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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2009

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

    A)   \[-0.32V\]

    B)  \[-1.20V\]

    C)  \[+1.20V\]

    D)  \[+0.32V\]

    Correct Answer: D

    Solution :

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


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