A) \[2.0\times {{10}^{11}}\]
B) \[4.0\times {{10}^{12}}\]
C) \[1.0\times {{10}^{2}}\]
D) \[1.0\times {{10}^{10}}\] (Given,\[F=96500\text{ }C\text{ }mo{{l}^{-1}};\] \[R=8.314\text{ }J{{K}^{-1}}\text{ }mo{{l}^{-1}}\])
Correct Answer: D
Solution :
\[E_{cell}^{o}=\frac{0.059}{n}\log \,{{K}_{c}}\] Given, \[n=2\] \[{{E}^{o}}=0.295V\] \[\Rightarrow \] \[0.295=\frac{0.059}{2}\log {{K}_{c}}\] \[\Rightarrow \] \[\frac{0.295\times 2}{0.059}=\log {{K}_{c}}\] \[\Rightarrow \] \[10=\log {{K}_{c}}\] \[\therefore \] \[{{K}_{c}}=1\times {{10}^{10}}\]
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