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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    Give that, \[E_{F{{e}^{2+}}/Fe}^{\circ }=-0.44V,\]\[E_{F{{e}^{2+}}/F{{e}^{3+}}}^{\circ }=-0.77V.\] if \[F{{e}^{2+}},F{{e}^{3+}}\] and Fe solid are kept together, then

    A) \[F{{e}^{3+}}\]decreases

    B) \[F{{e}^{3+}}\]increases

    C)  \[F{{e}^{2+}}\]decreases

    D) \[F{{e}^{2+}},F{{e}^{3+}}\]remain unchanged

    Correct Answer: A

    Solution :

    Cell reaction, \[2F{{e}^{3+}}+Fe\to 3F{{e}^{2+}}\] \[E_{Cell}^{\circ }=0.77-(-0.44)=1.21\text{V}\] \[\Rightarrow \]\[F{{e}^{3+}}\]and Fe will reduce.


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