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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer
    The standard Gibbs energy for the given cell reaction in \[kJ\,mo{{l}^{-1}}\] at 298 K is :
    \[Zn(s)+C{{u}^{2+}}(aq)\to Z{{n}^{2+}}(aq)+Cu(s),\]
    \[{{E}^{o}}=2\,V\,at\,298K\]
    (Faraday's constant, \[F=96000\,C\,mo{{l}^{-1}}\])
                [JEE Main 9-4-2019 Morning]

    A) -384

    B)               -192

    C) 192                 

    D) 384

    Correct Answer: A

    Solution :

    \[\Delta {{G}^{o}}=-nF{{E}^{o}}_{cell}\] \[=2\times 96000\times 2\] \[=384000\text{ }J\] \[=384\text{ }kJ\] \[\therefore \] Ans. is [a]


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