| For a cell involving one electron \[E_{cell}^{\Theta }=0.59\text{ }V\text{ }at\text{ }298\text{ }K\], the equilibrium constant for the cell reaction is: |
| [Given that \[\frac{2.303kT}{F}=0.059\text{ }V\text{ }at\text{ }T=298\text{ }K\]] [NEET 2019] |
A) \[1.0\times {{10}^{10}}\]
B) \[1.0\times {{10}^{30}}\]
C) \[1.0\times {{10}^{2}}\]
D) \[1.0\times {{10}^{5}}\]
Correct Answer: A
Solution :
| [a] \[E_{cell}^{o}=\frac{0.06}{n}{{\log }_{10}}k\] |
| \[0.6=\frac{0.06}{1}{{\log }_{10}}k\] |
| \[{{\log }_{10}}k=10\] |
| \[k={{10}^{10}}\] |
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