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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 39

  • question_answer
    The standard e.m.f. of a galvanic cell involving cell reaction with \[\operatorname{n} = 2\] is found to be 0.295 V at \[25{}^\circ C\]. The equilibrium constant of the reaction would be (Given\[F=96500\text{ }C\text{ }mo{{l}^{-}}^{1};\text{ }R=8.314\,J{{K}^{-1}}{{W}^{-}}^{1}\])

    A) \[2.0\times {{10}^{11}}\]         

    B)        \[4.0 \times 1{{0}^{12}}\]

    C) \[1.0\times 1{{0}^{2}}\]          

    D)        \[1.0 \times 1{{0}^{10}}\]

    Correct Answer: D

    Solution :

    \[E{}^\circ =\frac{0.0591}{n}\,\log \,K\] Here, \[\operatorname{n}=2, E{}^\circ  = 0.295\] \[\therefore \,\,\,\log \,\,K=\frac{2\times 0.295}{0.591}=9.8\approx 10\,\,or\,\,K={{10}^{10}}\]


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