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+