solution:} For the cell $\mathrm{Zn}|\mathrm{Zn}{}^{2+}||\mathrm{Cu}{}^{2+}|\mathrm{Cu}$ the Nernst equation ($n=2$) is: $E_{cell}=E_{cell}^{\circ }-\frac{0.059}{2}\log \frac{[\mathrm{Zn}{}^{2+}]}{[\mathrm{Cu}{}^{2+}]}$ Case 1: $\frac{[\mathrm{Zn}{}^{2+}]}{[\mathrm{Cu}{}^{2+}]}=\frac{0.01}{1.0}=10^{-2}$, so $\log =-2$ (negative) and $E_1>E_{cell}^{\circ }$. Case 2: $\frac{[\mathrm{Zn}{}^{2+}]}{[\mathrm{Cu}{}^{2+}]}=\frac{1.0}{0.01}=10^2$, so $\log =+2$ (positive) and $E_2<E_{cell}^{\circ }$. Therefore $E_1>E_{cell}^{\circ }>E_2$, i.e. $E_1>E_2$.