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Manipal Medical Manipal Medical Solved Paper-2015

  • question_answer
    The standard emf of the cell\[Cd(s)/CdC{{l}_{2}}(aq)(0.1M)|\,|\,AgCl(s)\,|\,Ag(s)\]in which the cell-reaction is \[Cd(s)+2AgCl(s)\,\,\to \]\[2Ag(s)+C{{d}^{2+}}(aq)+2C{{l}^{-}}(aq)\] is. 0.6915 V at 273 K and 0.6573 V at 298 K. The enthalpy change of the reaction at 298 K is

    A) \[-176.86\,\,kJ\]

    B) \[-245.6\,kJ\]

    C) \[+\,48.179\,kJ\]

    D) \[+\,223.5\,kJ\]

    Correct Answer: C

    Solution :

    Given, \[{{E}_{1}}=0.6915\,\,V,\,{{T}_{1}}=273\,K,\] \[{{E}_{2}}=0.6573\,V,\,{{T}_{2}}=298\,K\] Now, \[\Delta G=-nF{{E}^{\circ }}cell=-2\times 96500\times 0.6573\]                   \[=-126858.9\] \[\therefore \]      \[\Delta S=nF{{\left( \frac{\partial E}{\partial T} \right)}_{p}}\] Here, \[\frac{\partial E}{\partial T}=\]Temperature coefficient of emf                 \[{{\left( \frac{\partial E}{\partial T} \right)}_{p}}=\frac{{{E}_{2}}-{{E}_{1}}}{{{T}_{2}}-{{T}_{1}}}=\frac{0.6573-0.6915}{298-273}\]                    \[=-0.001368\] \[\therefore \]       \[\Delta S=2\times 96500\times (-0.001368)\]                   \[=-264.024\,J{{K}^{-1}}mo{{l}^{-1}}\] As we know that, \[\Delta G=\Delta H-T\Delta S\]              \[-126858.9=\Delta H-298\times (-264.024)\]                           \[\Delta H=48179.75\,J=48.179\,kJ\]


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