The Edision storage cell is represented as `Fe(s)|FeO(s)|KOH(aq)|Ni_(2)O_(3)(s)|Ni(s)`
The half cell reaction are
`Ni_(2)(s)+H_(2)O(l)+2e^(-)rarr2NIO(s)+2OH^(-),E^(@)=+0.40V` M
`FeO(s)+H_(2)O(l)+2e^(-)rarrFe(s)+2OG^(-),E^(@)=-0.87V`
What is th emaximum amount of electrical energy that can be obtained from one mole of `Ni_(2)O_(3)`?

Options

A. 127 kJ
B. 245.11 kj
C. 90.71 kj
D. 122.55 kj

Answer

A

Explanation

Given `E_("FeO"//"Fe")^(@)=-0.87 V,` and `E_(NiO_(2))^(@) =0.40 V`
At kanode
`Fe+ 2OH^(-) rarr FeO(s)+H_(2)O(l)+2e^(-)`
at cathode
`Ni_(2)O_(3)(s)+_(2)O(I)+2e^(-) rarr2NiO(s)+H_(2)O(I)+2e^(-)`
At cathode
`Fe(s)+H_(2)O(I)+2e^(-) rarr FeO(s)+2 NiO(s)`
Hence `E_(Cell)^(@) =E_(cell)^(@) -E_(cathode)^(@) =0.87 -(-0.40)`
or `E_(cell)^(@)==0.897 -(-0.40)=1.27 v`
and `triangle G^(@)=-nFE_(cell)^(2) =2xx1.27 xx96500`
`=245110 J =245.11 KJ`