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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

Verified by Experts

Actual half reactions are :
`Fe+2OH^(-) rarr FeO+H_(2)O+2e^(-)` Anode (Oxidation)
`Ni_(2)O_(3)+H_(2)O+2e^(-) rarr 2NiO+2OH^(-)` Cathode (Reduction)
Thus, the cell reaction is :
(a) `Fe+Ni_(2)O_(3) rarr FeO+2NiO`
(b) `E_(cell)=E_(cell)^(@)-0.0591/2"log" ([NiO]^(2)[FeO])/([Fe][Ni_(2)O_(3)])=E_(cell)^(@)`
`["Since, "([NiO]^(2)[FeO])/([Fe][Ni_(2)O_(3)])=1" as all are solids"]`
`=0.87+0.40=1.27` volt
The emf of the cell is independent of KOH concentration.
(c) Maximum amount of electrical energy
`=nFE^(@)=2xx96500xx1.27=245.11` kJ
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