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) \[{{E}_{2}}={{E}^{o}}_{cell}-\frac{0.0591}{2}\log \frac{1}{1}\] (iii) \[={{E}^{o}}_{cell}-\frac{0.0591}{2}\times 0={{E}^{o}}_{cell}\] \[\therefore \] \[{{E}_{3}}>{{E}_{2}}>{{E}_{1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
