| (i) \[Zn|Zn_{1M}^{2+}||Cu_{0.1M}^{2+}|Cu\] |
| (ii) \[Zn|Zn_{1M}^{2+}||Cu_{1M}^{2+}|Cu\] |
| (iii) \[Zn|Zn_{0.1\,M}^{2+}||Cu_{1M}^{2+}|Cu\] |
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, \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}\] (i) \[{{E}_{1}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{1}{0.1}\] \[E_{cell}^{o}-\frac{0.0591}{2}\] (ii) \[{{E}_{2}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{1}{1}\] \[=E_{cell}^{o}-\frac{0.0591}{2}\times 0\] \[=E_{cell}^{o}\] (iii) \[{{E}_{3}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{0.1}{1}\] \[=E_{cell}^{o}+\frac{0.0591}{2}\] \[\therefore \] \[{{E}_{3}}>{{E}_{2}}>{{E}_{1}}\]
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