Electrochemistry - Result Question 88
####36. The Edison storage cell is represented as:
$$ Fe(s) / FeO(s) / KOH(a q) / Ni _2 O _3(s) / Ni(s) $$
The half-cell reactions are :
$$ \begin{array}{r} Ni _2 O _3(s)+H _2 O(l)+2 e^{-} \rightleftharpoons 2 NiO(s)+2 OH^{-}, \ E^{\circ}=+0.40 V \ FeO(s)+H _2 O(l)+2 e^{-} \rightleftharpoons Fe(s)+2 OH^{-}, \ E^{\circ}=-0.87 V \end{array} $$
(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 $Ni _2 O _3$ ?
$(1994,4 M)$
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Solution:
- Given, $FeO(s) / Fe(s)$ and $\quad Ni _2 O _3 / NiO(s)$
$$ E^{\circ}=-0.87 V $$
Electrode at lower reduction potential act as anode and that at higher reduction potential act as cathode.
(i) Electrodes reaction :
$$ \begin{gathered} Fe(s)+2 OH^{-} \longrightarrow FeO(s)+H _2 O(l) \ E^{\circ}=+0.87 V \ Ni _2 O _3(s)+H _2 O(l)+2 e^{-} \longrightarrow 2 NiO(s)+2 OH^{-} E^{\circ}=0.40 V \ Net: Fe(s)+Ni _2 O _3(s) \longrightarrow 2 NiO(s)+FeO(s) \end{gathered} $$
(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.
