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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    Given the equilibrium constant \[{{K}_{c}}\]of the reaction: \[C{{u}_{(s)}}+2A{{g}^{+}}_{(aq)}\to C{{u}^{2+}}_{(aq)}+2A{{g}_{(s)}}\]is \[10\times {{10}^{15}},\]calculate the \[E_{cell}^{o}\]of this reaction at 298 K. \[\left[ 2.303\frac{RT}{F}at\,298\,K=0.059V \right]\] [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) 0.04736 V                    

    B) 0.4736 V

    C) 0.4736 mV       

    D)   0.04736 mV

    Correct Answer: B

    Solution :

    \[{{E}^{o}}_{cell}=\frac{2.303RT}{nF}\log {{K}_{c}}\] \[=\frac{0.059}{2}\log (10\times {{10}^{15}})=\frac{0.059\times 16}{2}\]\[=0.472V\]


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