NEET Solved Paper 2019 Question 18
Question: For a cell involving one electron $ E_cell^{\Theta }=0.59Vat298K $ , the equilibrium constant for the cell reaction is: [NEET 5-5-2019] [Given that $ \frac{2.303kT}{F}=0.059VatT=298K $ ]
Options:
A) $ 1.0\times 10^{10} $
B) $ 1.0\times 10^{30} $
C) $ 1.0\times 10^{2} $
D) $ 1.0\times 10^{5} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ 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} $
