-1-

CHAPTER I

TITRATIONS INVOLVING

OXIDATION REDUCTION REACTIONS

/ reactions involve transfer of electrons from one reactant to

another.

The oxidaion of Fe2+ with Ce

4+ is represented by the equation

Ce4+ + Fe

2+ Ce

3+ + Fe

3+

Ce4+ has strong affinity for electrons (electron acceptor) and is

called an oxidizing agent or oxidant. Fe2+ is ready to donate electrons

(electron donor) and is called reducing agent or reductant. Fe2+ is

oxidized by Ce4+ while Ce

4+ is reduced by Fe

2+.

The oxidation-reduction equation is made up of two half reactions.

Ce4+

+ e- Ce

3+ (reduction of Ce

4+)

Fe2+ Fe

3+ + e

- (oxidation of Fe

2+)

A half reaction cannot occur alone. It must be accompanied by a

second half reaction.

If chemical evidence shows that the equilibrium of the reaction lies

to the right, it can be stated that Ce4+ is a better electron acceptor

(stronger oxidant) than Fe3+ and that Fe

2+ is more effective electron donor

(better reductant) than Ce3+

or generally

Ox1 + Red2 Red1 + Ox2

To understand such tendency we have to study the electrochemical

cells and electrode potential.

ELECTROCHEMICAL CELLS

The electrochemical cell illustrates the fact that redox processes are

transfer of electrons.

The electrochemical cells include the galvanic (voltaic) and

electrolytic. In a voltaic cell, a spontaneous chemical reaction takes

place producing electricity. This occurs when the cell current is closed,

as when a flashlight appears. The cell voltage (e.g. in a battery) is

determined by the potential difference of the two half reactions. When

the reaction has gone to completion, the cell runs down, and the voltage is

zero (the battery is dead). In an electrolytic cell, however, the reaction is

-2-

forced to the opposite direction by applying an external voltage greater

than and opposite to the spontaneous voltage. In both types, the electrode

at which oxidation occurs is the anode and that at which reduction occurs

is the cathode.

Voltaic cells are of importance on discussing potentiometry while

the electrolytic cells in methods such as voltammetry.

The cell potential : Consider the following redox reaction in a voltaic cell

Fe2+ + Ce

4+ Fe

3+ + Ce

3+

Now assume that Fe2+ solution and Ce

4+ solution are placed in

separate beakers and connected by a salt bridge (it allows charge transfer

through the solution but prevents mixing of the solutions : i.e. maintains

electrical neutrality). Now put an inert Pt wire in each solution and

connect the two wires. The set up constitutes a voltaic cell. If a

microammeter is connected, current flow is indicated and the electron

directed from anode (at which Fe2+ is oxidized) to cathode (at which Ce

4+

is reduced).

Therefore, each electrode will adopt an electrical potential

(tendency to give off or take on electrons) and is called electrode

potential. The larger the potential difference the greater the tendency for

the reactions between Fe2+ & Ce

4+ to take place. This driving force

(potential difference) can be used to perform work such as lighting (e.g.

battery) or running motor (car). It should be noted that no half reaction

can occur by itself, but each will generate a definite potential.

-3-

HALF-REACTION POTENTIALS :

No way to measure individual electrode potentials (i.e. half

reaction potential), but the difference between two electrode potentials

can be measured. The electrode potential of the half-reaction.

H+ + e

- H

has arbitrarily been assigned a value of 0.00 V at standard conditions of

[H+], PH, and T and is called NHE (normal hydrogen electrode) or SHE

(standard hydrogen electrode). All other electrode potentials are

measured relative to it and are arranged in a decreasing order. In the

Gibbs-Stockholm convention, adopted at the 17th conference of the

International Union of Pure & Applied chemistry in Stockholm 1953,

half-reactions are written as reduction reactions and the potential

increases as the tendency for reduction (of the oxidized form of the half-

reaction) increases.

Table : Some Standard Potentials

Half-Reaction Eo (V)

H2O2 + 2H+ + 2e

- = 2H2O 1.77

MnO4- + 4H

+ + 3e

- = MnO2 + 2H2O 1.695

Ce4+ + e

- = Ce

3+ 1.61

MnO4- + 8H

+ + 5e

- = Mn

2+ + 4H2O 1.51

Cr2O72- + 14H

+ + 6e

- = 2Cr

3+ + 7H2O 1.33

MnO2 + 4H+ + 2e

- = Mn

2+ + 2H2O 1.23

2IO3- + 12H

+ + 10e

- = I2 + 6H2O 1.20

H2O2 + 2e- = 2 OH

- 0.88

Cu2 + I

- + e

- = CuI 0.86

Fe3+ + e

- = Fe

2+ 0.771

O2 + 2H+ + 2e

- = H2O2 0.682

I2(aq) + 2e- = 2I

- 0.6197

H3AsO4 + 2H+ + 2e

- = H3AsO3 + H2O 0.559

I3- + 2e

- = 3I

- 0.5355

Sn4+ + 2e

- = Sn

2+ 0.154

S4O62- + 2e

- = 2S2O3

2- 0.08

2H+ + 2e

- = H2 0.000

Zn2+ + 2e

- = Zn -0.763

2H2O + 2e- = H2 + 2OH

- -0.828

-4-

The general conclusions to be drawn are

(1) The more positive the electrode potential, the stronger the

oxidizing power.

(2) The more negative the electrode potential, the stronger the

reducing power.

Example :

a) Ce4+ (E

o = 1.61 V) is a good oxidizing agent but Ce

3+ is a poor

reducing agent

b) Zn is a good reducing agent but Zn2+ (E

o = -0.763 V) is a poor

oxidant.

(3) The reduced form in a half-reaction is capable of reducing the

oxidized form in another half reaction with a more positive

potential

e.g. Fe3+ + e

- = Fe

2+ (E

o = 0.771 V)

Sn4+ + 2e

- = Sn

2+ (E

o = 0.154 V)

Therefore 2Fe3+ + Sn

2+ = 2Fe

2+ + Sn

4+

Obviously no reaction possibility between two oxidants (Sn4+ & Fe

3+)

or between two reductants (Sn2+ & Fe

2+). But the reduced form of the

more negative potential (Sn2+) is reacting with the oxidized form of

the more positive potential (Fe3+). Note that the number of electrons

given from one Sn2+ must go to two Fe

3+ ions. If the two potentials

are substracted and a positive value is given, the reaction goes

spontaneously as written above i.e. 0.771 V – 0.154 V = 0.617 V. If it

is negative, the reaction will occur in the reverse direction*.

Example : List the oxidizing power in decreasing order and the

reducing capability in decreasing order among the following

MnO4-, Ce

3+, Cr

3+, IO3

-, Fe

3+, I

-, H

+, Zn

2+

* This is the result of the convention that, for a spontaneous reaction, the free energy

change is negative. At standard conditions

∆ Go = nF ∆ Eo

For example, for

Eo

Ce4+/Ce3+= 1.61 V and E

Fe3+/Fe2+o

= 0.771 V

∆Go for the former is more negative than the latter giving a negative value.

-5-

Solution : Referring to standard potentials table

- The most positive to least positive :

MnO4-, IO3

-, Fe

3+, H

+, Zn

2+ (MnO4

-, most powerful and Zn

2+, very poor).

- The remainder are reductants in order I-, Cr

3+, Ce

3+, (I

- is a reasonably

good reductant, Ce3+ is poor)

The Anode and the Cathode :

Anode/solution/cathode

By convention, the anode is written on the left, the single lines

represent boundary lines. In the above Figure the cell is written as

Pt / Fe2+ (C1), Fe

3+ (C2) // Ce

4+ (C3), Ce

3+ (C4) / Pt

C1, C2, C3 and C4 represent the concentrations of the different species, the

double line represents the salt bridge.

Since oxidation occurs at the anode and reduction occurs at the

cathode, the stronger reducing agent is placed on the left and the stronger

oxidizing agent is placed on the right.

The potential is given by

Ecell = Eright – Eleft = Ecathode - Eanode

= E(+) - E(-)

When the cell is set up properly, the calculated voltage will always

be positive, otherwise the reaction is proceeding in the reverse direction.

At standard conditions, in the above cell

Ecell

oECe / Ce

o=

4+ 3+EFe / Fe

o

3+ 2+_ = 1.61 - 0.77

= 0.84 V For a reaction to be complete enough to obtain a sharp end point in

a titration, there should be at least 0.2-0.3 V difference between the two

electrode potentials.

Example :

Determine the reaction between the following half reactions and

calculate the corresponding cell voltage.

Fe3+ + e

- = Fe

2+ E

o = 0.771 V

I3- + 2e

- = 3I

- E

o = 0.5355 V

-6-

Solution : 1

st is more +ve than 2

nd, Fe

3+ is a stronger oxidizing agent than I3

-

and Fe3+ oxidizes I

-.

∴ 2 Fe3+ + 3I

- = 2 Fe

2+ + I3

- & Ecell = 0.771 – 0.536 = 0.235 V.

Note that multiplying a half-reaction by any number does not

change its potential.

The Nernst Equation*

(Potential as a function of concentration) Eo listed in the table are determined when the [Ox] and [Red] and

all other species were at unit activity and they are called standard

potentials. Concentration rather than activity is used here because

titrations deal with large potential changes and errors are small by doing

so.

The potential dependence on concentration is described by the

Nernst equation. This equation is used to calculate the cell potential or

the electrode potential at concentrations other than 1.0 M.

aOx + ne- b Red

2.303 RT [Ox]a

E = Eo +

____________________ log

____________

nF [Red]b

Eo = standard electrode potential.

E = reduction potential under non-standard conditions.

n = number of electrons involved in the half-reaction.

R = gas constant (8.3145 JK-1 mol

-1).

T = absolute temp in Kelvins.

F = Faraday const (96485 coul eq-1)

- At 25oC the above constant is 0.05916, while at 30

oC it is equal to 0.06.

- The concentration of pure substances such as precipitates and liquids is

taken as unity.

Example :

A solution of 10-3 M Cr2O7

2- & 10

-2 M Cr

3+.

If pH is 2.0 what is the potential of the half reaction ?

* Walther Herman Nernst (1841-1941) : German physical chemist, Noble Prize 1920; The passage of

1 Faraday in a solution produces a chemical change of 1 eq wt of substance.

-7-

Solution : Cr2O7

2- + 14 H

+ + 6e

- 2Cr

3+ + 7H2O

0.06 (10-3) (10

-2)14

E = Eo +

_____________ log

________________________

6 (10-2)2

0.06

= 1.33 – 27 ( _____________

)

6

= 1.06 V

Theoretically, E would be infinity if there were no Cr3+ at all in

solution. Actually, the E is always finite, either there will be a small

amount of impurity (oxidized/reduced) or more probably the potential

will be limited by another half-reaction such as oxidation/reduction of

H2O that prevents it from going to infinity.

Remarks on Nernst Equation :

(1) In the following half reaction

aA + bB + ne- cC + dD

[C]c [D]

d

Keq = ________________

[A]a [B]

b

0.06 1

E = Eo +

___________ log

___________

n Keq

(2) [ ] = C = a , since a = fC & f = 1 for dilute solutions.

(3) ppt, solid and pure solvent have [ ] = 1

(4) E = Eo + 0 when [Ox] = [Red] or at unit activity.

(5) E is more +ve when [Ox] is high and vice versa E is more negative

when [Red] is high.

(6) In constructing a titration curve, we are interested in the equilibrium

electrode potential, Ecathode = Eanode and Ecell = 0, at which the titrant is

quantitatively reacted with the analyte (This happens when a battery

runs down).

-8-

e.g. : 2Fe3+ + Sn

2+ Sn

4+ + 2Fe

2+

Ox1 + Red2 Ox2 + Red1

[Ecathode = Eanode]

0.06 [Ox1] 0.06 [Ox2]

Eo1 +

________ log

___________ = E

o2 +

_________ log

____________

n [Red1] n [Red2]

Rearrangement

n ___________

. ∆Eo = log Keq

0.06

OR

0.06

∆Eo =

___________ log Keq

n

for the previous reaction

2 ________

( 0.77 – 0.15) = log Keq

0.06

Keq = 1021

Problem Calculate Keq for the reaction Cr2O7

2- (E

o = 1.33 V) and Fe

2+ (E

o2 =

0.77 V)

Note that : (i) log Keq = +ve → Keq > 1 → reaction is spontaneous as written

(ii) log Keq = -ve → Keq < 1 → reaction should be reversed (non-

spontaneous as written).

(iii) Keq ≥ 106 or ∆E ≥ n means that the reaction is quantitative.

(iv) The above reaction at equilibrium

2Fe3+ + Sn

2+ Sn

4+ + 2Fe

2+

a Ox1 + b Red2 b Ox2 + a Red1

Ecathode = Eanode = E

-9-

[Fe3+]

E = Eo1 + 0.06 log

____________

[Fe2+]

0.06 [Sn4+]

E = Eo2 +

_______ log

____________ , multiply by 2

2 [Sn2+]

[Sn4+]

2E = 2Eo2 + 0.06 log

____________

[Sn2+]

____________________________________________________________

[Fe3+] [Sn

4+]

3E = Eo1 + 2E

o2 + 0.06 log

_____________________

[Fe2+] [Sn

2+]

Since [Sn4+] = [Fe

2+] & [Fe

3+] = [Sn

2+] at equivalence.

Eo1 + 2E2

o

E = ___________________

3

or generally

bEo1 + aE2

o

E = ___________________

b + a

2 x 0.77 + 1 x 0.15

∴ E = ____________________________

3

= 0.563 V

Formal potential : Formal potential is designated as E

o`. It is the standard potential of

a redox couple at 1 M concentrations and with solution conditions

specified.

e.g. Eo` for Ce

4+/Ce

3+ couple in 1 M HCl is 1.28 V (note that E

o = 1.61 V);

in this case HCl complexes with cerium (IV and III) with different Kf.

The formal potential is used when not all species are known and the

Nernst equation is written as usual using Eo` instead of E

o.

-10-

Limitations for Eo uses :

(1) pH effect : Many redox reactions involve protons and their potentials are

influenced by pH.

Ox + m H+ + ne

- Red + m/2 H2O

(write equations for the couples Cr2O72-/Cr

3+; MnO4

-/Mn

2+; IO3

-/I-;

H3AsO4/H3 AsO3).

0.06 [Ox] [H+]m

E = Eo +

____________ log

_________________

ne [Red]

m 0.06 [Ox]

or E = Eo + 0.06

_________ log [H

+] +

___________ log

_________

n n [Red]

m

The term Eo + 0.06

_____ log [H

+] = E

0` = formal potential

n

Example : H3AsO4 + 2H

+ + 2e

- H3AsO3 + H2O

0.06 [H3AsO4] [H+]2

E = Eo +

_____________ log

_____________________

2 [H3AsO3]

0.06 [H3AsO4]

∴ E = Eo + 0.06 log [H

+] +

__________ log

________________

2 [H3AsO3]

0.06 [H3AsO4]

or E = Eo - 0.06 pH +

__________ log

________________

2 [H3AsO3]

The term (Eo –0.06 pH), where E

o is the standard potential, is

considered as Formal potential (Eo`). In neutral solution it is E

o – 0.06 (7)

= Eo – 0.42 = 0.559 – 0.42 = 0.139 V.

Therefore, in strongly acid medium H3AsO4, Eo = 0.559 V will

oxidize I-, but in neutral medium E

o` = 0.139 V less than (I3

-/3I

- = 0.535

-11-

V) the reaction goes in the reverse direction and I2 oxidizes H3AsO3.

The pH effect may be through secondary reactions as illustrated

here

(i) S + 2e- S

2-

H+

H2S

i.e. Acidity favors reduction and the reduced form is eliminated

increasing E measured or Eo`

(ii) I2 + 2e 2I-

+ OH-

IO

-

i.e. Alkalinity favors oxidation and eliminates the oxidant decreasing E

measured or Eo`.

In Fe(CN)63-/Fe(CN)6

4- couple; H

+ addition eliminates Fe(CN)6

4-

through weak acid formation, increasing thereby Eo`.

(2) Precipitation : Ag

+ + e

- Ag E

o = 0.799 V

E = Eo + 0.06 log [Ag

+]; X

- addition

Ag+ + X

- AgX

Ksp

[Ag+] =

____________

[X-]

Substitution gives:

Ksp

E = Eo + 0.06 log

__________

[X-]

= Eo + 0.06 log Ksp – 0.06 log [X

-]

E

o`

i.e. Eo` is going parallel to Ksp value

-12-

Eo` for AgCl/Ag = 0.222 V, for AgI/Ag = -0.151 V and the

measured

E is inversely proportional to [X-]

Example: Calculate E for Ag electrode immersed in 0.05 M NaCl

solution (Ksp = 1.8 x 10-10)

(a) EoAg+/Ag = 0.799 V (b) E

o`AgCl/Ag = 0.222 V

Solution : (a) Ag

+ + e

- Ag E

o` = 0.799

E = Eo + 0.06 log [Ag

+]

Ksp

= 0.799 + 0.06 log ___________

= 0.299 V

[Cl-]

(b) AgCl + e- Ag + Cl

- E

o` = 0.222 V

1

E = 0.222 + 0.06 log __________

= 0.299 V

0.05

* Other situations

- Addition of Zn2+ to Fe(CN)6

3-/Fe(CN)6

4- increases E due to

pptation of ferrocyanide.

- ECu2+/Cu

+ is increased also by addition of I- or NCS

- through

Cu+ removal.

(3) Complexation : Ag

+ + e

- Ag

o

Ag+ + 2X

- = Ag X2

-

[Ag X2-]

and Kf = _______________

[Ag+] [X

-]2

E = EoAg+/Ag + 0.06 log [Ag

+]

[Ag X2-]

= EoAg+/Ag + 0.06 log

_______________

Kf [X-]2

[Ag X2-]

E = EoAg+/Ag - 0.06 log Kf + 0.06 log

_______________

[X-]2

Eo`

-13-

[Ag X2-]

E = Eo` + 0.06 log

_____________

[X-]2

Example : Calculate E for Ag electrode in 1.0 M KCN to which sufficient Ag

+

is added to give Ag(CN)2- complex of 0.01 M using

Eo = 0.799 V or E

o` = -0.46 V (Answer = -0.58 V)

- Another situation is the F- addition to Fe

3+/Fe

2+ where its

potential decreases through FeF63- formation.

-14-

Titration Curves

100 ml of 0.1 M Fe2+ is titrated with 0.1 M Ce

4+ in dil H2SO4

Ce4+ + Fe

2+ = Ce

3+ + Fe

3+

(1) (2) E01 = 1.45 V; E

o2 = 0.75 V

Solution : 1

log Keq = ____________

(1.45 – 0.75); and Keq = 7 x 1011

0.06

0.36

& ∆E > ___________ ∴ Reaction is quantitative (refer p. 8)

1

(a) Before e.p. change will be in Fe3+/Fe2+ ratio * with 10 ml Ce

4+

10

E = 0.75 + 0.06 log _____

= 0.69 V

90

* with 50 ml Ce4+ E = 0.75 V

* with 90 ml Ce4+ E = 0.81 V

* with 99 ml Ce4+ E = 0.87 V

1Eo1 + 1E

o2

(b) At e.p. E = _________________

⇒ 1.10 V

1 + 1

(c) After e.p. change in Ce4+ / Ce

3+ ratio

[Ce4+]

* with 100.1 ml Ce4+ E = 1.45 + 0.06 log

___________

[Ce3+]

E = 1.27 V

* with 110 ml Ce4+

E = 1.39 V

End point detection : (1) Self indicator : MnO4

- and I2 titrations.

(2) External indicator :

Use of ferricyanide in Fe2+ titration with Cr2O7

2-

-15-

(3) Internal indicator :

Inox + ne In red

Requirements : (a) Inox and In red are of different distinctive colors.

(b) The reaction should be reversible, so that back titration is possible.

(c) Soluble in H2O to give stable solution and should not undergo side

reactions.

(d) The Transition Potential Range (TPR) E = Eo ± 0.06 should be

between Eo of sample and E

o titrant; note that (±1) from

Inox 10 1

log ___________

( _____

to ______

) ∴ i.e. +1 to –1.

In red 1 10

(e) TPR should not be affected by pH change otherwise a buffer is used

N

N

Fe2+e.g. (i) (Ph)3 Fe3+ + e- (Ph)3 Fe

2+

blue red

3

Ferroinor (1,10-phenanthroline)

(ii) Diphenylamine for Fe2+ vs Cr2O7

2-

NH NH H

N + 2H+ + 2e-

Diphenylaminecolorless

Diphenylbenzidinecolorless

N N + 2H+ + 2e-

VioletDiphenylbenzidine violet

Eo = 0.76 V

-16-

CHAPTER II Applications of Redox Titrations

Permanganate Titrations [A] Reactions : Mn

7+ is reduced to +2, +3, +4 and +6 states.

• Strong acid medium (+2 state)

MnO4- + 8H

+ + 5e

-Mn

2+ + 4H2O (E

o = 1.51 V)

• Moderate acidic medium (+3 state)

H2P2O72-

* MnO4- + 8H

+ + 4e

- Mn(H2P2O7)3

3- + 4H2O (E

o = 1.7 V)

(H2P2O72- stabilizes Mn

3+, otherwise 2Mn

3+ Mn

2+ + Mn

4+)

• Neutral (pH 4-7) medium (+4 state)

MnO4- + 4H

+ + 3e

- MnO2 + 2H2O (E

o = 1.69 V)

• Alkaline medium (+ 6 state)

(1 M OH-)

MnO4- + e

- MnO4

2- (E

o = 0.55 V)

[B] Stability : It is a SECONDARY STANDARD and should be standardized

- Unstable in presence of Mn2+

2 MnO4- + 3Mn

2+ + 2H2O 5 MnO2 + 4H

+

So addition of H+ favors the appearance of MnO4

- color towards the

end point.

- Tends to oxidize H2O

4MnO4- + 2H2O 4MnO2

+ 3O2 + 4OH

-

- Standardized by H2C2O4, 2 MnO2, As2O3 every 3 weeks

[C] Applications : (1) Direct titration :

Standard MnO4- is used to determine H2O2; Fe

2+ & Fe

3+

mixture, reduced iron (Fe2O3 + Feo), iron protoxalate (FeC2O4 &

H2C2O4); oxalic and oxalate mixture; Ferrocyanide and Ferricyanide

mixture.

(2) Back titration :

NO2- & HCOO

-

(3) Indirect titration

Ca2+; Pb subacetate & persulfate.

-17-

Applications :

I- Direct Titration

(1) H2O2 5 H2O2 + 6H

+ + 2MnO4

- = 2Mn

2+ + 8H2O + 5O2

2MnO4- ≡ 5H2O2 = 10e

-

Note : 2H2O2 = 2H2O + O2

2 x 34.02 → 22400 ml

1 g → 329.2 ml O2

(2) Fe2+ & Fe

3+

MnO4- + 5Fe

2+ + 8H

+ = 5Fe

3+ + 4H2O

MnO4- ≡ 5Fe2+ ≡ 5e-

In total, Fe3+ is reduced by Sn

2+ followed by HgCl2 addition & then

Zimmermann Reinhardt Reagent

(MnSO4, H3PO4, H2SO4) why ?

(3) Reduced Iron (Fe2O3 + Feo) contains not less than 80% Fe

through reduction of Fe2O3 by H2 gas.

Fe2O3 + Fe + Cu2+ = Fe2O3 + Cu

o + Fe

2+

↓ Filter Fe

2+, determined as above

(4) Iron protoxalate : (FeC2O4 & H2C2O4)

- 5H2C2O4 + 2MnO4- + 6H

+ = 10CO2 + 2Mn

2+ + 8H2O

V1 = Total

- 5FeC2O4 + 3MnO4- + 24H

+ = 5Fe

3+ 10CO2 + 3Mn

2+ + 12H2O

↓ Zn powder filter

V2 ≡ Fe3+ ← 5Fe

2+

MnO4-

∴ 3V2 ≡ FeC2O4

and V1 – 3V2 = H2C2O4

(5) H2C2O4 & Na2C2O4 : Acid by base; Total by MnO4-.

(6) Fe(CN)64- & Fe(CN)6

3-

2Fe(CN)63- + H2O2 = 2Fe(CN)6

4- + 2H

+ + O2

then 5Fe(CN)64- + MnO4

- + 8H

+ = 5Fe(CN)6

3- + Mn

2+ + 4H2O

(Total); New portion with MnO4-; V2 ≡ Ferrocyanide & ∆V ≡

Ferricyanide.

∴ V1 = total

{

-18-

II- Back Titration : (7) NO2

-

V2 MnO4-

NO2-

V1 MnO4-

MnO4-

C2O42- Vox

[(V1 + V2) MnO4- - VOX] = MnO4

- corresponding to NO2

-

∆ 5NO2

- + 2MnO4

- + 6H

+ 5NO3

- + 2Mn

2+ + 3H2O

50oC

(8) HCOO- Same procedure like (7)

5HCOO- + 2MnO4

- + 11H

+ = 5CO2 + 2Mn

2+ + 8H2O

HCOO-

1 ml 0.1 N MnO4- =

____________________

2 x 10 x 1000

III- Indirect Determination : (9) Ca

2+ :

Ca2+ + C2O4

2- = CaC2O4 ↓

↓ H+

H2C2O4

Then Free oxalic is titrated with 0.1 N MnO4-

Ca2+

1 ml 0.1 N MnO4- =

_____________________

2 x 10 x 1000

(10) PbAc and PbO (Pb subacetate)

- -

PbA + H2C2O4 = PbC2O4 + 2HA MnO4-

PbO + H2C2O4 = PbC2O4 + H2O

Portion ≠ NaOH → alkalinity H2C2O4

Portion ≠ MnO4- → total PbO PbAc OH

-

2 HA

Base

(11) S2O82-

C2O42- + S2O8

2- = 2SO4

2- + 2CO2

5C2O42- + 2MnO4

- + 16H

+ = 10CO2 + 2Mn

2+ + 8H2O

H2C2O4

S2O82-

MnO4-

-19-

Dichromate titrations :

Cr2O72- + 14H

+ + 6e

- 2Cr

3+ + 7H2O (E

o = 1.33 V)

a- PRIMARY STANDARD Why ? Its solution is stable indefinitely

others?

b- Due to its lower Eo; compared with MnO4

-; cannot oxidize H2C2O4,

HCl and Fe(CN)64- quantitatively and therefore it is of limited use.

c- Cannot be used as self-indicator.

Ferroin is used as indicator for Fe2+ in Cr2O7

2- titration.

d- Used in the determination of

(i) reducing agents as glycerol

(ii) oxidants by addition of excess Fe2+ and back-titrate Fe

2+

Cerric titrations :

Ce4+ + e

- Ce

3+ E

o (V)

More stable complexes [Ce(ClO4)63-] 1.70 in HClO4

Less stable complexes [Ce(NO3)63-] 1.61 in HNO3

1.45 in H2SO4

1.23 in HCl

Remarks : (a) has different E

o on variation of acids, may be due to complexation of

Ce3+ species.

(b) oxidizing power ranges from stronger than MnO4- (E

o = 1.51 V) to

slightly less than Cr2O72- (E

o = 1.33 V).

(c) cannot be used in alkaline/neutral medium otherwise Ce(OH)3 ↓ will precipitate.

(d) Compared with MnO4- & Cr2O7

2-

(1) Solutions are stable even on boiling. (2) No intermediate product, it is a one-step reaction

Ce4+ + e

- Ce

3+

(3) Ce3+ is colorless (i.e. does not obscure end point). (4) Ce4+ does not oxidize Cl- even in 2M H

+.

(5) It is available in high purity and is used as a PRIMARY

STANDARD.

(6) Indicator used is Ferroin.

-20-

Applications :

Direct titration : Fe(CN)6

4- , Fe

2+ , H2O2, I

- using Ferroin as indicator

Back titration : Polyhydroxyalcohols (glycerol), aldehydes, hydroxy acids.

The excess is titrated with oxalate or AsO33-

2Ce4+ + C2O4

2- = 2Ce

3+ + 2CO2

I Cl is catalyst, ferroin indicator

C2O42- + I

+ = 2CO2 + I

-

2Ce4+ + I

- = I

+ + 2Ce

3+

____________________________________________

C2O42- + 2Ce

4+ = 2CO2 = 2Ce

3+

or 2Ce4+ + (H3 AsO3) As

3+ = 2Ce

3 + As

5+

I Cl catalyst and ferroin indicator

As3+ + I

+ = As

5+ + I

-

2Ce4+ + I

- = 2Ce

3+ + I

+

_____________________________________________

2Ce4+ + As

3+ = 2Ce

3+ + As

5+

-21-

IODINE TITRATIONS

The intermediate Eo is behind the wide use of iodine titration

I2 + 2e- 2I

- (E

o = 0.54 V)

Two Types are known (A)Direct (Iodimetry) B) Indirect (Iodometry)

Sample :

Reducing agents Oxidizing agents

(S2-, S2O3

2-, SO3

2- … etc) (MnO4

-, Ce

4+, Fe

3+ … etc)

Titrant I2 S2O32-

Indicator appearance of disappearance of

blue starch complex blue starch complex

Reaction I2 + 2e- 2I

- 2I

- 2e

- + I2

(A) (B)

Example (A) * I2 + 2S2O32- = S4O6

2- + 2I

- ;

(B) * 6I- + BrO3

- + 6H

+ = Br

- + 3I2 + 3H2O

Factors affecting Eo of the system. These are

(1) pH effect : as described earlier under AsO43-/AsO3

3- system (P 10).

(2) Precipitating agent :

(a) Although EoCu2+/Cu

+ = 0.17 V, is less than EoI2/I

-, yet it oxidizes I

-.

Cu2+ + 2I

- CuI + I

Ksp = 10-11

Applying Nernst Eo = 0.83 V > E

oI2/I

- system

(b) Fe(CN)63- + e

- Fe(CN)6

4- (E

o = 0.36 V)

Nevertheless, it oxidizes I- in presence of Zn

2+ ion. The latter

removes Fe(CN)64- increasing its potential.

(3) Complexing agent :

- Hg2+ addition increases I2/I

- potential through removal of I

-

as HgI42-. The E

o for I2/I

- is increased to 1.0 V. So it is easy

to oxidize AsO33- in the presence of Hg

2+ (or as before in

neutral medium, HCO3-).

- EDTA or H2P2O72- complexes Fe

3+ decreasing thereby its

potential and, therefore, Fe2+ could be titrated with I2 in

presence of any stated complexing agent.

-22-

End point detection (a) Starch solution :

Colloidal dispersed starch particles form a dark blue adsorbate that

appears at the end point of direct titration (iodimetry) and disappears at

the end point of indirect titration.

Remarks :

(1) Starch solution is added towards the end point in

iodometry. The reversibility in colour appearance is

attained when iodine concentration is low.

(2) The color stability is decreased at high temperature.

(3) Starch should not be used in high acid medium. Why?

(4) Starch is unstable on storage due to bacterial

decomposition.

(b) Extraction method :

CCl4 or HCCl3 dissolves iodine giving pink color. It is less

convenient and more time consuming than starch solution. But

sometimes both methods give excellent results.

ERROR SOURCES :

1- Due to pH consideration : (a) pH and I2

I2 + 2OH- = I

- + IO

- + H2O

3IO- = 2I

- + IO3

-

Therefore I2 titration cannot be used satisfactorily in alkaline

medium.

(b) pH & S2O32-

H+ + S2O3

2- = HS2O3

- = HSO3

- + S

xss = SO2 + H2O + S

The reaction is not rapid, so S2O32- could be used safely in

moderate acid medium as long as there is –at any time – insignificant

excess of S2O32-.

I2 reaction with thiosulphate is not quantitative in alkaline medium

since S2O32- is partially converted to SO4

2-.

(c) pH & I-

4I- + 4H

+ + O2 = 2I2 + 2H2O

Therefore considerable O2 error is introduced giving in iodometry

positive error.

-23-

2- Due to I2 (Secondary standard) : - Volatilizes easily at high temperature and low I

-

concentration giving positive error.

- Its concentration changing with time as it reacts with

reducing agents (organic matter, SO2, H2S) giving positive

error.

- Disproportionates with H2O.

(I2 + H2O = HIO + I- + H

+) giving positive error.

3- Due to starch : Easily decomposable with bacteria or high acidity.

SELECTED APPLICATIONS

(I) IODIMETRY

(A) Direct titration : * (1) As2O3 / Sb2O3

As2O3 + 2I2 + 2H2O = As2O5 + 4I- + 4H

+

N.B. : * using OH-/CO3

2- gives positive error due to side reactions

6 OH- (CO3

2-) + 3I2 = 5I

- + IO3

- + 3H2O (CO2)

Therefore, use of HCO3- is appropriate to make the reaction

quantitative.

(B) Back titration of excess I2 : OH

- H

+

* (2) 2 Ampicillin sod : → 2 penicilloic acid →

I2

2(CH3)2 C – CHCOOH → (CH3)2 C – CHCOOH

| | | | SH NH2 S NH2

(D-penicillamine) | S + 2HI

| (CH3)2 C – CH COOH + 2HI

| NH2

Disulphide

Excess I2 is back titrated with S2O32-

(∴ 2 ampicillin + I2 ≡ 2e-)

-24-

* (3) HgCl2

a. HgCl2 + HCHO + 3NaOH → Hg + 2NaCl + HCOONa + 2H2O

b. Hg + I2 = HgI2

c. HgI2 + 2KI = K2HgI4 soluble (∴ HgCl2 ≡ I2 ≡ 2e-)

* (4) SO32- + I2 + H2O = SO4

2- + 2H

+ + 2I

-

II- IODOMETRY (5) Chlorinated lime : ( % of Cl2 available)

CaCl (OCl) + 2CH3COOH = Ca (CH3COO)2 + HCl + HClO

HCl + HClO = H2O + Cl2

2HI + Cl2 = 2HCl + I2

I2 + 2S2O32- = 2I

- + S4O6

2-

(∴ ClO- ≡ Cl2 ≡ I2 ≡ 2e

- )

(6) CuSO4

2Cu2+ + 4I

- = Cu2I2 + I2

(∴ 2Cu2+ ≡ I2 ≡ 2e

- )

(7) Phenol salicylic acid, Aspirin, phenylepherine (any of them gives the

same reaction)

COOH

O - COCH3

COOH

OH

Br

OH

Br

Br

H+ Br2

aspirin Br2 + 2I

- = 2Br

- + I2

(∴ Aspirin ≡ 3Br2 ≡ 6e-)

(8) Glycerol

CH2OH

| 3 CHOH + 7 Cr2O7

2- + 56H

+ = 14Cr

3+ + 9CO2 + 40H2O

| CH2OH

Cr2O72- 3 glycerol glycerol

(∴ _________________

≡ ___________________ ≡ _______________ ≡ 1e-) 6 7 x 2 14

-25-

(9) Organically combined I : SO3

N

OH

I

a. NaIfuse

Na2CO3

* I

- + 3Br2 + 3H2O = IO3

- + 6HBr (remove excess Br2)

* IO3- + 5I

- + 6H

+ = 3I2 + 3H2O

(∴ Org I ≡ IO3- ≡ 6e-; increasing sensitivity)

(10) Polyhydroxy Alcohols / Aldehydes / Aldoses

(a) e.g. Mannitol

CH2OH (CHOH)n CH2OH consumes n + 1 HIO4

(mannitol consumes 5 HIO4)

2CH2O + 4 HCOOH + 5HIO3 + H2O

(from terminal OH) (from 2 ry OH)

Both IO3- & IO4

- react with I

- to give I2

IO3- + 5I

- + 6H

+ = 4I2 + 4H2O

IO4- + 7I

- + 6H

+ = 3I2 + 4H2O

VBlank – VExp. = Vmannitol

(∴ I M IO4- = Mannitol)

(b) Aldehydes and Aldoses (e.g. glucose)

(i) I2 + OH- = OI

- + I

-

RCHO + OI- = RCOOH + I

-

(ii) OI- + I

- + 2H

+ = I2 + H2O

∴ RCHO ≡ IO- ≡ I2 ≡ 2e

-

(11) Karl Fischer Reagent : (for H2O determination)

KFR is a mixture of pyridine, I2, SO2 & CH3OH and has a brown

color. Methanol is present in excess to prevent side reactions. It is

secondary standard and should be frequently standardized using dist H2O.

-26-

Main Reaction : End point from yellow (product) to brown color excess titrant

N

N

N N

+ I2 + 2

SO2 +

H

O

H

+

+ NO3S

N

H

CH3SO4

yellow

NSO3 + H2O(side reaction)

N

HSO4; prevented by additon of more CH3OH

H

+

excess CH3OH

This side reaction consumes more H2O, thereby giving negative

error.

KIO3 Titrations

(i) Primary standard, (ii) prepared in molar solution, (iii) solution

has different Eq. Wt according to the reaction involved.

1 N H+

Reaction 1: IO3- + 6H

+ + 5e

- I + 3H2O [E

o = 1.178 V)

4 N H+

Reaction 2: IO3- + Cl

- + 6H

+ + 4e

-ICl + 3H2O (E

o = 1.24 V)

-27-

• In Andrew’s reaction (reaction 2) the HCl normality should

be more than 4N, otherwise I+ will hydrolyze to give free I2

which consumes more IO3- , (i.e. positive error)

I+ + H2O HOI + H

+

• Lang’s modification In HCN solution, lower acidity (1-2 N HCl) is sufficient for

quantitation because ICN is more stable than ICl.

IO3- + 2I

- + 3CN

- + 6H

+ = 3ICN + 3H2O

• End point detection :

Lang’s vs Andrew’s; in HCN starch can be used, while in 4N HCl

it cannot be used. Why ?

KBrO3 Titrations

It is a primary standard

BrO3- + 6H

+ + 6e

- = Br- + 3H2O (E

o = 1.44 V)

The end point could be detected by irreversible indicator Methyl-

orange (Methyl Red) which is bleached by xss Br2.

Applications : It is used in the following determinations (1) Strong reductants :

As3+ in 2-3 N HCl

(2) Weak reductants

Ions (Mn2+, Cr

3+, H2C2O4) reduce BrO3

- partly to Br

- and partly to Br2.

BrO3- + 6H

+ + 5e

- ½ Br2 + 3H2O

Addition of Hg2+ removes Br

- as HgBr2, thus increasing the

tendency for reduction to Br- only.

-28-

(3) Indirect determination

(a) Phenol, Salicylic acid, Aspirin. (b) Cations e.g. Mg

2+, Al

3+

N

OH

N

OH

Br

Br

Mg2+ +

2HBr +

chelate (ppt)

H+

2Br 2

free ligand(8-hydroxyquinoline)

(∴ 1 Mg

2+ = 2 ligand = 4Br = 4e

-)

-29-

PRIOR OXIDATION and PRIOR REDUCTION

• This is carried out before the actual analysis to change the

sample component to the desired oxidation state. Iron salts,

for example, usually contain both Fe(II) & Fe(III). These

should quantitatively be converted to either Fe(II) through

prereductant addition or Fe(III) by preoxidant addition.

• Limitations :

1- Reaction should be quantitative and reasonably rapid.

2- Reaction must not interfere with other sample components.

3- Easy to remove the excess completely through

a- boiling (e.g. H2S, H2O2, S2O82-)

b- changing condition (e.g. HClO4 is oxidant when

hot, but not on cold).

c- 2nd phase use (e.g. metal, amalgam, BiO3-)

I- PREOXIDATIONS :

A. Gases : (1) (Ozone) :

O3 + 2H+ + 2e

- O2 + H2O (E

o = 2.07 V)

2H+ + O3 + Ag

+ Ag

3+ + O2 + H2O

3Ag3+ + 2Cr

3+ 3Ag

+ + 2Cr

6+

Used : In acid medium (Ce3+, Cr

3+, Mn

2+) and in alkaline medium (NO2

-,

I-, PO3

3-) to increase oxidation state. Excess O3 is removed by boiling or

by passing CO2. Ag+ is used as catalyst.

(2) (Bromine) :

Br2 + 2e- 2Br

- (E

o = 1.07 V)

Used : I- → I2, remove excess by adding phenol or passing inert gas

B. Homogenous (HClO4, S2O82-, H2O2)

(3) HClO4 : ∆ HCIO4 + 7H

+ + 7e

- → ½ Cl2 + 4H2O (E

o = 2.0 V)

- Used for Fe2+, Cr

3+, org. S, org. As oxidation

- Removed through dilution or cooling.

-30-

(4) (NH4)2 S2O8 : (strong in acid medium and in presence of Ag+ as

catalyst)

S2O82- + 2e

- 2SO4

2- (E

o = 2.01 V)

S2O82- + Ag

+ 2SO4

2- + Ag

3+ (slow)

3Ag3+ + 2Cr

3+ 3Ag

+ + 2Cr

6+ (fast)

Excess Ag+ is removed as AgCl & S2O8

2- by boiling

2S2O82- + 2H2O = 2SO4

2- + 4H

+ + O2

(5) H2O2 :

H2O2 + 2H+ + 2e

- 2H2O (E

o = 1.77 V)

Used in acid medium (Fe2+, Sn

2+ oxidation) or in alkaline medium

(Mn2+, Cr

3+) and removed by boiling

2H2O2 H2O + O2

C. Solid peroxidant (6) Na BiO3

BiO3- + 6H

+ + 2e

- = Bi

3+ + 3H2O (E

o = 1.7 V)

Used for Mn2+ & Cr

3+ oxidation and removed by filtration through

sintered glass filter.

II- PREREDUCTANTS

A. Gases : (1) H2S :

2H+ + S + 2e

- H2S (E

o = 0.14 V)

Used for Fe3+ reduction; removed by boiling or bubbling CO2. Its

drawback is the precipitation of cations.

(2) SO2 : pH dependent

SO42- + 4H

+ + 2e

- = H2SO3 + H2O (E

o = 0.17 V)

S2O62- + 4H

+ + 2e

- = 2H2SO3 (E

o = 0.15 V)

Used for Fe3+, As

5+, Sb

5+, Cr

6+; Cu

2+ (in SCN); removed by boiling

or CO2 bubbling.

B. Homogenous : Sn

4+ + 2e

- Sn

2+ (E

o = 0.154 V)

Used for Fe(III) reduction, keeping slight Sn2+ excess and high

[Cl-]. Excess Sn

2+ is destroyed by Hg

2+ addition giving Hg2Cl2. Large

[Sn2+] gives Hg

o, the latter consumes more titrant producing positive

error.

-31-

C. Metallic :

* Jones Reductor : contains Zn-Hg packed into a column 36-cm long, prepared through Zn/HgCl2 boiling.

Used : Reduction of Fe(III), Ti(IV), Ce(IV), Mn(VII) and Cr(VI).

Limitations : (1) Its reducing action is non-selective. (2) Cannot be used for solutions, containing NO3

-. This is

converted to NH3, exhibiting interference.

(3) Air is reduced, therefore vacuum must be drawn into

column.

(4) H2SO4

2Zn + O2 + 2H2O = H2O + 2Zn (OH)2

is the appropriate acid used since HCl liberates H2 reacting with Zn.

* Walden Reductor : Less powerful but more selective reductant

Ag metal is used.

Ag+ + e

- Ag (E

o = 0.77 V)

or AgCl + e- Ag + Cl

- (E

o = 0.222 V)

Therefore, it is possible to decrease the reductant power on

changing [Cl-]; AgCl ppt on Ag metal could be washed out with NH3.

Again air could be reduced to H2O2.

Used : for Fe(III), Ce(IV), Mn(VII) and Cr(VI) reduction.

Standard Solutions : * Oxidants :

Permanganate, Dichromate, Bromate, Ceric, Iodine and Iodate.

* Reductants :

Oxalates, thiosulphates, asrenite, ferrous ammonium sulphate.

Questions :

1- Calculate the equivalent weight for each.

2- Which is used as primary standard, and which as secondary

standard ?

3- How can you standardize the secondary standard solution ?