Question

The Edison storage cell is represented as:
$Fe(s)/FeO(s)/KOH(aq)/N{i}_2{O}_3(s)/Ni(s)$
The half-cell reactions are :
$N{i}_2{O}_3(s)+H_{2}O(l)+2e^{-}\rightleftharpoons 2NiO(s)+2O{H}^{-},$
$E^{\circ }=+0.40V$
$FeO(s)+H_{2}O(l)+2e^{-}\rightleftharpoons Fe(s)+2O{H}^{-},$
$E^{\circ }=-0.87V$
(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 $N{i}_2{O}_3$?

Answer 1

Given, $FeO(s)/Fe(s)E^{\circ }=-0.87V$
and $N{i}_2{O}_3/NiO(s)E^{\circ }=+0.40V$
Electrode at lower reduction potential act as anode and that at higher reduction potential act as cathode.
(i) Electrodes reaction :
$Fe(s)+2O{H}^{-}⟶FeO(s)+H_{2}O(l)E^{\circ }=+0.87V$
$N{i}_2{O}_3(s)+H_{2}O(l)+2e^{-}⟶2NiO(s)+2O{H}^{-}{E}^{\circ }=0.40V$
Net: $Fe(s)+N{i}_2{O}_3(s)⟶2NiO(s)+FeO(s)E^{\circ }=1.27V$
(ii) Emf is independent of concentration of $KOH$.
(iii) Maximum amount of energy that can be obtained $=\Delta G^{\circ }$
$\Rightarrow \Delta G^{\circ }=-n{E}^{\circ }F=-2\times 1.27\times 96500J$
$=-245.11kJ$
i.e. $245.11kJ$ is the maximum amount of obtainable energy.