• # question_answer 30) The e.m.f of a Daniell cell, $Zn\left| \underset{(0.01M)}{\mathop{ZnS{{O}_{4}}}}\, \right|\left| \underset{(1.0M)}{\mathop{CuS{{O}_{4}}}}\, \right|Cu,$, at   $298\text{ }K$ is ${{E}_{1}}$. When the concentration of ZnS04 is 1.0 M and  that of $CuS{{O}_{4}}$ is $0.01\text{ }M,$ the e.m.f. 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)  ${{E}_{2}}=0\ne {{E}_{1}}$            D)  ${{E}_{1}}>{{E}_{2}}$

Cell reaction $Zn+C{{u}^{++}}\xrightarrow{{}}Z{{n}^{++}}+Cu$ ${{E}_{1}}=E_{cell}^{o}-\frac{0.059}{2}\log \frac{0.01}{1.0}$ $\therefore$ ${{E}_{1}}=(E_{cell}^{o}+0.059)V$ ${{E}_{2}}=E_{cell}^{o}-\frac{0.059}{2}\log \frac{1.0}{0.01}$ $\therefore$ ${{E}_{2}}=(E_{cell}^{o}-0.059)V.$ Thus ${{E}_{1}}>{{E}_{2}}$