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NEET Chemistry Electrochemistry / विद्युत् रसायन Question Bank Self Evaluation Test - Electrochemistry

  • question_answer
    Following cell has EMF 0.7995V. \[Pt|{{H}_{2}}(1\,atm)|HN{{O}_{3}}(1M)||AgN{{O}_{3}}(1M)|Ag\] If we add enough \[KCl\] to the Ag cell so that the final \[C{{l}^{-}}\] is 1M. Now the measured emf of the cell is 0.222V. The \[{{K}_{sp}}\] of \[AgCl\] would be -

    A) \[1\times {{10}^{-9.8}}\]

    B) \[1\times {{10}^{-19.6}}\]

    C) \[2\times {{10}^{-10}}\]          

    D) \[2.64\times {{10}^{-14}}\]

    Correct Answer: A

    Solution :

    [a] \[2A{{g}^{+}}+{{H}_{2}}\xrightarrow{{}}2{{H}^{+}}+2Ag\] \[E=E{}^\circ -\frac{0.0591}{2}\log \frac{{{[{{H}^{+}}]}^{2}}}{{{P}_{{{H}_{2}}}}\times {{[A{{g}^{+}}]}^{2}}}\] \[0.222=0.7995-\frac{0.0591}{2}log\frac{1}{{{[A{{g}^{+}}]}^{2}}}\] \[[A{{g}^{+}}]={{10}^{-9.8}}\] \[{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]=({{10}^{-9.8}})\times (1)={{10}^{-9.8}}\]


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