1.

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 0.830V

Answer»

0.101
0.924
0.12
None of these

Solution :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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