The Edison storage cell is symbolized: Fe(s)|FeO(s)|KOH(aq, a)||Ni_2O_3(s)|NiO(s

Question

The Edison storage cell is symbolized: Fe(s)|FeO(s)|KOH(aq, a)||Ni_2O_3(s)|NiO(s)|Ni(s) The half-cell reactions are: Ni_2O_3(s) + H_2O(l) + 2e^- rightarrow 2NiO(s) + 2OH^- FeO(s) + H_2O(l) + 2e^- rightarrow Fe(s) + 2OH^- standard electrode potentials: E^0 = 0.45V E^0 = -0.87V What is the initial electrochemical cell potential (EMF)? If a miniaturized version containing 1.5mg of the active materials were assembled, what would its initial electrode potential be? (ignore the iron and nickel electrodes).

Explanation / Answer

E° = Ereduction - Eoxidation

Reduction is Ni2O3 to NiO

oxidation is Fe(s) to FeO(s)

so

E° = 0.45 - -0.87

E° = 1.32 V

b)

m = 1.5 mg of active materials...

the initial electrode potential...

The EMF is constant independent of amount of material.

It will be constant as standard conditions ( 1M, T = 298K and P = 1 atm) are constant

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