A) 1.10 V
B) + 0.110 V
C) -1.10 V
D) -0.11 V
Correct Answer: A
Solution :
Key Idea: \[Zn+C{{u}^{2+}}\xrightarrow{{}}Z{{n}^{2+}}+Cu\] \[\therefore \] \[Zn\] 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, \[E_{cell}^{o}=1.1\,\,V,\,n=2\] \[[Z{{n}^{2+}}]=0.1\,M,\,\,[C{{u}^{2+}}]=0.1\,M\] \[\therefore \] \[{{E}_{cell}}=1.1-\frac{0.592}{2}\log \left[ \frac{0.1}{0.1} \right]\] or \[{{E}_{cell}}=1.10-0\] \[=1.10\,V\]You need to login to perform this action.
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