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In the electrochemical cell: Zn|$ZnSO_4$(0.01M)||$CuSO_4$(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 following, which one is the relationship between $E_1$ and $E_2$? (Given, $\frac{RT}{F} = 0.059$)
Explanation
Smaller $[Zn^{2+}]/[Cu^{2+}]$ gives larger emf.
Detailed Solution
$E_1 = E^\circ_{cell} - \frac{2.303RT}{2F}\log\frac{(0.01)}{1}$
When the concentrations are changed: $E_2 = E^\circ_{cell} - \frac{2.303RT}{2F}\log\frac{1}{0.01}$
i.e. $E_1 > E_2$