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EAMCET Medical EAMCET Medical Solved Paper-2003

  • question_answer
    If the standard electrode potential of \[\text{C}{{\text{u}}^{\text{2+}}}\text{/Cu}\]electrode is 0.34 V, what is the electrode potential at 0.01 M concentration of \[\text{C}{{\text{u}}^{2+}}\]? \[(T=298{{\,}^{o}}K)\]

    A)  0.399 V               

    B)  0.28 IV

    C)  0.222V                                

    D)  0.176V

    Correct Answer: B

    Solution :

                     It can be calculated with the help of Nernst equation. \[E={{E}^{o}}+\frac{0.059}{n}\log [{{M}^{n+}}]\] Where, E = emf of cell \[{{E}^{o}}=\]Standard emf o fcell n = number of \[{{e}^{-}}\]transferred \[{{\text{M}}^{\text{n+}}}\text{=}\]Concentration \[{{E}^{o}}=0.34\,\,V\] \[n=2\] \[[C{{u}^{2+}}]={{10}^{-2}}M\] So,   \[E=0.34+\frac{0.059}{2}\log {{10}^{-2}}\] \[=0.34+\frac{0.059}{2}\times -2\] \[=0.34-0.059\] \[=+\,0.281\,V\]


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