Electrochemistry - Result Question 37

####36. The Edison storage cell is represented as:

Fe(s)/FeO(s)/KOH(aq)/Ni2O3(s)/Ni(s)

The half-cell reactions are :

Ni2O3(s)+H2O(l)+2e2NiO(s)+2OH, E=+0.40V FeO(s)+H2O(l)+2eFe(s)+2OH, E=0.87V

(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 Ni2O3 ?

(1994,4M)

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Solution:

  1. Given, FeO(s)/Fe(s) and Ni2O3/NiO(s)

E=0.87V

Electrode at lower reduction potential act as anode and that at higher reduction potential act as cathode.

(i) Electrodes reaction :

Fe(s)+2OHFeO(s)+H2O(l) E=+0.87V Ni2O3(s)+H2O(l)+2e2NiO(s)+2OHE=0.40V Net:Fe(s)+Ni2O3(s)2NiO(s)+FeO(s)

(ii) Emf is independent of concentration of KOH.

(iii) Maximum amount of energy that can be obtained =ΔG ΔG=nEF=2×1.27×96500J=245.11kJ

i.e. 245.11kJ is the maximum amount of obtainable energy.



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