Question

The Edison storage cell is represented as:
$Fe (s)/ FeO(s)/ KOH (aq)/N i_2O_3(s)/Ni(s)$
The half-cell reactions are :
$N i_2O_3(s)+H_2 O(l)+2e^- \rightleftharpoons 2NiO(s)+2OH^-,$
$ E^\circ=+0.40\, V $
$ FeO(s)+H_2 O(l)+2e^- \rightleftharpoons Fe(s)+2OH^-,$
$ E^\circ=-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 $Ni_2O_3$?

Answer 1

Given, $ FeO (s) / Fe (s) E^{\circ}=-0.87\, V$
and $ Ni _{2} O _{3} / NiO (s) E^{\circ}=+0.40\, V$
Electrode at lower reduction potential act as anode and that at higher reduction potential act as cathode.
(i) Electrodes reaction :
$Fe (s)+2 OH ^{-} \longrightarrow FeO (s)+ H _{2} O (l)E^{\circ}=+0.87 \,V$
$Ni _{2} O _{3}(s)+ H _{2} O (l)+2 e^{-} \longrightarrow 2 NiO (s)+2 OH ^{-}E^{\circ}=0.40\, V$
Net: $Fe (s)+ Ni _{2} O _{3}(s) \longrightarrow 2 NiO (s)+ FeO (s)E ^{\circ}=1.27\, V$
(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 96500\, J $
$=-245.11\, kJ$
i.e. $245.11\, kJ$ is the maximum amount of obtainable energy.