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CMC Medical CMC-Medical Ludhiana Solved Paper-2011

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

    A)  \[-\,0.32V\]                      

    B)  \[-\,1.20\,V\]

    C)  \[+1.20\text{ }V\]         

    D)  \[+\,0.32\text{ }V\]

    Correct Answer: D

    Solution :

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


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