Electrochemistry 1 Question 37
36. The Edison storage cell is represented as:
$$ \mathrm{Fe}(s) / \mathrm{FeO}(s) / \mathrm{KOH}(a q) / \mathrm{Ni}{2} \mathrm{O}{3}(s) / \mathrm{Ni}(s) $$
The half-cell reactions are :
$$ \begin{array}{r} \mathrm{Ni}{2} \mathrm{O}{3}(s)+\mathrm{H}{2} \mathrm{O}(l)+2 e^{-} \rightleftharpoons 2 \mathrm{NiO}(s)+2 \mathrm{OH}^{-}, \ E^{\circ}=+0.40 \mathrm{~V} \ \mathrm{FeO}(s)+\mathrm{H}{2} \mathrm{O}(l)+2 e^{-} \rightleftharpoons \mathrm{Fe}(s)+2 \mathrm{OH}^{-}, \ E^{\circ}=-0.87 \mathrm{~V} \end{array} $$
(i) What is the cell reaction?
(ii) What is the cell emf ? How does it depend on the concentration of $\mathrm{KOH}$ ?
(iii) What is the maximum amount of electrical energy that can be obtained from one mole of $\mathrm{Ni}{2} \mathrm{O}{3}$ ?
$(1994,4 \mathrm{M})$
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Solution:
- Given, $\mathrm{FeO}(s) / \mathrm{Fe}(s)$ and $\quad \mathrm{Ni}{2} \mathrm{O}{3} / \mathrm{NiO}(s)$
$$ E^{\circ}=-0.87 \mathrm{~V} $$
Electrode at lower reduction potential act as anode and that at higher reduction potential act as cathode.
(i) Electrodes reaction :
$$ \begin{gathered} \mathrm{Fe}(s)+2 \mathrm{OH}^{-} \longrightarrow \mathrm{FeO}(s)+\mathrm{H}{2} \mathrm{O}(l) \ E^{\circ}=+0.87 \mathrm{~V} \ \mathrm{Ni}{2} \mathrm{O}{3}(s)+\mathrm{H}{2} \mathrm{O}(l)+2 e^{-} \longrightarrow 2 \mathrm{NiO}(s)+2 \mathrm{OH}^{-} E^{\circ}=0.40 \mathrm{~V} \ \mathrm{Net}: \mathrm{Fe}(s)+\mathrm{Ni}{2} \mathrm{O}{3}(s) \longrightarrow 2 \mathrm{NiO}(s)+\mathrm{FeO}(s) \end{gathered} $$
(ii) Emf is independent of concentration of $\mathrm{KOH}$.
(iii) Maximum amount of energy that can be obtained $=\Delta G^{\circ}$ $\Rightarrow \Delta G^{\circ}=-n E^{\circ} F=-2 \times 1.27 \times 96500 \mathrm{~J}=-245.11 \mathrm{~kJ}$
i.e. $245.11 \mathrm{~kJ}$ is the maximum amount of obtainable energy.
