A) \[{{E}_{2}}=0\ne {{E}_{1}}\]
B) \[{{E}_{1}}={{E}_{2}}\]
C) \[{{E}_{1}}<{{E}_{2}}\]
D) \[{{E}_{1}}>{{E}_{2}}\]
Correct Answer: D
Solution :
\[Zn|ZnS{{O}_{4}}(0.01\,M)||CuS{{O}_{4}}(1.0\,M)|Cu\] \[\therefore \] \[{{E}_{1}}=E_{cell}^{0}-\frac{2.303RT}{2\times F}\times \log \frac{(0.01)}{1}\] When concentrations are changed \[\therefore \] \[{{E}_{2}}=E_{cell}^{o}-\frac{2.303RT}{2F}\times \log \frac{1}{0.01}\] i.e., \[{{E}_{1}}>{{E}_{2}}\]You need to login to perform this action.
You will be redirected in
3 sec