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BCECE Medical BCECE Medical Solved Papers-2004

  • question_answer
    The standard reduction potential \[{{E}^{o}}\] for the half reactions are as : \[Zn\xrightarrow{{}}Z{{n}^{2+}}+2{{e}^{-}},\,\,{{E}^{o}}=0.76\,\,V\] \[Cu\xrightarrow{{}}C{{u}^{2+}}+2{{e}^{-}},\,\,{{E}^{o}}=0.34\,\,V\] The emf for the cell reaction: \[Zn+C{{u}^{2+}}\xrightarrow{{}}Z{{n}^{2+}}+Cu\]

    A)  0.42 V          

    B)  - 0.42 V

    C)  - 1.1 V        

    D)  1.1 V

    Correct Answer: A

    Solution :

    Key Idea : (i) Decide cathode and anode (ii) \[{{E}^{o}}_{cell}={{E}^{o}}_{C}-{{E}^{o}}_{A}\] when values of electrode potential are substituted in terms of reduction potential. Given,   \[E_{Zn/Z{{n}^{2+}}}^{o}=0.76\,\,V\]                 \[E_{Cu/C{{u}^{2+}}}^{o}=0.34\,\,V\] \[\therefore \] Zn is anode (\[\because \] it has higher oxidation potential) \[\therefore \] \[E_{Z{{n}^{2+}}/Zn}^{o}=-0.76\,\,V\] and        \[E_{C{{u}^{2+}}/Cu}^{o}=-0.34\,\,V\] \[{{E}^{o}}_{cell}={{E}^{o}}_{cell}-{{E}^{o}}_{A}\]


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