A) \[-\text{ }0.637\]
B) + 0.637
C) > 0.637
D) + 0.889
Correct Answer: C
Solution :
\[{{E}_{cell}}=E_{P{{b}^{2+}}/Pb}^{o}-E_{Z{{n}^{2+}}/Zn}^{o}\] \[=-0.126-(-0.763)\] \[=+\,0\,.637\,\text{V}\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \frac{[Z{{n}^{2+}}]}{[P{{b}^{2+}}]}\] \[=0.637-\frac{0.0591}{2}\,\log \,0.1\] \[=0.637+0.02955=0.667\,\text{V}\]You need to login to perform this action.
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