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BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    \[Zn(s)+C{{u}^{2+}}(aq)\xrightarrow{{}}Cu(s)+Z{{n}^{2+}}(aq)\] The emf of above reaction is 1.10 V at \[\text{25}{{\,}^{\text{o}}}\text{C}\text{.}\] Find emf of cell well when \[[Z{{n}^{2+}}]=0.1\,M\]\[[C{{u}^{2+}}]=0.1\,M\]

    A) \[1.10\text{ V}\]                  

    B)        \[~-1.10\text{ V}\]

    C) \[~0.110\text{ V}\]        

    D)        \[~-0.110\text{ V}\]

    Correct Answer: A

    Solution :

    Key Idea: Use following relationship to find emf of cell \[E_{cell}^{o}={{E}^{o}}-\frac{0.059}{n}\log \frac{[prdouct]}{[Reactant]}\] Given,     \[{{E}^{o}}=1.10\,V\] \[[Z{{n}^{2+}}]=0.1\,M\] \[[C{{u}^{2+}}]=0.1\,M\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.059}{2}\log \left[ \frac{Z{{n}^{2+}}}{C{{u}^{2+}}} \right]\] (\[\because \,\] 2 electrons are involved in cell reaction and concentration of solids is taken as unity) \[\therefore \]      \[{{E}_{cell}}=1.10-\frac{0.059}{2}\log \left[ \frac{0.1}{0.1} \right]\]                 or            \[=1.10-\frac{0.059}{2}\log 1\]                                 \[=1.10-\frac{0.059}{2}\times 0\]                 \[\therefore \]      \[{{E}_{cell}}=1.10\,V\]


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