A) \[{{E}_{1}}<{{E}_{2}}\]
B) \[{{E}_{1}}={{E}_{2}}\]
C) \[{{E}_{2}}=0\ne {{E}_{1}}\]
D) \[{{E}_{1}}>{{E}_{2}}\]
Correct Answer: D
Solution :
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 \[{{E}_{1}}>{{E}_{2}}\]
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