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AFMC AFMC Solved Paper-2009

  • question_answer
    Given, \[P{{b}^{2+}}/Pb=-0.126V;\]\[Z{{n}^{2+}}/Zn=-0.763\,\,V\] Find the emf of the following cell \[Zn|Z{{n}^{2+}}(0.1\,M)||P{{b}^{2+}}(1M)Pb.\]

    A) \[-\text{ }0.637\]             

    B) + 0.637

    C) > 0.637                                 

    D) + 0.889

    Correct Answer: C

    Solution :

    \[{{E}_{cell}}=E_{P{{b}^{2+}}/Pb}^{o}-E_{Z{{n}^{2+}}/Zn}^{o}\] \[=-0.126-(-0.763)\] \[=+\,0\,.637\,\text{V}\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \frac{[Z{{n}^{2+}}]}{[P{{b}^{2+}}]}\] \[=0.637-\frac{0.0591}{2}\,\log \,0.1\] \[=0.637+0.02955=0.667\,\text{V}\]


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