The Edison storage cell is represented as `:`
`Fe(s)|FeO(s)|KOH(aq)|Ni_(2)O_(3)(s)|Ni(s)`
The half`-` cell reactions are `:`

`a.` What is the cell reaction ?
`b.` What is the cell `EMF` ? How does it depend on the concentration of `KOH` ?
`c.` What is maximum amount of electrical energy that can be obtained from `1 mol ` of `Ni_(2)O_(3)` ?

Explanation

Given, `E_("FeO"//Fe)^(@) = -0.87, E_(Ni_(2)O_(3)//NiO) = +0.40 V`
`:. E_("Fe"//FeO)^(@) = +0.87 V, E_(NiO_(3)//Ni_(2)O_(3)) = -0.40 V`
Since , `E_(OP)^(@)` for `Fe//FeOgtE_(OP)^(@)` for `NiO // Ni_(2)O_(3)`, and thus, redox changes are,
At anode: `Fe_((s)) + 2OH^(-) rarr FeO_((s)) + underset(("oxidation"))(H_(2)O_((l)) + 2e)`
At cathode: `Ni_(2)O_(3_((s))) + H_(2)O_((l)) + 2e rarr 2NiO_((s)) + underset(("reducation"))(2OH^(-))`
Redox reaction:
`Fe_((s)) + Ni_(2)O_(3(s)) rarr FeO_((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_(OP_(Fe//FeO))^(@) + E_(RP_(Ni_(2)O_(3)//NiO))^(@) = 0.87 + 0.40 = 1.27 V`
(ii) The `E_(cell)` is indepdent of `OH^(-)` ion concentration.
(iii) `-DeltaG^(@) = nE^(@)F = 2 xx 1.27 xx 96500 = 245110J`
`= 245.11 kJ`