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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Electrode potential Ecell Nernst equation and ECS

  • question_answer
    The \[{{E}^{0}}_{{{M}^{3+}}/{{M}^{2+}}}\]values for \[Cr,\ Mn,\ Fe\]and \[Co\]are \[-0.41,\ +1.57,\ +0.77\]and \[+1.97\ V\] respectively. For which one of these metals the change in oxidation state from \[+2\] to \[+3\]is easiest                                                [AIEEE 2004]

    A)                 \[Fe\]   

    B)                 Mn

    C)                 Cr           

    D)                 Co

    Correct Answer: C

    Solution :

              
    Reduction \[{{E}_{0}}{{M}^{3+}}/{{M}^{2+}}\] Cell reaction \[{{E}_{0}}{{M}^{2+}}/{{M}^{3+}}\](Oxidation)
    ? .41 V \[C{{r}^{2+}}\] \[C{{r}^{3+}}\] + . 41 V
    + 1.57 V \[M{{n}^{2+}}\] \[M{{n}^{3+}}\] ? 1.57 V
    + 0.77 V \[F{{e}^{2+}}\] \[F{{e}^{3+}}\] ? 0.77 V
    + 1.97 V \[C{{o}^{2+}}\] \[C{{o}^{3+}}\] ? 1.97 V
                                    As \[Cr\] has maximum oxidation potential value, therefore its oxidation should be easiest.


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