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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2009

  • question_answer
    For the redox reaction \[Zn(s)+C{{u}^{2+}}(0.1\,\,M)\xrightarrow{{}}Z{{n}^{2+}}(1M)+Cu(s)\] taking place in a cell, \[E_{cell}^{\text{o}}\] is\[1.10\,\,V\]. \[{{E}_{cell}}\] for the cell will be\[\left( 2.3303\frac{RT}{F}=0.0591 \right)\]

    A) \[2.14\,\,V\]                      

    B) \[1.80\,\,V\]      

    C)        \[0.82\,\,V\]      

    D)        \[1.07\,\,V\]

    Correct Answer: D

    Solution :

    Cell reaction of given cell is                 \[Zn\xrightarrow{{}}Z{{n}^{2+}}(1M)+2{{e}^{-}}\]                 \[\underline{C{{u}^{2+}}(0.1M)+2{{e}^{-}}\xrightarrow{{}}Cu}\] Overall reaction\[=Zn+C{{u}^{2+}}(0.1M)\xrightarrow{{}}\]                                                 \[Cu+Z{{n}^{2+}}(1M)\]                 \[{{E}_{cell}}=E_{cell}^{\text{o}}+\frac{2.303RT}{nF}\log \frac{[C{{u}^{2+}}]}{[Z{{n}^{2+}}]}\]                         \[=1.10+\frac{0.0591}{2}\log \frac{0.1}{1}\]                         \[=1.10+0.0295\times (-1)\] \[\therefore \]  \[{{E}_{cell}}=1.0705\,\,V\]


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