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KVPY Sample Paper KVPY Stream-SX Model Paper-11

  • question_answer
    You are given the following cell at 298 K, \[Zn\,\,\left| \begin{matrix}    Z{{n}^{++}}_{(aq.)}  \\    0.01\,\,M  \\ \end{matrix} \right|\,\,\left| \begin{matrix}    HC{{l}_{(aq.)}}  \\    1.0\,\,\text{lit}  \\ \end{matrix} \right|\,\,\left| \begin{matrix}    {{H}_{2}}\,(g)  \\    1.0\,\,\,\text{atm}  \\ \end{matrix} \right|\,\,Pt\] with \[{{E}_{\text{cell}}}=0.701\] and \[E_{Z{{n}^{2\,+}}/Zn}^{0}=-\,0.76\,V.\] Which of the following amounts of NaOH  \[(\text{equivalent }\,\text{weight}=\text{40})\] will just make the pH of cathodic compartment to be equal to 7.0:

    A) 0.4 gms

    B) 4 gms

    C) 10 gms

    D) 2 gms

    Correct Answer: A

    Solution :

    Anode               \[Zn\to Z{{n}^{2+}}+2{{e}^{-}}\]
    Cathode \[2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}\]
    Cell:                  \[Zn+2{{H}^{+}}{{H}_{2}}+Z{{n}^{2+}}\]
    \[;\,\,E_{\text{cell}}^{0}=0-(-\,0.76)=0.76\,V\]
    \[\therefore \]      \[0.701=0.76-\frac{0.059}{2}\log \frac{0.01\times 1}{{{[{{H}^{+}}]}^{2}}}\]
    \[\therefore \]      \[[{{H}^{+}}]={{10}^{-\,2}}M\]
    \[\therefore \]      NaOH required is 0.01 mole = 0.4 gms.


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