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( λ ) + 2e – 
2NiO(s) + 2OH – Eº = +0.40 V
FeO(s) + H 2 O( λ ) + 2e
– 
Fe(s) + 2OH – Eº = –0.87 V
(i) What is the cell reaction ?
(ii) What is the cell emf ? How does it 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 ExpertsC
Given that,
Eº FeO/Fe = –0.87 V and 
= +0.40 V
In these electrode Eº of FeO/Fe is greater than that of Ni 2 O 3 /NiO
So, following reaction are possible at anode and cathode :
At anode
Fe(s) + 2OH – ( λ ) → FeO(s) + H 2 O ( λ ) + 2e – Eº = +0.87 V
At cathode
Ni 2 O 3 (s) + H 2 O( λ ) + 2e – ⎯→ 2NiO(s) + 2OH – ( λ ) Eº = +0.40 V
(i) Hence cell reaction
Fe(s) + Ni 2 O 3 ⎯→ FeO(s) + 2NiO(s)
(ii) emf of the cell = Eº cathode – Eº anode
= 
ss
= 0.40 –(–0.87) = +1.27 V
This emf is not based upon the conc. of KOH.
(iii) Produced electrical energy with one mole
of Ni 2 O 3 = n × Eº cell × F
= 2 × 1.27 × 96500 J
= 245.11 kJ
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