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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

  • question_answer
    \[{{E}_{1}},{{E}_{2}},{{E}_{3}}\]are the emf values of the three galvanic cells respectively. (i) \[Zn|Zn_{1M}^{2+}||Cu_{0.1M}^{2+}|Cu\] (ii) \[Zn|Zn_{1M}^{2+}||Cu_{1M}^{2+}|Cu\] (iii) \[Zn|Zn_{0.1M}^{2+}||Cu_{1M}^{2+}|Cu\] Which one of the following is true?

    A)  \[{{E}_{2}}>{{E}_{3}}>{{E}_{1}}\]

    B)  \[{{E}_{3}}>{{E}_{2}}>{{E}_{1}}\]

    C)  \[{{E}_{1}}>{{E}_{2}}>{{E}_{3}}\]

    D)  \[{{E}_{1}}>{{E}_{3}}>{{E}_{2}}\]

    Correct Answer: B

    Solution :

    For the given cell, \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}\] (i) \[{{E}_{1}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{1}{0.1}\] \[=E_{cell}^{o}-\frac{0.0591}{2}\] (ii) \[{{E}_{2}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{1}{1}\] \[=E_{cell}^{o}-\frac{0.0591}{2}\times 0\] \[=E_{cell}^{o}\] (iii) \[{{E}_{3}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{0.1}{1}\] \[=E_{cell}^{o}+\frac{0.0591}{2}\] \[\therefore \] \[{{E}_{3}}>{{E}_{2}}>{{E}_{1}}\]


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