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KVPY Sample Paper KVPY Stream-SX Model Paper-31

  • question_answer
    You are given the following cell at 298 K,  \[Zn\left| \,\begin{matrix}    Z{{n}^{++}}\,(aq.)  \\    00.1\,M  \\ \end{matrix}\, \right|\,\,\,\left| \,\begin{matrix}    HCl\,(aq.)  \\    1.0\,lit  \\ \end{matrix}\, \right|\,\,\,\left| \,\begin{matrix}    {{H}_{2}}(g)  \\    1.0\,atm  \\ \end{matrix}\, \right|Pt\] With \[{{E}_{cell}}=0.701\] and \[E_{z{{n}^{2+}}/Zn}^{0}=-0.76\,\,V.\] Which of the following amounts of NaOH (equivalent weight = 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 : \[Zn+2{{H}^{+}}\,\,\,\,{{H}_{2}}+Z{{n}^{2+}}\]
    \[E_{cell}^{0}=0-(-0.76)=0.76\,V\] \[\therefore \,\,\,0.701=0.76-\frac{0.059}{2}\,\,\log \,\,\frac{0.01\times 1}{{{[{{H}^{+}}]}^{2}}}\]
    \[\therefore \,\,\,[{{H}^{+}}]={{10}^{-2}}M\]
    \[\therefore \,\,\,NaOH\]Required is 0.01 mole = 0.4 gms.
     


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