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JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    Given the data at\[25{}^\circ C,\] \[Ag+{{I}^{-}}\xrightarrow{{}}AgI+{{e}^{-}};\]                     \[E{}^\circ =0.152\text{ }V\] \[Ag\xrightarrow{{}}A{{g}^{+}}+{{e}^{-}};\]                     \[E{}^\circ =-0.800\text{ }V\] What is the value of\[\log \text{ }{{K}_{sp}}\]for\[AgI\]? \[\left( 2.303\frac{RT}{F}=0.059\,V \right)\]     AIEEE  Solved  Paper-2006

    A) \[\text{ }8.12\]                 

    B)        \[+\text{ }8.612\]            

    C)        \[-37.83\]                           

    D) \[-16.13\]

    Correct Answer: D

    Solution :

    \[Agl(s)+{{e}^{-}}AG(s)+l;\]      \[{{E}^{o}}=-0.152\] \[\begin{align}   & \underline{Ag(s)\xrightarrow{{}}A{{g}^{+}}+{{e}^{-}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{E}^{o}}=-0.8} \\  & Agl(s)\xrightarrow{{}}A{{g}^{+}}+{{l}^{-}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{E}^{o}}=-0.952 \\ \end{align}\] \[E_{cell}^{o}=\frac{0.059}{1}\log {{K}_{sp}}\] \[-0.952=\frac{0.059}{1}\log {{K}_{sp}}\] \[\log {{K}_{sp}}=\frac{-0.952}{0.059}=-16.135\]


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