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Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    If three faradays of electricity is passed through the solutions of\[AgN{{O}_{3}},CuS{{O}_{4}}\]and \[AuC{{l}_{3}},\]the molar ratio of the cations deposited at the cathode will be

    A)  \[1:1:1\]        

    B)  \[1:2:3\]

    C)  \[3:2:1\]       

    D)  \[6:3:2\]

    Correct Answer: D

    Solution :

     \[A{{g}^{+}}+{{e}^{-}}\xrightarrow[{}]{{}}Ag\] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Cu\] \[A{{u}^{3+}}+3{{e}^{-}}\xrightarrow[{}]{{}}Au\] There faradays will deposit 3 moles of Ag, 1.5 moles of Cu and 1 mole of Au. Hence, molar ratio\[=3:1.5:1=6:3:2\]


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