A) \[1.10\text{ V}\]
B) \[~-1.10\text{ V}\]
C) \[~0.110\text{ V}\]
D) \[~-0.110\text{ V}\]
Correct Answer: A
Solution :
Key Idea: Use following relationship to find emf of cell \[E_{cell}^{o}={{E}^{o}}-\frac{0.059}{n}\log \frac{[prdouct]}{[Reactant]}\] 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 \[=1.10-\frac{0.059}{2}\log 1\] \[=1.10-\frac{0.059}{2}\times 0\] \[\therefore \] \[{{E}_{cell}}=1.10\,V\]You need to login to perform this action.
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