CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    Given the standard reduction potentials \[Z{{n}^{2+}}/Zn=-0.74\,V,\]       \[C{{l}_{2}}/C{{l}^{-}}=1.36\,V,\] \[{{H}^{+}}/1/2{{H}_{2}}=0\,V\]and \[F{{e}^{2+}}/F{{e}^{3+}}=0.77\,V\] The order of increasing strength as reducing agent is:

    A) \[C{{l}^{-}},Zn,{{H}_{2}},F{{e}^{2+}}\]

    B)        \[{{H}_{2}},Zn,F{{e}^{2+}},Cl\]

    C) \[C{{l}^{-}}F{{e}^{2+}},Zn,{{H}_{2}}\]     

    D) \[{{H}_{2}}F{{e}^{2+}},C{{l}^{-}},Zn\]

    E) \[C{{l}^{-}},F{{e}^{2+}},{{H}_{2}},Zn\]

    Correct Answer: E

    Solution :

    The substances which have lower reduction potentials are stronger reducing agents. Hence the order of strength of reducing agent is increases in the following order: \[\xrightarrow[strength\text{ }of\text{ }reducing\text{ }agent-increases]{Cl<F{{e}^{2+}}<{{H}_{2}}<Zn}\]


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