InterviewSolution
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InterviewSolution
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Consider the cell `Pt|H_(2(g,1atm))|H_((aq.1M))^(+)||Fe_((aq))^(3+),Fe_((aq))^(2+)||Pt_((s))` Given that `E_(Fe^(3+)|Fe^(2+))^(o)=0.771V` the ratio of conc. Of `Fe_((aq))^(2+)` to `Fe_((aq))^(3+)` is, when the cell potential is 0.830VA. 0.101B. 0.924C. 0.12D. None of these |
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Answer» Correct Answer - A For oxidation-reduction in half-cell: The half-cell reaction is `Fe_((aq))^(3+)+etoFe_((aq))^(2+)` `Fe_(Fe^(3+)//Fe^(2+))=E_(Fe^(3+)//Fe^(2+))^(o)-(0.059)/(1)log(([Fe^(2+)])/([Fe^(3+)]))` `0.83=0.771-(0.0591)/(1)log(([Fe^(2+)])/([Fe^(3+)]))` `implies([Fe^(2+)])/([Fe^(+3)])=0.10039~0.1004` |
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