The Edison strong cell is represented as :
`Fe(s)|FeO(s)|KOH(aq)|Ni_(2)O_(3)(s)|Ni(s)`
The half - cell reactions are as follows:
`Ni_(2)O_(3)(s)+H_(2)O(l)+2e hArr 2NiO(s)+2OH^(-), E^(@)=+0.40V`
`FeO(s)+H_(2)O(l)+2e hArr Fe(s)+2OH, E^(@)=-0.87V`
(2) What is the cell EMF? How does it depend on the concentration of KOH?

Explanation

Given, `E_(FeO|Fe)^(@)=-0.87V, E_(Ni_(2)O_(3)|NiO)^(@)=+0.40V`
`"At anode :"Fe(s)+2OH^(-)(l)rarr FeO(s)+H_(2)O(l)+2e`
`"At cathode : "Ni_(2)O_(3)(s)+H_(2)O(l)+2e rarr 2NiO(s)+2OH^(-)(aq)`
(2) `E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)=0.40-(-0.87)=+1.27V`
`E_("cell")=E_("cell")^(@)-(0.059)/(2)log.([FeO][NiO]^(2))/([Fe][Ni_(2)O_(3)])`
`=E^(@)-(0.059)/(2)log.(1xx1^(2))/(1xx1)=E^(@)`
`" "[because " Molar concentration of pure solid = 1 "]`
The expression for `E_("cell")` does not contain concentration term of `OH^(-)`, so `E_("cell")` is independent of `OH^(-)` concentration.