A) \[{{E}_{1}}={{E}_{2}}\]
B) \[{{E}_{2}}=0\ne {{E}_{2}}\]
C) \[{{E}_{1}}>{{E}_{2}}\]
D) \[{{E}_{1}}<{{E}_{2}}\]
Correct Answer: C
Solution :
[c] Using the relation \[{{E}_{cell}}=E_{cell}^{0}-\frac{0.0591}{n}\log \frac{[anode]}{[cathode]}\] \[=E_{cell}^{0}-\frac{0.0591}{n}\log \frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}\] Substituting the given values in two cases. \[{{E}_{1}}={{E}^{0}}-\frac{0.0591}{2}\log \frac{0.01}{1.0}\] \[={{E}^{0}}-\frac{0.0591}{2}\log {{10}^{-2}}\] \[={{E}^{0}}+\frac{0.0591}{2}\times 2\] or \[({{E}^{0}}+0.0591)V\] \[{{E}_{2}}={{E}^{0}}-\frac{0.0591}{2}\log \frac{1}{0.01}\] \[={{E}^{0}}-\frac{0.0591}{2}\log {{10}^{2}}\] \[={{E}^{0}}-\frac{2\times 0.0591}{2}\] or \[({{E}^{0}}-0.0591)\,V\] Thus, \[{{E}_{1}}>{{E}_{2}}\]You need to login to perform this action.
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