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)` ?

Explanation

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):}`