Question

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)` ?

Answer 1

Correct Answer - `b. 1.27V`,`c. 245.11kJ`
Given that,
`E^(c-)._(FeO|Fe)=-0.87V` and `E^(c-)._(Ni_(2)O_(3)|NiO)=+0.40V`
In these electrodes, `E^(c-)` of `FeO|Fe` is greater than that of `Ni_(2)O_(3)|NiO`.
So, following reactions are possible at anode and cathode `:`
At anode `:`
`Fe(s)+2overset(c-)(O)H(l) rarr FeO(s)+H_(2)O(l)+2e^(-) " "E^(c-)=+0.87V`
At cathode`:`
`Ni_(2)O_(3)(s)+H_(2)O(l)+2e^(-) rarr 2NiO(s)+2overset(c-)(O)H(l),,,,,,E^(c-)=+0.40V`
`a.` Hence, cell reaction
`Fe(s)+Ni_(2)O_(3) rarr FeO(s)+2NiO(s)`
`b. EMF` of the cell `=E^(c-)._(cathode )-E^(c-)._(anode)`
`=E^(c-)._(Ni_(2)O_(3)|NiO)-E^(c-)._(FeO|Fe)`
`=0.40-(-0.87)=+1.27V`
This `EMF` is not based upon the concentration of `KOH`.
`c.` Produced electrical energy with one mole of
`Ni_(2)O_(3)=nxxE^(c-)._(cell)xxF`
`=2xx1.27xx96500J`
`=245.11kJ`