A) \[\frac{0.32}{{{e}^{0.0295}}}\]
B) \[\frac{0.32}{{{10}^{0.0295}}}\]
C) \[\frac{0.26}{{{10}^{0.0295}}}\]
D) \[\frac{0.32}{{{10}^{0.0591}}}\]
Correct Answer: B
Solution :
For this cell, reaction is: \[Zn+F{{e}^{2+}}\to Z{{n}^{2+}}+Fe\] \[E={{E}^{0}}-\frac{0.0591}{n}\log \frac{{{c}_{1}}}{{{c}_{2}}}\]; \[{{E}^{0}}=E+\frac{0.0591}{n}\log \frac{{{c}_{1}}}{{{c}_{2}}}\] \[=0.2905+\frac{0.0591}{2}\log \frac{{{10}^{-2}}}{{{10}^{-3}}}=0.32\ V\]. \[{{E}^{0}}=\frac{0.0591}{2}\log {{K}_{c}}\]; \[\log {{K}_{c}}=\frac{0.32\times 2}{0.0591}=\frac{0.32}{0.0295}\] \[\therefore \ \ {{K}_{c}}=\frac{0.32}{{{10}^{0295}}}\].
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