A) 1.10 V
B) -1.10 V
C) 0.110 V
D) -0.110 V
Correct Answer: A
Solution :
Key Idea: Use following relationship to find emf of cell \[{{E}_{cell}}={{E}^{o}}-\frac{0.059}{n}\log \frac{[Product]}{[Reactant]}\] Given, \[{{E}^{o}}=1.10\,\,V\] \[[Z{{n}^{2+}}]=0.1\,\,M\] \[[C{{u}^{2+}}]=0.1\,\,M\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.059}{2}\log \left[ \frac{Z{{n}^{2+}}}{C{{u}^{2+}}} \right]\] (\[\because \,\,2\] electrons are involved in cell reaction and concentration of solids is taken as unity) \[\therefore \] \[{{E}_{cell}}=1.10-\frac{0.059}{2}\log \left[ \frac{0.1}{0.1} \right]\] or \[=1.0-\frac{0.059}{2}\log \,\times 0\] \[=1.10-\frac{0.059}{2}\times 0\] \[\therefore \] \[{{E}_{cell}}=1.10\,\,V\]
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