A) 1.10V
B) \[-1.10V\]
C) 2.20V
D) \[-2.20V\]
Correct Answer: A
Solution :
\[{{E}_{cell}}=E_{cell}^{o}-\frac{0.059}{2}\log \frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}\] \[=1.10-\frac{0.059}{2}\log \frac{0.1}{0.1}\] \[=1.10V\]You need to login to perform this action.
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