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Ther Edson storage cell is represented a...

Ther Edson storage cell is represented as
`Fe_((s))|FeO_((s))|koh_((aq))|N_(2)O_(3_((s)))|N_((s))`
The half cell reaction are
`Ni_(2)O_(a_((s))) +H_(2)O_((l))+2e to 2NiO_((s))+2OH " "E^(0)=0.40V`
`FeO_((s))+H_(2)O_((l))+2e to Fe_((s))+2OH" "E^(0)=-0.87V`
(a) What is the cell reaction?
(b) What is the cell emf ? How does it depend on the concentration of KOH?
(c) 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

The correct Answer is:
`Fe(s)+Ni_(2)O_(3) (g) to FeO(s)+2NiO(s)`
(b) Sicnce net cell reaction does not contain `OH^(-)`, hence cell emf does not depend on the concentration of KOH.
(c) `2.45 xx 10^(2)"Kj mol"^(-1)`
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Knowledge Check

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