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BCECE Engineering BCECE Engineering Solved Paper-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: Decide cathode and anode \[{{E}^{o}}_{cell}={{E}^{o}}_{C}-{{E}^{o}}_{A}\]of electrode  potential are substituted in terms of reduction potential. Given,   \[E_{Zn/Z{{n}^{2+}}}^{o}=0.76V\] \[E_{Cu/C{{u}^{2+}}}^{o}=0.34\,V\]        \[\therefore \]Zn is anode (\[\because \] it has higher oxidation potential)                 \[\therefore \]  \[E_{Z{{n}^{2+}}/Zn}^{o}=-0.76\,V\]                 and        \[E_{C{{u}^{2+}}/Cu}^{o}=-0.34\,V\]                                 \[{{E}^{o}}_{cell}={{E}^{o}}_{C}-{{E}^{o}}_{A}\]                                 \[=-0.34\,V-(-0.76\,V)\]                                 \[=0.34\,V+0.76\,V\]                                 \[=0.42\,V\]


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