2. Solved Papers
  3. AMU Medical

AMU Medical AMU Solved Paper-2015

  • question_answer
    The emf of the following three cells I. \[Zn\left| Z{{n}^{2+}}\,\,(1M) \right|\left| C{{u}^{2}}(1M)\,\, \right|Cu\] II. \[Zn\left| Z{{n}^{2+}}\,\,(0.1M) \right|\left| C{{u}^{2}}(1M)\,\, \right|Cu\] III. \[Zn\left| Z{{n}^{2+}}\,\,(1M) \right|\left| C{{u}^{2+}}(0.1M)\,\, \right|Cu\] are represented by \[{{E}_{1}},{{E}_{2}}\] and \[{{E}_{3}}\]. Which of the following statement is correct?

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

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

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

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

    Correct Answer: D

    Solution :

                    \[Zn+C{{u}^{2+}}\xrightarrow{{}}Z{{n}^{2+}}+Cu\]From Nernst equation, \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.591}{n}\log \frac{\left[ Product \right]}{\left[ Reactant \right]}\]                 \[{{E}_{1}}=-\frac{0.0591}{2}\log \frac{1\,M}{1\,M}=0\]                 \[{{E}_{2}}=-\frac{0.0591}{2}\log \frac{0.1\,M}{1\,M}\]                 \[=+0.02955\]                 \[{{E}_{3}}=-\frac{0.0591}{2}\log \,\frac{1\,M}{0.1M}=-0.0591\]                 Thus, \[{{E}_{2}}>{{E}_{1}}>{{E}_{3}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner