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JIPMER Jipmer Medical Solved Paper-1999

  • question_answer
    What is the standard free-energy change at \[25{}^\circ C\] for the reaction: \[Cd(s)+P{{b}^{2+}}(aq.)\xrightarrow{{}}C{{d}^{2+}}(aq.)+Pb(s)\] Given:   \[E_{C{{d}^{2+}}/Cd}^{\text{o}}=-\,0.403\,V,\] \[E_{P{{b}^{2+}}/Pb}^{\text{o}}=-\,0.126\,V\]

    A)  \[-\,5.346\text{ }kJ\]       

    B)         \[-\,53.46\text{ }kJ\]    

    C)         \[-\,106.92\text{ }kJ\]      

    D)         \[-\,10.69\text{ }kJ\]

    Correct Answer: B

    Solution :

    \[Cd(s)+P{{b}^{2+}}(aq.)\xrightarrow{{}}C{{d}^{2+}}(aq.)+Pb(s)\] From the reaction, it is clear that transfer of \[2{{e}^{-}}\] is taking place i.e., \[n=2\] Free energy change \[(\Delta G)=-n\,.\,F\,.\,E\] \[=-\,2\times 96500\times (E_{cell}^{0})\] \[=-\,2\times 96500\times (E_{cathode}^{\text{o}}-E_{anode}^{\text{o}})\] \[=-193000\,(-0.126+0.403)\] \[=-\,193000\times 0.277\] \[=-\,53461\,joules\] \[=-\,53.46\,kJ\]


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