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KVPY Sample Paper KVPY Stream-SX Model Paper-18

  • question_answer
    Calculate the potential of a half cell having reaction:\[A{{g}_{2}}S\,(s)+2e-2\,Ag\,(s)+{{S}^{2\,-}}\,(aq)\] in a solution buffered at pH = 3 and which is also saturated with 0.1 M \[{{H}_{2}}S\,(aq).\] \[({{K}_{{{a}_{1}}}}.{{K}_{{{a}_{2}}}}{{H}_{2}}S={{10}^{-24}},\] \[{{K}_{sp}}(A{{g}_{2}}S)={{10}^{-49}}\]

    A)  1.18

    B) 0.19

    C) \[-\,0.19V\]

    D) None

    Correct Answer: C

    Solution :

    \[{{H}_{2}}S2{{H}^{+}}{{S}^{2-}}\]
    \[{{K}_{{{a}_{1}}}}.{{K}_{{{a}_{2}}}}=\frac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{[{{H}_{2}}S]}=\frac{{{({{10}^{-3}})}^{2}}[{{S}^{-2}}]}{0.1}\]
    \[\Rightarrow [{{S}^{2-}}]=\frac{{{10}^{-21}}\times 0.1}{{{({{10}^{-3}})}^{2}}}={{10}^{-16}}\]
    \[{{E}_{{{S}^{2-}}}}|A{{g}_{2}}S|Ag\]\[=E_{A{{g}^{+}}/Ag}^{{}^\circ }-\frac{0.0591}{2}\,\log \,\frac{[{{S}^{2-}}]}{[{{K}_{sp}}]}\]
    \[0.8-\frac{0.0591}{2}\log \frac{{{10}^{-\,16}}}{{{10}^{-\,49}}}=-\,0.19\,\,V\]


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