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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Answer:
Correct Answer: 36. $-245.11 \mathrm{kJ}$
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} \\ \hline \mathrm{Net}: \mathrm{Fe}(s)+\mathrm{Ni} _{2} \mathrm{O} _{3}(s) \longrightarrow 2 \mathrm{NiO}(s)+\mathrm{FeO}(s) E^{\circ}=1.27 \mathrm{V} \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.
