JEE Advance - Chemistry (1994 - No. 13)
Fe(s) | FeO(s) | KOH (aq) | Ni2O3(s) | Ni(s)
The half-cell reactions are:
Ni2O3 + H2O (l) + 2e- $$\leftrightharpoons$$ 2NiO(s) + 2OH-; Eo = +0.40V
FeO(s) + H2O(l) + 2e- $$\leftrightharpoons$$ Fe(s) + 2OH-; Eo = -0.87V
(i) What is the cell reaction?
(ii) What is the cell e.m.f? 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 Ni2O3?
Explanation
Given, $$E_{FeO|Fe}^0 = - 0.87\,V$$, $$E_{N{i_2}{O_3}|NiO}^0 = + 0.40\,V$$
At anode : $$Fe(s) + 2O{H^ - }(l) \to FeO(s) + {H_2}O(l) + 2e$$
At cathode : $$N{i_2}{O_3}(s) + {H_2}O(l) + 2e \to 2NiO(s) + 2O{H^ - }(aq)$$
(1) Cell reaction : $$Fe(s) + N{i_2}{O_3}(s) \to FeO(s) + 2NiO(s)$$
(2) $$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = 0.40 - ( - 0.87) = + 1.27\,V$$
$${E_{cell}} = E_{cell}^0 - {{0.059} \over 2}\log {{[FeO]{{[NiO]}^2}} \over {[Fe][N{i_2}{O_3}]}}$$
$$ = {E^0} - {{0.059} \over 2}\log {{1 \times {1^2}} \over {1 \times 1}} = {E^0}$$ [$$\because$$ Molar concentration of pure solid = 1]
The expression for Ecell does not contain concentration term of OH$$-$$, so Ecell is independent of OH$$-$$ concentration.
(3) Electrical energy obtained from 1 mole of
$$N{i_2}{O_3} = n \times E_{cell}^0 \times F = 2 \times 1.27 \times 96500\,J$$ = 245110 J = 245.11 kJ.
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