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 :
| [d] 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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