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The Edison storage cell is represented a...

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)` ?

Text Solution

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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.
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