Question

Consider the cell S.C.E.|| cell solution| Pt(s), whose voltage is -0.126 V. the cell contains 2 mmol Fe(NH4)2(SO4)2 1 mmol fecl3 4 mmol Na2EDTA and lots of buffer, pH 6.78, in a volume of 1.00L

a)write a reaction for the right half cell

b) Find quotient [Fe^2+]/[Fe^3+] in the cell solution

c) find the quotient of formation constants: (K_f for FeEDTA-)/(K_f for FeEDA^2+)

Im having truble getting the reaction set up, help is greatly appreciated

Answer 1

At or below near neutral pH both Fe(II) and Fe(III) form stable complexes with EDTA.

So in the solution almost all of the ions [Fe(III) and Fe(III)] will remain as EDTA complex.

Fe(II) + Na₂EDTA = [FeEDTA]^2- + 2H⁺ ; K_II = [FeEDTA]^2- * [H⁺]² / [Na₂EDTA] [Fe²⁺]

Fe(III) + Na₂EDTA = [FeEDTA]^- + 2H⁺ ; K_III = [FeEDTA]^- * [H⁺]² / [Na₂EDTA] [Fe³⁺]

Buffer is added to the solution to keep the pH of the solution constant.

a. The redox reaction at the right electrode(cathode) will be:

[FeEDTA]^- + e = [FeEDTA]^2-

b. If we take the reaction at right electrode Fe³⁺ + e = Fe²⁺

We get, E_cell = E_R - E_L

-0.126 = E_R - 0.241

E_R = 0.115 V = 0.77 - 0.059log[Fe²⁺]/[ Fe³⁺].....{As E⁰_Fe3+/Fe2+ = 0.77 V}

or, [Fe²⁺]/[ Fe³⁺] = 1.26*10¹¹

c. K_III / K_II = {[FeEDTA]^- * [H⁺]² / [Na₂EDTA] [Fe³⁺]} / { [FeEDTA]^2- * [H⁺]² / [Na₂EDTA] [Fe²⁺]}

= {[FeEDTA]^- / [Fe³⁺]} / { [FeEDTA]^2- / [Fe²⁺]}

= [FeEDTA]^- * [Fe²⁺] / [Fe³⁺] * [FeEDTA]^2- ......[As 1 mmol FeCl₃ will from 1mmol complex with EDTA and 2 mmol Fe(NH4)2(SO4)2 complex will form 2 mmol complex with EDTA]

= 1 mmol * 1.26*10¹¹ / 2 mmol

= 6.3 * 10¹⁰