The Edison storage cell is represented as follows: Fe(s | FeO(s) | KOH( aq , C )

Question

The Edison storage cell is represented as follows:

Fe(s | FeO(s) | KOH( aq , C ) | Ni2 O3 (s) | NiO(s) | Ni(s) ,

where C indicates the concentration in mol/L of the aqueous solution of KOH. All the other cell elements are pure solids. The half-cell reactions (written as reductions), and their corresponding standard reduction potentials are:

Ni2O3 (s) + H2O(liq) + 2e- --> 2NiO(s) + 2OH- (aq) , E0( Ni2 O3 (s) / NiO(s) ) = +0.4 V

FeO(s) + H2O(liq) + 2e- --> Fe(s) + 2OH- (aq) , E0( FeO(s) / Fe(s) ) = -0.87 V

(a) What is the cell reaction?

(b) What is the standard electromotive force E 0 of the cell?

(c) What is the electromotive force of the cell when the molar concentration of the

KOH solution is C = 0.050 mol/L?

(d) What is the electromotive force of the cell when the molar concentration of the KOH solution is C = 2.50 mol/L?

Explanation / Answer

The half-cell reaction at the cathode:

2 NiOOH + 2 H2O + 2 e%u2212 %u2194 2 Ni(OH)2 + 2 OH%u2212

and at the anode:

Fe + 2 OH%u2212 %u2194 Fe(OH)2 + 2 e%u2212

Standard EMF E0 = 0.4-(-0.87)=1.27

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