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, `FeO(s)//Fe(s) E^(@)=-0.87 V` and `Ni_2O_2//NiO(s) E^(@)=+0.40 V`
Electrode at lower reduction potential act as anode and that at higher reduction potential act as cathode
(i)Electrodes reaction:
`Fe(s)+2oH^(-)to FeO(s)+H_2o(l) E^(@)=+0.87V`
`Ni_(2)o_3(s)+H_(o)(1)+2e^(-)to 2Nio(s)+2oH^(-)` `E^(@)=1.27 V`
`overline("Net reaction": Fe(s)+N i_2o_3(s)to2Nio(s)+Feo (s)` `E^(@)=1.27 V`
(ii) Emf is independent of concentration of KOH.
(iii) Maximum amount of energy that can be obtained `=DeltaG^(@)`
`implies DeltaG^(@)=-nFE=-2xx1.27xx96500 J=245.11 kJ`
.e. 245.11 kJ is the maximum amount of obtainable energy.