Redox Titrations
-the oxidation/reduction reaction between analyte and titrant
-titrants are commonly oxidizing agents, although reducing titrants can be used -the equivalence point is based upon:
Aox + Bred → Ared + Box
Rx’n goes to completion after each addition of titrant – Potentiometric Titration:
Titration reaction:
Ce4+ + Fe2+ → Ce3+ + Fe3+ (1) Reference half-reaction:
2Hg(l) + 2Cl- ⇔ Hg2Cl2(s) + 2e- At the Pt indicator electrode (Indicator half-reaction)
Fe3 + e- ⇔Fe2+ E0 = 0.767 V (2) Ce4+ + e- ⇔Ce3+ E0 = 1.70V (3)
Cell reactions (in 1 M HClO4):
2Fe3+ + 2Hg(l) + 2Cl- ⇔2Fe2+ + Hg2Cl2(s) (4) 2Ce4+ + 2Hg(l) + 2Cl- ⇔ 2Ce3+ + Hg2Cl2(s) (5)
Relationships
- Cell reactions are not the same as the titration reaction - May describe the cell voltage with either (4) or (5) or
both
Balancing Redox Reactions
Balance: -atoms -# of electrons transferred Example:
Cr(s) + Ag+ → Cr3+ + Ag(s)
1. Write the half reactions: 2. Balance the electrons: 3. Recombine
Equilibrium constants for oxidation-reduction reactions
Cu(s) + 2Ag ⇔ Cu2+ + 2 Ag(s)
Keq = 2
2
]Ag[][Cu
+
+
Galvanic cell:
Ecell = Ecathode – Eanode = EAg+ - ECu2+
Under equilibrium conditions, the potential of the cell becomes zero, thus can write:
Ecell = O = Ecathode – Eanode = EAg+ - ECu2+
or
Ecathode = Eanode = EAg+ = ECu2+
-Also when in equilibrium, electrode potentials of all
systems are identical:
EOx1 = EOx2 = EOx3 = EOx4
Where EOx1…..are electrode potentials for the four half-
reactions
Calculating Equilibrium Constants
Cu(s) + 2Ag ⇔ Cu2+ + 2 Ag(s)
E0Ag+ -
20592.0 log =+ 2]Ag[
1 E0Cu2+-
20592.0 log
]Cu2[1
+
E0
Ag+ - E0Cu2+ =
]Cu[1log
20592.0
][1log
20592.0
22 +−+Ag
= eq2
2Cu2
00
K log][Ag][Culog
0592.0)2(
=+
=− +
++ EE Ag
Ex: Calculate the equilibrium constant:
0.05920.337)-2(0.799
]Ag[][CulogKlog 2
2
eq =+
=+
= 15.6
Keq = antilog 15.6 = 4.1 x 1015 = 4 x 1015
Redox Titration Curves
Fe2+ + Ce4+ ↔ Fe3+ + Ce3+
ECe4+ = EFe3+ = Esystem
EIn = ECe4+ = EFe3+ = Esystem
Equivalence Point Potentials
Fe3 + e- ⇔Fe2+ Ce4+ + e- ⇔Ce3+
1. ]4[Ce
]3[Celog1
0592.00Ce4eq +
+−+= EE
2. ]3[Fe
]2[Felog1
0592.00Fe3eq +
+−+= EE
2]Fe][[Ce]][Fe[Ce0
Fe0Ce eq 34
23
34 ++
++
++−+= EEE (1)
Definition of e.p. requires that: [Fe3+] = [Ce3+] [Fe2+] = [Ce4+]
2]3Ce][4[Ce
]4][Ce3[Celog1
0592.00Fe3
0Ce4 eq ++
++−+++= EEE = 0
Fe0Ce 34 ++ + EE
eqE = 2
EE 03Fe
0Ce4 ++
(2)
The Derivation of Titration Curves Titration of 50.00 mL of 0.05000 M Fe2+ with 0.1000 M Ce4+ in a solution that is 1.0 M in H2SO4 at all times.
Ce4+ + e- ↔Ce3+ Ef = 1.44V Fe3+ + e- ↔Fe2+ Ef = 0.68V
1. Initial potential – Ce and Fe3+ only present in very small amounts. 2. Potential after addition of 5.00 mL of Ce4+
[Fe3+]=
00.55500.0]Ce4[
5.00 50.000.1000 x 5.00 ≈+−+
[Fe2+] = 00.55
000.2]Ce[55.00
0.1000 x 5.00 - 0.0500 x 50.00 4 ≈+ +
Substitution into Nernst equation:
systemE = +0.68- V64.000.55/500.000.55/00.2log
10592.0
=
E.P. potential
VEE
Ef
Fef
Ceeq 06.1
268.044.1
234 =+== +
++
3. Potential after addition of 25.10 mL of Ce4+
[Fe2+] = amt of Ce4+ left unreacted, therefore added to CCe4+ calculated from the volumes of the two solutions and subtracted from CCe3+
Conc of two cerium ion species:
[Ce3+]=10.75
500.2]275.10
0.1000 x 25.00 Fe[ ≈+−
][Ce][Celog
10592.044.1
4
3
+
+−+=E =+1.44-
10.75/010.000.75/500.2log1
0592.0
= +1.30 V
Effect of system variables on redox titration curves Concentration – independent of analyte and reagent concentrations. Exception: Electrode potentials dependent upon dilution
I −3 +2e-
↔3 I-
]-I[]-[Ilog
20592.0
3
30 −= EE
num-mol/L3, denom-mol/L
Completeness of reaction – the change in Esystem in the e.p. region becomes larger as the reaction becomes more concentrated.
Redox indicators
a. specific indicators – react with one of the participants in the titration to produce a color, e.g. thiocyanate
b. Oxidation-reduction indicators- respond to the potential of
the system rather than to the appearance or disappearance of some species during the course of the titration, e.g. methylene blue
Color changes will occur over the range:
Voltsn
EE )05916.0( 0 ±=
where n= # of electrons in the indicator half-reaction
-larger diff in std potential between titrant and analyte, the sharper
the break in the titration curve at the e.p.
≥0.2 V, best detected potentiometrically
Gran plot - more accurate way to use potentiometric data - uses data well before e.p. (Ve) to locate Ve For the oxidation of Fe2+ to Fe3+, the potential prior to Ve is:
ref)]Fe[
[Felog(05916.0'[3
]20 EEE −−=
+
+
where, '0E = formal potential for Fe3+Fe2+ and Eref is the potential of the reference electrode. If vol of analyte = V0 and the vol of titrant = V, and if reaction goes to completion with each addition of titrant:
[Fe2+] / [Fe3+] = (Ve-V)
V 10-nE/0.05916 = Ve10-n(Eref –E0’)/0.05916 - V 10-n(Eref –E0’)/0.05916 y b x m
Adjustment of Analyte Oxidation State -before titration, e.g. Mn2+ preoxidized to MnO4
-
-excess preadjustement reagent must be destroyed so that it will not interfere in subsequent titration
Preoxidation -powerful oxidants can be removed after preoxidation, e.g. peroxydisulfate (S2O8
2-) – requires Ag+ as a catalyst.
++++ →+ 2-4
-24 -2
82 AgSOSOAgOS
Excess reagent destroyed:
+++ →+ 4H2O4SOO2H 2O2S -24 -2
82boiling
Prereduction -Stannous chloride (SnCl2) will reduce Fe3+ to Fe2+ in hot HCl Excess reductant is then destroyed:
Sn2+ + 2HgCl2 → Sn4+ + HgCl2 + 2 Cl-
Oxidation with Potassium Permanganate -strong oxidant, violet color In strongly acidic solutions, reduced to colorless Mn2+:
MnO4- + 8H+ + 5e- ↔Mn2+ + 4 H20
In neutral or alkaline solution, the product is the brown solid, MnO2:
MnO4- + 4H+ + 3e- ↔MnO2(s) + 2H2O
In strongly alkaline solution (2 M NaOH), green manganate is produced:
MnO4- + e- ↔ MnO4
2-
Tales 16.3……..see below
Note: permanganate solutions are unstable, therefore not a primary standard.
4MnO4- + 2H2O > 4MnO2 + 4OH- + 3O2 (MnO2 catalyses this
reaction)
Permanganate must be standardized for example with oxalate;
H2C2O4 > 2H+ + CO2 + 2e-
Overall: . 2MnO4
- + 5H2C2O4 + 16H+ > 2Mn2+ + 10CO2 + 8H2O
Initially the reaction is slow but is catalyzed by Mn2+ so becomes more rapid.
Can also standardize with arsenic (III) oxide
As(III) > As(V) + 2e-
The reaction of As (III) with permanganate ion takes place without complications in acidic medium if a trace of an iodine compound (for example potassium iodate) is added as a catalyst.
The reaction generally carried out in HCl rather than H2SO4 .... in the latter a brown green coloration occurs due to formation of a manganese arsenate compound
KMnO4 can serve as own indicator, since product Mn2+ is colorless.
Cerium(IV)
Strong oxidant > Ce(III)
Ce4+ [Yellow ]+ e- > Ce3+ [Colorless]
Note however that the color change not good enough for it to act as self indicator.
Ce(IV) not found in acid solution as simple aqua ion .. forms complexes.
Dichromate reactions
Dichromate ion is an oxidizing agent
Cr2O72- + 14H+ + 6e- > 2Cr3+ + 7H2O E = +1.33V
Dichromate has replace permanganate in many analyses ... notably iron (II)... it can be prepared as a standard solution and so avoids the need to standardize as is the case with permanganate.
Iodine Methods
I2 + 2e- > 2I- E = +0.54V
Value for E is intermediate can therefore be reduced or oxidized...iodine can be reduced to iodide by for example As(III), Sn(II) whilst iodide can be oxidized to iodine by for example permanganate. Use of iodide as titrant..practical problems ..so add excess potassium iodide and titrate the liberated iodine with for example standard thiosulphate solution
Miscellaneous Oxidizing agents
Sodium bismuthate and lead (IV) oxide are strong oxidizing agents.
NaBiO3 + 6H+ + 2e- > Na+ + Bi3+ + 3H2O E = +1.6V
PbO2 + 4H+ + 2e- > Pb2+ + 2H2O E = +1.5V
Hydrogen peroxide
... strong oxidant even in alkaline conditions.
H2O2 + 2H+ + 2e- > 2H2O E = +1.77V
Excess peroxide may be removed by boiling...decomposes
Ex: Derive a titration curve for the titration of 50.00 mL of 0.02500 M U4+ with 0.1000 M Ce4+