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MGIMS WARDHA MGIMS WARDHA Solved Paper-2003

  • question_answer
    The\[E{}^\circ \]for half cells\[Fe/F{{e}^{2+}}\]and\[Cu/C{{u}^{2+}}\] are\[-0.44\text{ }V\]and\[+0.32\text{ }V\]respectively. Then:

    A) \[C{{u}^{2+}}\]oxidises\[Fe\]

    B) \[C{{u}^{2+}}\]oxidises \[F{{e}^{2+}}\]

    C) \[Cu\]oxidises\[F{{e}^{2+}}\]

    D) \[Cu\]reduces\[F{{e}^{2+}}\]

    Correct Answer: A

    Solution :

                     The standard reduction potential of\[F{{e}^{2+}}/Fe,\]is less than that of\[C{{u}^{2+}}/Cu,\]hence Fe is more electropositive than copper. Hence,\[C{{u}^{2+}}\]can oxidise\[Fe\].


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