A) \[2.0\times {{10}^{11}}\]
B) \[\text{4}.0\times \text{l}{{0}^{\text{12}}}\]
C) \[\text{1}.0\times \text{l}{{0}^{\text{2}}}\]
D) \[\text{1}.0\times \text{l}{{0}^{\text{1}0}}\] (Given: \[F=96500\,\,C\,mo{{l}^{-1}};\] \[R=8.314\,J{{K}^{-1}}mo{{l}^{-1}})\]
Correct Answer: D
Solution :
By Nernst equation, \[{{E}_{cell}}={{E}^{0}}_{cell}-\frac{2.303RT}{nF}{{\log }_{10}}K\] At equilibrium \[{{E}_{cell}}=0\] Given that \[\therefore \,\,\,\,\,\,\,\,\,\,R=8.315\,J{{K}^{-1}}\,mo{{l}^{-1}}\] \[\text{T}=\text{25}{}^\circ \text{C}+\text{273}=\text{298 K}\] \[\text{F}=\text{965}00\text{ C and n}=\text{2}\] \[\therefore \]\[{{E}^{0}}_{cell}=\frac{2.303\times 8.314\times 298}{2\times 96500}\]\[{{\log }_{10}}K\] \[=\frac{0.0591}{2}{{\log }_{10}}K\] \[\because \]Given that \[{{E}^{0}}_{cell}\,\,=0.295\,\,V\] \[\therefore \]\[0.295=\frac{0.0591}{2}{{\log }_{10}}K\] \[{{\log }_{10}}K=\frac{0.295\times 2}{0.0591}=10\] or antilog of\[{{\log }_{10}}K=anti\log \,10\] \[K=1\times {{10}^{10}}\]
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