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J & K CET Medical J & K - CET Medical Solved Paper-2012

  • question_answer
    Standard electrode potential of half-cell reactions are given below \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Cu;\] \[E{}^\circ =0.34V\] \[Z{{n}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Zn;\] \[E{}^\circ =-0.76V\] What is the emf of the cell?

    A)  \[+1.10V\]        

    B)  \[-1.10V\]

    C)  \[-0.42V\]         

    D)  \[+0.42V\]

    Correct Answer: A

    Solution :

     \[Zn\xrightarrow[{}]{{}}Z{{n}^{2+}}+2{{e}^{-}}\] (oxidation half-reaction) \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Cu\](reduction half-reaction) Emf of the cell = standard oxidation potential of zinc electrode + standard reduction potential of Cu electrode \[=0.76\text{ }V+0.34V\] \[=1.10V\]


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