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:\[F=96500\text{ }C\text{ m}o{{l}^{-1}};\] \[R=8.314\text{ }J{{K}^{-1}}mo{{l}^{-1}}\])
Correct Answer: D
Solution :
By Nernst equation, \[{{E}_{cell}}=E_{cell}^{o}-\frac{2.303\,RT}{nF}{{\log }_{10}}K\] At equilibrium\[{{E}_{cell}}=0\] Given that \[\therefore \] \[R=8.315J{{K}^{-1}}mo{{l}^{-1}}\] \[T=25{}^\circ C+273=298\text{ }K\] \[F=96500\text{ }C\text{ }and\text{ }n=2\] \[\therefore \]\[E_{cell}^{o}=\frac{2.303\times 8.314\times 298}{2\times 96500}{{\log }_{10}}K\] \[=\frac{0.0591}{2}{{\log }_{10}}K\] \[\because \]Given that \[E_{cell}^{o}=0.295V\] \[\therefore \] \[0.295=\frac{0.0591}{2}{{\log }_{10}}K\] \[{{\log }_{10}}K=\frac{0.295\times 2}{0.0591}=10\] or antilog of \[{{\log }_{10}}K=anti\log \,10\] \[K=1\times {{10}^{10}}\]
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