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BCECE Engineering BCECE Engineering Solved Paper-2014

  • question_answer
    \[{{E}_{1}},{{E}_{2}},{{E}_{3}}\]are the emf values of the three galvanic cells respectively. (i) \[\text{Zn }\!\!|\!\!\text{ Zn}_{\text{1M}}^{\text{2+}}\text{ }\!\!|\!\!\text{  }\!\!|\!\!\text{ Cu}_{\text{0}\text{.1M}}^{\text{2+}}\text{ }\!\!|\!\!\text{ Cu}\]                     (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,\[{{E}_{cell}}={{E}^{o}}cell-\frac{0.0591}{2}\log \frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}\] (i)\[{{E}_{1}}={{E}^{o}}cell-\frac{0.0591}{2}\log \frac{1}{0.1}={{E}^{o}}cell-\frac{0.0591}{2}\] (ii) \[{{E}_{2}}={{E}^{o}}cell-\frac{0.0591}{2}\log \frac{1}{1}\] \[={{E}^{o}}cell-\frac{0.0591}{2}\times 0={{E}^{o}}cell\] (iii) \[{{E}_{3}}={{E}^{o}}cell-\frac{0.0591}{2}\log \frac{0.1}{1}\] \[={{E}^{o}}cell+\frac{0.0591}{2}\] \[\therefore \]  \[{{E}_{3}}>{{E}_{2}}>{{E}_{1}}\]


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