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JEE Main & Advanced Sample Paper JEE Main Sample Paper-37

  • question_answer
    What would be the reduction potential of an electrode at \[298\text{ }K,\]which originally contained \[1M\,\,{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] solution in acidic buffer solution of \[pH=1.0\]and which was treated with 50% of the \[Sn\] necessary to reduce all \[C{{r}_{2}}{{O}_{7}}^{2-}\] to \[C{{r}^{3+}}\] Assume pH of solution remains constant. Given: \[E_{C{{r}_{2}}O_{7}^{2-}/C{{r}^{3+}},{{H}^{+}}}^{0}=1.33V,\,\log 2\,=0.3,\] \[\frac{2.303\,RT}{F}=0.06\]

    A)  \[1.285V~\]                 

    B)  \[1.193\,V\]

    C)  \[1.187\,V\]                  

    D)  None of these

    Correct Answer: C

    Solution :

          \[C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O\]
    Initial \[co{{n}^{n}}\] \[1\] \[0.1\] \[0\]
    After reaction \[0.5\] \[0.1M\] \[1M\]
    \[{{E}_{RP}}=E_{RP}^{0}-\frac{0.06}{6}\log \frac{{{[C{{r}^{3+}}]}^{2}}}{[C{{r}_{2}}O_{7}^{2-}]{{[{{H}^{+}}]}^{14}}}\] \[{{E}_{RP}}=1.33-\frac{0.06}{6}\log \frac{1}{(0.5)\,{{(0.1)}^{14}}}\]             \[1.33-\frac{0.06}{6}\log (2\times {{10}^{14}})=1.187\,V\]


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