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WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    The standard reduction potential \[{{\text{E}}^{\text{o}}}\]for half- reaction are \[Zn\to Z{{n}^{2+}}+2{{e}^{-}};{{E}^{o}}=+0.76\,V\] \[Fe\to F{{e}^{2+}}+2{{e}^{-}};{{E}^{o}}=+\,0.41\,V\] The EMF of the cell reaction \[F{{e}^{2+}}+Zn\to Z{{n}^{2+}}+Fe\]is

    A)  \[-0.35\,V\]                      

    B)  \[+\,0.35\text{ }V\]

    C)  \[+1.17\,V\]                     

    D)  \[-1.17\,V\]

    Correct Answer: B

    Solution :

                     \[Zn\xrightarrow{{}}Z{{n}^{2+}}+2{{e}^{-}};E_{oxi}^{o}=+0.76\,V\]or \[E_{red}^{o}=-0.76\,V\] \[Fe\xrightarrow{{}}F{{e}^{2+}}2{{e}^{-}};E_{oxi}^{o}=+0.41\,V\] or \[E_{red}^{o}=-0.41\,V\] \[\therefore \]  \[E_{cell}^{o}=E_{cathode}^{o}(RP)-E_{anode}^{o}(RP)\]                 \[=-0.41-(0.76)\] \[=+\,0.35\,V\]


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