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KVPY Sample Paper KVPY Stream-SX Model Paper-20

  • question_answer
    An alloy weighing 1.05 g of \[\operatorname{Pb}-Ag\] was dissolved in desired amount of \[{{\operatorname{HNO}}_{3}}\] and the volume was made 350 mL. An Ag electrode was dipped in solution   and   \[{{E}_{cell}}\] of the cell \[Pt,\,\underset{1\,atm}{\mathop{{{H}_{2}}}}\,\underset{1M}{\mathop{\left| {{H}^{+}} \right|}}\,A{{g}^{+}}|Ag\] was 0.503 V at 298 K. The percentage of lead in alloy is: [Given\[E_{Ag+/Ag}^{o}~= 0.80 V\]]

    A) 99.97

    B) 98.50

    C) 0.033

    D) 0.33

    Correct Answer: A

    Solution :

    The cell reaction are:
    Anode: \[{{H}_{2}}\xrightarrow{{}}2{{H}^{+}}+2{{e}^{-}}\]
    Cathode: \[2A{{g}^{+}}+2{{e}^{-}}\xrightarrow{{}}2Ag\]
    Thus, \[{{E}_{cell}}=E_{O}^{{}^\circ }+E_{R}^{{}^\circ }+\frac{0.059}{2}{{\log }_{10}}\frac{{{[A{{g}^{+}}]}^{2}}{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}\] or \[0.503=0+0.80+\frac{0.059}{2}{{\log }_{10}}{{[A{{g}^{+}}]}^{2}}\] or \[[A{{g}^{+}}]=9.25\times {{10}^{-6}}M\]
    \[\therefore \]Mole of \[{{\operatorname{Ag}}^{+}}\] in 350mL\[=9.25\times {{10}^{-6}}\times \frac{350}{1000}\]
    \[\therefore \] Weight of\[{{\operatorname{Ag}}^{+}}\]in 350mL\[=9.25\times {{10}^{-6}}\times \frac{350}{1000}\times 108=3.497\times {{10}^{-4}}g\]
    \[\therefore \]\[%\]Of Ag in 1.05g alloy\[=\frac{3.497\times {{10}^{-4}}}{1.05}\times 100=0.033%\]
    \[\therefore \]\[%\]of lead in alloy \[=99.97%~\]


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