A) \[-0.637\]
B) \[+0.637\]
C) \[>0.637\]
D) \[+0.889\]
Correct Answer: C
Solution :
\[{{E}_{cell}}=E_{P{{b}^{2+}}/Pb}^{o}-E_{Z{{n}^{2+}}/Zn}^{o}\] \[=-0.126-(-0.763)\] \[=+0.637\text{ }V\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \frac{[Z{{n}^{2+}}]}{[P{{b}^{2+}}]}\] \[=0.637-\frac{0.0591}{2}\log 0.1\] \[=0.637+0.02955=0.667\text{ }V\]
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