A) +0.73
B) \[-0.79\]
C) \[-0.82\]
D) \[-0.70\]
Correct Answer: B
Solution :
[b] For \[Z{{n}^{2+}}\xrightarrow{{}}Zn\] \[{{E}_{Z{{n}^{2+}}/Zn}}=E{{{}^\circ }_{Z{{n}^{2+}}/Zn}}-\frac{2.303RT}{nF}\log \frac{[Zn]}{[Z{{n}^{2+}}]}\] \[=-0.76-\frac{0.06}{2}log\frac{1}{[0.1]}=-0.76-0.03\] \[{{E}_{Z{{n}^{2+}}/Zn}}=-0.79V\]You need to login to perform this action.
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