NEET NEET SOLVED PAPER 2019

  • question_answer
    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: [NEET 5-5-2019] [Given that \[\frac{2.303kT}{F}=0.059\text{ }V\text{ }at\text{ }T=298\text{ }K\]]

    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 :

    \[E_{cell}^{o}=\frac{0.06}{n}{{\log }_{10}}k\]             \[0.6=\frac{0.06}{1}{{\log }_{10}}k\]             \[{{\log }_{10}}k=10\]             \[k={{10}^{10}}\]      


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