Question

The Edison storage cell is represented as:
$Fe\left(s\right)|FeO\left(s\right)|KOH(aq.)|N{i}_2{O}_3\left(s\right)|Ni\left(s\right)$
The half-cell reactions are:
$N{i}_2{O}_3\left(s\right)+H_{2}O\left(l\right)+2e^{-}\to 2NiO\left(s\right)+2O{H}^{-};E^{\circ }+0.40$
$FeO\left(s\right)+H_{2}O\left(l\right)+2e^{-}\to Fe\left(s\right)+2O{H}^{-};E^{\circ }=-0.87$,
(a) What is the cell reaction?
(b) What is the emf of the cell? How does it depend on concentration of $KOH$?
(c) What is the maximum amount of energy that can obtained from one mole of $N{i}_2{O}_3$?

Answer 1

Actual half reactions are:
$Fe+2O{H}^{-}\to FeO+H_{2}O+2e^{-}$ Anode (Oxidation)
$N{i}_2{O}_3+H_{2}O+2e^{-}\to 2NiO+2O{H}^{-}$ Cathode (Reduction)
Thus, the cell reaction is:
(a) $Fe+N{i}_2{O}_3\to FeO+2NiO$
(b) $E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}\log \frac{\left[NiO{]}^2\right[FeO]}{\left[Fe\right]\left[N{i}_2{O}_3\right]}=E_{cell}^{\circ }$
[Since, $\frac{\left[NiO{]}^2\right[FeO]}{\left[Fe\right]\left[N{i}_2{O}_3\right]}=1$ as all are solid.
$=0.87+0.40=1.27volt$
The emf of the cell is independent of $KOH$ concentration.
(c) Maximum amount of electrical energy
$=nF{E}^{\circ }=2\times 96500\times 1.27=245.11kJ$