| In the electrochemical cell [NEET-2017] |
| \[Zn|ZnS{{O}_{4}}(0.01)||CuS{{O}_{4}}(1.0\,M)Cu,\] the emf of this Daniel cell is \[{{E}_{1}}\]. When the concentration of \[ZnS{{O}_{4}}\] is changed to 1.0 M and that of \[CuS{{O}_{4}}\] changed to 0.01 M, emf changes to \[{{E}_{2}}\]. From the following, which one is the relationship between\[{{E}_{1}}\] and \[{{E}_{2}}\]? |
| (Given, \[\frac{RT}{F}=0.059\]) |
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 :
| [d] \[Zn|ZnS{{O}_{4}}(0.01\,M)||CuS{{O}_{4}}(1.0\,M)|Cu\] |
| \[\therefore \] \[{{E}_{1}}=E_{cell}^{0}-\frac{2.303RT}{2\times F}\times \log \frac{(0.01)}{1}\] |
| When concentrations are changed |
| \[\therefore \] \[{{E}_{2}}=E_{cell}^{o}-\frac{2.303RT}{2F}\times \log \frac{1}{0.01}\] |
| i.e., \[{{E}_{1}}>{{E}_{2}}\] |
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