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NEET Chemistry Electrochemistry / विद्युत् रसायन Question Bank Self Evaluation Test - Electrochemistry

  • question_answer
    A solution contains \[F{{e}^{2+}},F{{e}^{3+}}\] and \[{{I}^{-}}\] ions. This solution was treated with iodine at\[35{}^\circ C\]. \[E{}^\circ \] for \[F{{e}^{3+}}/F{{e}^{2+}}\] is + 0.77 V and \[E{}^\circ \] for \[{{I}_{2}}/2{{I}^{-}}=0.536V.\] he favourable redox reaction is:

    A) \[{{I}_{2}}\] will be reduced to \[{{I}^{-}}\]

    B) There will be no redox reaction

    C) \[{{I}^{-}}\] will be oxidised to \[{{I}_{2}}\]

    D) \[F{{e}^{2+}}\] will be oxidised to \[F{{e}^{3+}}\]

    Correct Answer: C

    Solution :

    [c] Given \[F{{e}^{3+}}/F{{e}^{2+}}=+0.77V\] and \[{{I}_{2}}/2{{I}^{-}}=0.536V\] \[2({{e}^{-}}+F{{e}^{3+}}\xrightarrow{{}}F{{e}^{2+}})\,E{}^\circ =0.77V\] \[2{{I}^{-}}\xrightarrow{{}}\,{{I}_{2}}+2{{e}^{-}}\,\,\,\,E{}^\circ =-0.536V\] \[2F{{e}^{3+}}+2{{I}^{-}}\xrightarrow{{}}2F{{e}^{2+}}+{{I}_{2}}\] \[E{}^\circ =E{{{}^\circ }_{ox}}+E{{{}^\circ }_{red}}\] \[=0.77-0.536=0.164V\] So, reaction will take place.


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