A) \[{{E}_{1}}={{E}_{2}}\]
B) \[{{E}_{2}}=0\,\ne {{E}_{1}}\]
C) \[{{E}_{1}}>E{{ & }_{2}}\]
D) \[{{E}_{1}}<{{E}_{2}}\]
Correct Answer: C
Solution :
For Daniell cell \[Zn\left| \begin{align} & ZnS{{O}_{4}} \\ & 0.01\,M \\ \end{align} \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\left| \begin{align} & CuS{{O}_{4}} \\ & 1.0\,M \\ \end{align} \right|\,\,Cu\] Cell reaction is \[Zn(s)+C{{u}^{2+}}(aq)\rightleftharpoons Cu(s)+Z{{n}^{2+}}(aq)\] For above cell \[{{E}_{1}}={{E}^{o}}_{cell}-\frac{0.0591}{n}{{\log }_{10}}\frac{[Z{{n}^{2+}}]}{[Cu]}\] \[{{E}_{1}}=E_{cell}^{o}-\frac{0.0591}{2}{{\log }_{10}}\frac{0.01}{1.0}\] \[{{E}_{1}}=E_{cell}^{o}-\frac{0.0591}{2}{{\log }_{10}}\frac{1}{100}\] \[=E_{cell}^{o}+0.0591\,\,\,{{\log }_{10}}\,10\] \[=E_{cell}^{o}+0.0591\] ....(i) When the concentration of \[Z{{n}^{2+}}\] is 1.0 M and concentration of \[C{{u}^{2+}}\,\]is 0.01 M \[{{E}_{2}}=E_{cell}^{o}-\frac{0.0591}{2}{{\log }_{10}}\frac{1}{0.01}\] \[=E_{cell}^{o}-\frac{0.0591}{2}{{\log }_{10}}{{10}^{2}}\] \[=E_{cell}^{o}-0.0591....(ii)\] From Eqs. (i) and (ii) \[{{E}_{1}}>{{E}_{2}}\]
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