A) \[{{E}_{2}}>{{E}_{3}}>{{E}_{1}}\]
B) \[{{E}_{3}}>{{E}_{2}}>{{E}_{1}}\]
C) \[{{E}_{1}}>{{E}_{2}}>{{E}_{3}}\]
D) \[{{E}_{1}}>{{E}_{3}}>{{E}_{2}}\]
Correct Answer: B
Solution :
For the given cell,\[{{E}_{cell}}={{E}^{o}}cell-\frac{0.0591}{2}\log \frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}\] (i)\[{{E}_{1}}={{E}^{o}}cell-\frac{0.0591}{2}\log \frac{1}{0.1}={{E}^{o}}cell-\frac{0.0591}{2}\] (ii) \[{{E}_{2}}={{E}^{o}}cell-\frac{0.0591}{2}\log \frac{1}{1}\] \[={{E}^{o}}cell-\frac{0.0591}{2}\times 0={{E}^{o}}cell\] (iii) \[{{E}_{3}}={{E}^{o}}cell-\frac{0.0591}{2}\log \frac{0.1}{1}\] \[={{E}^{o}}cell+\frac{0.0591}{2}\] \[\therefore \] \[{{E}_{3}}>{{E}_{2}}>{{E}_{1}}\]You need to login to perform this action.
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