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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)|Ni(s)`
the half-cell reactions are
`Ni_(2)O_(3)(s)+H_(2)O(l)+2e^(-) hArr 2NiO(s)+ 2OH^(-), E^(@)= + 0.40 V`
`FeO(s)+H_(2)O(l) +2e^(-) hArr Fe(s)+2OH^(-), E^(@)=- 0.87 V`
What is the maximum amount of electrical energy that can be obtained from one mole of `Ni_(2)O_(3)`?

A

127 kJ

B

245.11 kJ

C

90.71 kJ

D

122.55 kJ

Text Solution

Verified by Experts

The correct Answer is:
B
Given, `E_(FeO //Fe)^(@)=-0.87 V` and `E_(Ni_(2)O_(3)//NiO)^(@)=0.40 V`
`:. E_(Fe//FeO)^(@)=+0.87 V`
and `E_(NiO //Ni_(2)O_(3))^(@)=0.40 V`
Since, `E_("oxi")^(@)` for `Fe//FeO gt E_("oxi")^(@)` for `NiO //Ni_(2)O_(3)`
`:.` Redox changes
At anode `Fe+2OH^(-) rarr FeO (s)+H_(2)O(l) +2e^(-)`
At cathode `Ni_(2)O_(3)(s)+H_(2)O(l) +2e^(-) rarr NiO(s)+2OH^(-)`
Overall reaction `Fe(s)+Ni_(2)O_(3)(s) rarr FeO(s) +2NiO(s)`
Hence, `E_("cell")^(@)=E_("anode")^(@)-E_("cathode")^(@)=0.87-(-0.40)`
or `E_("cell")^(@)=1.27 V`
and `DeltaG^(@)=nFE_("cell")^(@)=2xx1.27xx96500`
`=245110 J`
`=245.11 kJ`
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