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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    The standard electrode potential for the half-cell reactions are \[Z{{n}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Zn;\,\,{{E}^{o}}=-0.76\,\,V\] \[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Fe;\,\,{{E}^{o}}=-0.44\,\,V\] The emf of the cell reaction, \[F{{e}^{2+}}+Zn\xrightarrow{{}}Z{{n}^{2+}}+Fe\]is

    A) \[-0.32\,\,V\]                    

    B) \[-1.20\,\,V\]

    C) \[+1.20\,\,V\]   

    D)        \[+0.32\,\,V\]

    Correct Answer: D

    Solution :

    \[Z{{n}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Zn;\,\,{{E}^{o}}=-0.76\,\,V\] \[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Fe;\,\,{{E}^{o}}=-0.44\,\,V\] Cell reaction is \[F{{e}^{2+}}+Zn\xrightarrow{{}}Z{{n}^{2+}}+Fe\]                 \[{{E}_{cell}}={{E}_{cathode}}-{{E}_{anode}}\]                          \[=-0.44-(-0.76)\]                          \[=-0.44+0.76\]                          \[=\text{ }0.32V\]


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