VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    For the following cell reaction, \[911\] \[104\] \[52153\] \[Alkyne\xrightarrow[Lindlars\,\,catalyst]{{{H}_{2}}}A\xrightarrow{Ozonolysis}\underset{only}{\mathop{B}}\,\] \[\xleftarrow[\Pr ocess]{Wac\ker }C{{H}_{2}}=C{{H}_{2}}\] of the cell is

    A)  \[{{H}_{3}}C-C\equiv C-C{{H}_{3}}\]

    B)  \[{{H}_{3}}C-C{{H}_{2}}-C\equiv CH\]

    C)  \[{{H}_{2}}C=CH-C\equiv CH\]

    D)  None of these

    Correct Answer: A

    Solution :

     For the given cell, \[Ag|A{{g}^{+}}|AgCl|C{{l}^{O-}}|C{{l}_{2}}Pt\] the cell reactions are as follows At anode: \[Ag\xrightarrow{{}}A{{g}^{+}}+{{e}^{-}}\] At cathode: \[\underline{AgCl+{{e}^{-}}\xrightarrow{{}}Ag(s)+C{{l}^{-}}}\] Net cell reaction: \[AgCl\xrightarrow{{}}A{{g}^{+}}+C{{l}^{-}}\] \[\therefore \] \[\Delta G_{reaction}^{o}=\Sigma \Delta G_{p}^{o}-\Sigma \Delta G_{R}^{o}\] \[=(78-129)-(-109)\] \[=+58kJ/mol\]           \[\Delta {{G}^{o}}=-nF{{E}^{o}}\] \[58\times {{10}^{3}}J=-1\times 96500\times E_{cell}^{o}\] \[E_{cell}^{o}=\frac{-58\times 1000}{96500}\] \[=-0.6V\]


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