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NEET Sample Paper NEET Sample Test Paper-59

  • question_answer
    The standard emf of the \[Cd\,(s)/Cd\text{ }Cl,(0.1)\,M//\text{ }Ag\,Cl\,(s)/Ag\,(s)\] in which cell reaction is \[+\,\,2C{{l}^{-}}\,(aq.)\text{ }\]is \[0.6915v\text{ }at\text{ }0{}^\circ C\] and\[0.6753v\text{ }at25{}^\circ .C\]. The enthalpy change of the reaction at\[25{}^\circ C\]is:

    A) (a)\[-176kJ\]                  

    B) \[-\,334.7kJ\]

    C) \[+123.5kJ\]                 

    D) \[-\,167.6kJ\]

    Correct Answer: D

    Solution :

    [d] ? 167.6 kJ \[\Delta G=-\,n\,F{{E}_{cell}}\] \[=-2\times 96500\times 0.6753\text{ (at}\,\text{25}{}^\circ \text{C)}\] \[=-130333J\] \[\Delta S=nf{{\left( \frac{\partial E}{\partial T} \right)}_{p}}\left( \frac{\partial E}{\partial T}=temperature\,coeff.\,of\,e.m.f \right)\]\[{{\left( \frac{\partial E}{\partial T} \right)}_{p}}=\frac{{{E}_{2}}-{{E}_{1}}}{{{T}_{2}}-{{T}_{1}}}\] \[~=\frac{0.6753-0.6915}{298-273}\] \[=-\,6.48\times {{10}^{4}}\] \[\therefore \,\,\Delta \,S=2\times 96500\times (-6.48\times {{10}^{-4}})\] \[=-125.064J{{k}^{-1}}\,mo{{l}^{-1}}\] \[\Delta G=\Delta H-T\Delta S\] \[\Delta H=\Delta H+T\Delta S\] \[=-13033+T(-125.064)\] \[=-167.6KJ\]


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