InterviewSolution
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InterviewSolution
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An excess of liquid mercury is added to an acidicfied solution of `1.0xx10^(-3) M Fe^(3+)` . It is found that `5%` of `Fe^(3+)` remains at equilibrium at `25^(@)C`. Calculate `E^(c-)._((Hg_(2)^(2+)|Hg))` assuming that the only reaction that occurs is `2Hg+2Fe^(3+) rarr Hg_(2)^(2+)+2Fe^(2+)` Given `: E^(c-)._((Fe^(3+)|Fe^(2+)))=0.77V` |
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Answer» For `2Hg+2Fe^(3+)hArrHg_(2)^(2+)+2Fe^(2+)` `{:("before reaction Excess 10"^(-3)),("after reaction Excess 10"^(-3)xx(5)/(100)):}` `{:(" 0 0"),((95)/(2xx100)xx10^(-3)" "(95)/(100)xx10^(-3)):}` For cell at equilibrium `E_(cell)=0=E_(OP_(Hg//Hg_(2)^(2+)))+E_(RP_(Fe^(3+)//Fe^(2+)))` `0=E_(OP_(Hg//Hg_(2)^(2+)))^(0)-(0.059)/(2)log_(10)[Hg_(2)^(2+)]+E_(RP_(Fe^(3+)//Fe^(2+)))^(0)+(0.059)/(2)log_(10).([Fe^(3+)]^(2))/([Fe^(2+)]^(2))` `0=E_(OP_(Hg//Hg_(2)^(2+)))^(0)+0.77+(0.059)/(2)log_(10).([Fe^(3+)])/([Fe^(2+)]^(2)[Hg_(2)^(2+)])` `(becauseE_(OP_(Fe^(2+)//Fe^(3+)))^(0)=-0.77VthereforeE_(RP_(Fe^(3+)//Fe^(2+)))^(0)=+0.77V)` or `E_(OP_(Hg//Hg_(2)^(2+)))^(0)=-0.771-(0.059)/(2)` `([(5)/(100)xx10^(-3)])/([(95xx10^(-3))/(100)]^(2)[(95xx10^(-3))/(2xx100)])=-0.793V` |
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