The e.m.f. of the following Daniell cell at 298 K is $\text{ is }{\mathrm{E}}_1. \mathrm{Zn}/{\mathrm{ZnSO}}_4\left(0.01\mathrm{M}\right)//{\mathrm{CuSO}}_4\left(1.0\mathrm{M}\right)/\mathrm{Cu}$ When the concentration of ${\mathrm{ZnSO}}_4$ is 1.0 M and that of CuSO₄ is 0.01M, the e.m.f. changed to E₂. What is the relationship between E₁ and E₂ ?

For the given red-ox reaction , the Nernst equation is written as ,

Substituting the given values in given two Daniel cells . 

$$\begin{gathered} {\mathrm{E}}_1={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{2}\log \frac{0.01}{1.0} \\ \mathrm{on} \mathrm{solving} \\ {\mathrm{E}}_1= {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}+0.0591\mathrm{V} \end{gathered}$$

and 

$$\begin{gathered} {\mathrm{E}}_2={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{2}\log \frac{1}{0.01} \\ \mathrm{on} \mathrm{solving} \\ {\mathrm{E}}_2= {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-0.0591 \mathrm{V}\hspace{0.17em} \end{gathered}$$

Thus E₁> E₂