A) \[2.0\times {{10}^{11}}\]
B) \[4.0\times {{10}^{12}}\]
C) \[1.0\times {{10}^{2}}\]
D) \[1.0\times {{10}^{10}}\]
Correct Answer: D
Solution :
By Nernst equation, \[{{E}_{cell}}=E{{{}^\circ }_{cell}}-\frac{2.303RT}{nF}{{\log }_{10}}K\] At equilibrium \[{{E}_{cell}}=0\] Given that \[\therefore \] \[R=8.314\,J{{K}^{-1}}\,mo{{l}^{-1}}\] \[T=25{}^\circ C+273=298K\] \[F=96500\text{ }C\] and \[n=2\] \[\therefore \] \[E{{{}^\circ }_{cell}}=\frac{2.303\times 8.314\times 298}{2\times 96500}{{\log }_{10}}K\] \[\because \] Given that \[E{{{}^\circ }_{cell}}=0.295\,V\] \[\therefore \] \[0.295=\frac{0.0591}{2}{{\log }_{10}}K\] or \[K=\text{antilog}\,10\] or \[K=1\times {{10}^{10}}\] (Given F = 96500 C\[mo{{l}^{-1}},\]R= 8.314\[J{{K}^{-1}}mo{{l}^{-1}}\])You need to login to perform this action.
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