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AMU Medical AMU Solved Paper-2012

  • question_answer
    Consider two half-cells based on the reaction\[A{{g}^{+}}(aq)+{{e}^{-}}\xrightarrow{{}}Ag\,(s)\] The left half-cell contains \[A{{g}^{+}}\] ions at unit concentration, and the right half-cell initially had the same concentration of \[A{{g}^{+}}\] ions, but just enough \[NaCl\,\,(aq)\] had been added to completely precipitate the \[A{{g}^{+}}\,\,(aq)\] as \[AgCl\]. If the emf of the cell is 0.29 V, the \[{{\log }_{10}}\,\,{{K}_{sp}}\]would have been

    A)  9.804                   

    B)  - 9.804

    C)  - 4.902                 

    D)  10.004

    Correct Answer: B

    Solution :

                    Anode half reaction                 \[Ag\xrightarrow{{}}A{{g}^{+}}+{{e}^{-}}\] Cathode half reaction                 \[A{{g}^{+}}\xrightarrow{{}}Ag+{{e}^{-}}\] \[NaCl\] is added to precipitate \[A{{g}^{+}}\] and \[AgCl\], thus \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \frac{[A{{g}^{+}}]\,[Cl]}{[A{{g}^{+}}]}\]                 \[{{E}_{cell}}=0-\frac{0.0591}{2}\log \,\,{{K}_{sp}}\]                                                 \[\because [A{{g}^{+}}]=1\] (Given)]                 \[0.29\times 2=-0.0591\,\,\log \,\,{{K}_{sp}}\]                 \[{{\log }_{10}}\,\,{{K}_{sp}}=-\frac{0.29\times 2}{0.0591}=-4.906\times 2\]                                 = - 9.804


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