Question

7. Given the standard reduction potential for the following two half-cell reactions (in the presence of 1.00 M HCl):

Fe3+ + e <==> Fe2+, E0 = + 0.68 V

AsO4 - + 2H+ + 2e <==> AsO3 - + H2O, Eo = +0.559 V.

Please calculate the system potential at the equivalence point when Fe3+ was used to titrate AsO3 - in the presence of 1.00 M HCl.

Answer: 0.60 V

Answer 1

since the concentraion is not given ,

i will be using conceot for this ,ie..

from the reduction potentails, Fe³⁺ gets reduced easily when compared to AsO₄^- . hence Fe³⁺ will go under reduction and AsO₃^- will go undr reduction

also, the half cells AsO₃^- takes 2 e^- , and Fe 3+ gives 1 e^-

from this i can say that twice teh amount of Fe3+ is required so as to oxidize AsO₃^- at eqivalance point.

reaction

Fe₃₊ +e ----> Fe²⁺ (reductio half)

AsO₃^- + H₂O ---AsO₄^- + 2H⁺ +2e (oxidation half)

E _{(Fe³⁺/Fe²⁺)} =E ⁰ _{(Fe³⁺/Fe²⁺)} - (0.0591/n) X log ([Fe2+]/ [Fe3+])

multiplying ny n throughout

n X E _{(Fe³⁺/Fe²⁺)} = n X E ⁰ _{(Fe³⁺/Fe²⁺)} - 0.0591 X log ([Fe2+]/ [Fe3+]); n=1

E _{(AsO4^-/AsO3^-)} =E ⁰_{(AsO4-/AsO3-)} - (0.0591/n) X log ([AsO₃^-]/ [AsO₄^-] X [H⁺])

multiplying ny n throughout

n X E _{(AsO4^-/AsO3^-)} =n X E ⁰_{(AsO4-/AsO3-)} - 0.0591/ X log ([AsO₃^-]/ [AsO₄^-] X [H⁺]) ; n =2

at euquivalence point

E _{(Fe³⁺/Fe²⁺)} = E _{(AsO4^-/AsO3^-) =} E_sys

and [Fe2+] = [Fe3+] , [AsO₃^-] = [AsO₄^-] and we know that [H+] = 1

thus adding both the equations

we have,

3E_sys = E ⁰ _{(Fe³⁺/Fe²⁺)} + 2E ⁰_{(AsO4-/AsO3-)} -  (0.0591/n) X log ([Fe2+] X[AsO4^-] X [H⁺]/ [Fe3+] [AsO₃^-])

or

3E_sys = E ⁰ _{(Fe³⁺/Fe²⁺)} + 2E ⁰_{(AsO4-/AsO3-)}

hence

E_sys = 0.6 v