A) 1.10V
B) \[+\text{ }0.110\text{ }V\]
C) \[-1.10V\]
D) \[-0.11\,V\]
Correct Answer: A
Solution :
\[Zn+C{{u}^{2+}}\xrightarrow[{}]{{}}Z{{n}^{2+}}+Cu\] \[\therefore \]Zn is anode and Cu is cathode \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.591}{n}\log \left[ \frac{Z{{n}^{2+}}}{C{{u}^{2+}}} \right]\] Given, \[E_{cell}^{o}=1.1V,n=2\] \[[Z{{n}^{2+}}]=0.1M,[C{{u}^{2+}}]=0.1M\] \[\therefore \] \[{{E}_{cell}}=1.1-\frac{0.592}{2}\log \left[ \frac{0.1}{0.1} \right]\] Or \[{{E}_{cell}}=1.10-0\] \[=1.10V\]
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