ElectrochemistryVery Hard
Question
Standard electrode potentil data are useful for understanding the suitability of an oxidant in a redox titration. Some half-cell reaction and their standard potentials are give below :
MnO4- (aq) + 8H+ (aq) + 5e- → Mn2+(aq) + 4H2O(l), Eo = 1.51V
Cr2O72- (aq) + 14H+ (aq) + 6e- → 2Cr3+ (aq) + 4H2O(l), Eo = 1.38V
Fe3+(aq) + e- → Fe2+(aq) Eo = 0.77V
Cl2(g) + 2g- → Cl- (aq) Eo = 1.40V
Identify the incorrect statement regarding the quantitative estimation of aqueous Fe (NO3)2 :
MnO4- (aq) + 8H+ (aq) + 5e- → Mn2+(aq) + 4H2O(l), Eo = 1.51V
Cr2O72- (aq) + 14H+ (aq) + 6e- → 2Cr3+ (aq) + 4H2O(l), Eo = 1.38V
Fe3+(aq) + e- → Fe2+(aq) Eo = 0.77V
Cl2(g) + 2g- → Cl- (aq) Eo = 1.40V
Identify the incorrect statement regarding the quantitative estimation of aqueous Fe (NO3)2 :
Options
A.MnCO4- can be used in aqueous HCl
B.Cr2O72- can be used in aqueous HCl
C.MnCO4- can be used in aqueous H2SO4
D.Cr2O72- can be used in aqueous H2SO4
Solution
MnCO4- will oxidise Cl- ion according to equation.
Mn7+ + 5e → Mn2+
2Cl- → Cl2 + 2e
Thus Eocell = EoO.P.Cl-/Cl2 + EoR.P.Mn7+/Mn2+
= - 1.40 + 1.51 = 0.11V
or reaction is feasible.
MnO4- will oxidise Fe2+ to Fe3+
Fe2+ → Fe3+ + e
Eocell = EoO.P.Fe2+/Fe3+ + EoR.P.Mn2+/Mn2+
= - 0.77 + 1.51 = 0.74V
or reaction is feasible.
Thus MnO4- will not oxidise only Fe2+ to Fe3+ in aqueous HCl but it will also oxidise Cl- to Cl2. Suitable oxidant should not oxidise Cl- to Cl2 and should oxidise only Fe2+ to Fe3+ in redox titration.
Mn7+ + 5e → Mn2+
2Cl- → Cl2 + 2e
Thus Eocell = EoO.P.Cl-/Cl2 + EoR.P.Mn7+/Mn2+
= - 1.40 + 1.51 = 0.11V
or reaction is feasible.
MnO4- will oxidise Fe2+ to Fe3+
Fe2+ → Fe3+ + e
Eocell = EoO.P.Fe2+/Fe3+ + EoR.P.Mn2+/Mn2+
= - 0.77 + 1.51 = 0.74V
or reaction is feasible.
Thus MnO4- will not oxidise only Fe2+ to Fe3+ in aqueous HCl but it will also oxidise Cl- to Cl2. Suitable oxidant should not oxidise Cl- to Cl2 and should oxidise only Fe2+ to Fe3+ in redox titration.
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