JEE Main & Advanced Sample Paper JEE Main - Mock Test - 40

  • question_answer
    The standard emf of a cell, involving one electron change is found to be \[0.591\text{ }V\]at\[25{}^\circ C\]. The equilibrium constant of the reaction is \[(F=96500\,\,C\,\,mo{{l}^{-1}})\]

    A) \[1.0\times {{10}^{1}}\]                      

    B) \[1.0\times {{10}^{5}}\]

    C) \[1.0\times {{10}^{10}}\]           

    D)       \[1.0\times {{10}^{30}}\]

    Correct Answer: C

    Solution :

    \[E{{{}^\circ }_{cell}}=\frac{0.0591}{n}\log \,{{K}_{eq}}\] \[\therefore \,\,0.591=\frac{0.0591}{1}\log {{K}_{eq}}\] or \[\log \,{{K}_{eq}}=\frac{0.591}{0.0591}=10\] or \[{{K}_{eq}}=1\times {{10}^{10}}\]


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