lecture10-redox-titrations.pdf, Summaries of Analytical Chemistry

Redox Titrations. -the oxidation/reduction reaction between analyte and titrant. -titrants are commonly oxidizing agents, although reducing.

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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)
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff

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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:

A

ox

+ B

red

!

A

red

+ B

ox

Rx’n goes to completion after each addition of titrant –

Potentiometric Titration:

Titration reaction:

Ce

4+

+ Fe

2+

!

Ce

3+

+ Fe

3+

Reference half-reaction:

2Hg(l) + 2Cl-! Hg

2

Cl

2

(s) + 2e-

At the Pt indicator electrode (Indicator half-reaction)

Fe

3

+ e- !Fe

2+

E

0

= 0.767 V (2)

Ce

4+

+ e- !Ce

3+

E

0

= 1.70V (3)

Cell reactions (in 1 M HClO

4

2Fe

3+

+ 2Hg(l) + 2Cl

!

2Fe

2+

+ Hg

2

Cl

2

(s) (4)

2Ce

4+

+ 2Hg(l) + 2Cl

!

2Ce

3+

+ Hg

2

Cl

2

(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+! Cr

3+

+ Ag(s)

E

cell

= O = E

cathode

– E

anode

= E

Ag+

- E

Cu2+

or

E

cathode

= E

anode

= E

Ag+

= E

Cu2+

- Also when in equilibrium, electrode potentials of all

systems are identical:

E

Ox

= E

Ox

= E

Ox

= E

Ox

Where E

Ox

…..are electrode potentials for the four half-

reactions

Calculating Equilibrium Constants

Cu(s) + 2Ag ! Cu

2+

+ 2 Ag(s)

E

0

Ag+

2

  1. 0592

log =

  • 2 [Ag ]

1

E

0

Cu2+

2

  1. 0592

log

[Cu2]

1

E

0

Ag+

- E

0

Cu2+

[Cu ]

1

log

2

  1. 0592

[ ]

1

log

2

  1. 0592

2 2 +

!

Ag +

2 eq

2 Cu

0 0

log K

[Ag ]

[Cu ]

log

  1. 0592

2( )

=

=

!

E (^) Ag+ E +

Ex: Calculate the equilibrium constant:

2(0.799- 0.337)

[Ag ]

[Cu ]

logK log 2

2

eq

=

=

= 15.

K eq = antilog 15.6 = 4.1 x 10

15 = 4 x 10

15

Redox Titration Curves

Fe

2+

  • Ce

4+! Fe

3+

  • Ce

3+

E

Ce

4+ = E Fe3+

= E

system

E

In

= E

Ce4+

= E

Fe3+

= E

system

Equivalence Point Potentials

Fe

3

+ e- !Fe

2+

Ce

4+

+ e- !Ce

3+

]

[Ce

]

[Ce

log

Ce eq

E = E

]

[Fe

]

[Fe

log

Fe eq

E = E

[Ce ][Fe ]

[Ce ][Fe ] 0

Fe

Ce eq 4 3

3 2

4 3 + +

E = E + E! (1)

Definition of e.p. requires that:

[Fe

3+

] = [Ce

3+

]

[Fe

2+

] = [Ce

4+

]

V

E E

E

f

Fe

f

Ce

eq

4 3

  1. Potential after addition of 25.10 mL of Ce

4+

[Fe

2+ ] = amt of Ce

4+ left unreacted, therefore added to C Ce4+

calculated from the volumes of the two solutions and subtracted from

C

Ce3+

Conc of two cerium ion species:

[Ce

3+ ]=

  1. 10

  2. 500

]

2

25.00x 0.

[Fe!

"

[Ce ]

[Ce ]

log

4

3

E=+! =+1.44-

  1. 010 / 75. 10

  2. 500 / 75. 00

log

1

  1. 0592

= +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

]

[I

log

3

3

0 E= E!

num-mol/L

3 , denom-mol/L

Completeness of reaction – the change in E system

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:

Volts

n

E E )

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

  • before titration, e.g. Mn

2+ preoxidized to MnO 4

  • 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 (S 2

O

8

2 - ) – requires Ag+ as a catalyst.

+! + +

4

2 -

4

2 -

2 8

S O Ag SO SO Ag

Excess reagent destroyed:

+ !!! !!" + +4H

O 4SO O

2S O 2 H

2 -

4

2 -

2 8

boiling

Prereduction

  • Stannous chloride (SnCl 2

) will reduce Fe

3+ to Fe

2+ in hot HCl

Excess reductant is then destroyed:

Sn

2+

  • 2HgCl 2

! Sn

4+

  • HgCl 2

  • 2 Cl

Oxidation with Potassium Permanganate

  • strong oxidant, violet color

In strongly acidic solutions, reduced to colorless Mn

2+ :

MnO 4

    • 8H
  • 5e
  • ! Mn

2+

  • 4 H 2

In neutral or alkaline solution, the product is the brown solid, MnO 2

MnO 4

    • 4H
  • 3e
  • ! MnO 2

(s) + 2H 2

O

In strongly alkaline solution (2 M NaOH), green manganate is

produced:

MnO 4

    • e-! MnO 4

2 -

Tales 16.3……..see below

Note: permanganate solutions are unstable, therefore not a primary

standard.

4MnO 4

    • 2H 2

O > 4MnO 2

+ 4OH

    • 3O 2

( MnO 2

catalyses this

reaction )

Permanganate must be standardized for example with oxalate;

H

2

C

2

O

4

> 2H

  • CO 2

  • 2e

Overall:

. 2MnO 4 - + 5H 2

C

2

O

4

+ 16H

> 2Mn

2+

  • 10CO 2

+ 8H

2

O

KMnO4 can serve as own indicator, since product Mn

2+ is colorless.

Cerium(IV)

Strong oxidant > Ce(III)

Ce

4+ [Yellow ]+ e

  • > Ce

3+ [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

Cr 2

O

7

2 -

  • 14H

  • 6e

  • > 2Cr

3+

  • 7H 2

O 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

I

2

  • 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.