InterviewSolution
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InterviewSolution
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The Edison storage cell is represented as, `Fe_((s))|FeO_(S)||KOH_((aq.))||Ni_(2)O_(3(S))|No_((S))` the half-cell reactions are : `Ni_(2)O_(3(S))+H_(2)O_((l))+2e^(-)rarr 2NiO_((S))+2OH^(-) , E^(@)=+0.40V` `FeO_((S))+H_(2)O_((I))+2e^(-)rarr Fe_((S))+2OH^(-) , E^(@)=-0.87V` (i) What is the cell reaction ? (ii) What is the cell e.m.f. ? How does ir depend on the concentration of KOH ? (iii) What is the maximum amount of electrical energy that can be obtained from one mole of `Ni_(2)O_(3)` ? |
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Answer» Given, `E_(FeO //Fe)^(@)=-0.87V,E_((Ni_(2)O_(3)//NiO))=+0.40V` `thereforeE_(FeO //Fe)^(@)=+0.887V,E_(NiO //Ni_(2)O_(3))^(@)=-0.40V` Since, `E_(OP)^(@)" for Fe"//FeOgtE_(Op)^(@)" for NiO/Ni"_(2)O_(3)`, and thus, redox change are, At anode `Fe_((S))+2OH^(-)rarrFeO_((S))+underset(("oxidation"))(H_(2)O_((l)))+2e` At cathode `Ni_(2)O_(3(S))+H_(2)O_((l))+2erarrunderset(("reduction"))(2NiO_((S))+2OH^(-))` Redox reaction: `Fe_((S))+Ni_(2)O_(3(S))rarrFeO_((S))+2NiO_((S)` i) `E_(cell)==E_(OP_(Fe//FeO))^(@)-(0.059)/(2)log_(10).([H_(2)O])/([OH^(-)]^(2))+E_(RP_(Ni_(2)O_(3)//NiO))^(@)+(0.059)/(2)log_(10).([H_(2)O])/([OH^(-)]^(2))` `E_(RP_(Ni_(2)O_(3)//NiO))^(@)=0.87+0.40=1.27V` ii) The `E_(cell)` is independent of `OH^(-)` ion concentration. iii) `{:(-DeltaG^(0)=nE^(@)F,=2xx1.27xx96500),(,=245.11J),(,=245.11kJ):}` |
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