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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2009

  • question_answer
        Electrode potential data are given below \[F{{e}^{3+}}(aq)++{{e}^{-}}F{{e}^{2+}}(aq)\]    \[E{}^\circ =+0.77V\] \[A{{l}^{3+}}(aq)+3{{e}^{-}}\xrightarrow[{}]{{}}Al(s)\]    \[E{}^\circ =-1.66\,V\] \[B{{r}_{2}}(aq)+2{{e}^{-}}\xrightarrow[{}]{{}}2B{{r}^{-}}(aq)\]   \[E{}^\circ =+1.08\,V\] Based on the data given above, reducing power of\[F{{e}^{2+}},Al\]and\[B{{r}^{-}},\]will increase, in the order

    A)  \[B{{r}^{-}}<F{{e}^{2+}}<Al\]    

    B)  \[F{{e}^{2+}}<Al<B{{r}^{-}}\]

    C)  \[Al<B{{r}^{-}}<F{{e}^{2+}}\]    

    D)  \[Al<F{{e}^{2+}}<B{{r}^{-}}\]

    Correct Answer: A

    Solution :

                    Reducing power ie, the tendency to lose electrons increases as the reduction potential decreases so the increasing order of reducing power of the given ions is\[B{{r}^{-}}~<F{{e}^{2+}}<Al\].


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