1

Chapters 19 & 20

Applications of

Standard Electrode Potentials

&

Redox Titrations

• Part 1: Calculating Potentials of Electrochemical Cells (Examples 19-1~10)--Nernst equation for half-cells and chemical cells

0 0

cell cathode(+, right) anode(-,left) cell

(1) For an half cell: Ox + ne = Red

0.0592ln lo g (@ 298 )

(2) For an electrochemical cell:

= - ;

(3) At chemical equilibriu

OX OX

Red Red

a aRTE E E K

nF a n a

E E E G nF E

cell cathode(+, right) anode(-,left)

0 0

m:

= - 0

When the reaction reaches equalibrium

ln , or log (at 298K)0.0592eq eq

E E E

nF E n EK K

RT

1

2

3

2

(1 calorie = 4.18 Joules)

4

5

6

3

7

8

9

4

Part 2: Constructing Redox Titration Curves

2+ 4+ 3+ 3+Fe + Ce Fe + Ce (rapid and reversible)

3+ 2+ 4+ 3+

2+ 4+ 3+ 3+

systemFe /Fe Ce /Ce

Fe + Ce Fe + Ce (rapid and reversible)

Reaction is rapid and reversible, hence at equilibrium

= =

)

( During the entire titration proc

E E E

ess

Analyte Titrant

How to Measure the Electrode Potential?

3+ 2+ 4+ 3+

3+ 2+ 4+ 3+

cell system SCE

system cell SCE Fe /Fe Ce /Ce

Fe , Fe ,Ce , Ce | PtRef. E (SCE)

=

=

||

E E E

E E E E E

SCE: Saturated calomel electrode

10

11

12

5

Equivalence-point potentials

4+ 3+ 4+ 3+

3+ 2+ 3+ 2+

4+ 3+ 3+ 2+

4+0

3+Ce /Ce Ce /Ce

3+0

2+Fe

e

/Fe Fe /Fe

4+ 3+0 0

3+ 2+e

eq

eq

Ce /C e /q e F F

[Ce ]0.0592lo g (1)

[Ce ]

[Fe ]0.0592lo g (2)

[Fe ]

Adding eqs. (

=

e

and

1) and (2)

[Ce ] [Fe ]2 ) 0.0592lo g{

[C ] [=(

Fe

=E E E

E E

E E

E

E

4+ 3+ 3+ 2+

4+ 3+ 3+ 2+

2+ 4+ 3+ 3+

3+ 3+ 2 4

0 0

Ce /Ce

e

Fe /Fe

0 0

Ce /Ce Fe /Fe

+ +

eq

q

}]

For at equilibrium

[ ]=[ ],

Fe + Ce Fe + Ce

2

F [ ]=[ ]

2 ) 0.0592lo g{1=

e

(

Ce F Ce

}

e

=

E

E

E E

E E

2+ 4+ 4+ 3+2

0 0'

UO /U Ce /Ceeq

2

4+ 4+ 2+ 3+ +2 2

4

2+ + 4+ 02 2

4+ 3+ 0'

l

For system:

U + 2Ce + 2H O UO + 2Ce + 4H

in 1.0 M H SO

UO + 4H +2e U + 2H O = 0.

2 0.0592=

334 V

Ce + e Ce

3

= 1.44 V

3

E

E EE

E

4

eq

og[H ]

is pH dependent.E

Generally,

1

2

1 1 1 2 2 2

1 2

1 Ox 1

2 Ox 2

1 2 1 2

0 0Ox Ox /Red Ox Ox /Red

eqOx Ox

For Ox + e Red

Ox + e Red

Ox Red Red Ox

n

n

n E n EE

n n

13

14

15

6

Redox Titration Curve

Derivation of a titration curve

Fe2+ + Ce4+ Ce3+ + Fe3+

EXAMPLE: Derive the titration curve for50.00 mL of 0.0500 M Fe2+ with 0.1000 MCe4+ in a medium that is 1.0 M in H2SO4.

Fe2+ + Ce4+ Ce3+ + Fe3+

E = E0 + (0.0592/n) log([Ox]/[Red])

(1) At 0.00 mL of Ce4+ added, initial point, noCe4+ present; minimal, unknown [Fe3+]; thus,insufficient information to calculate E.

(2) At equivalence point, how many mLs ofCe4+ are required?

4+ 2+ 4+ 2+

4 2+

4

Ce Fe Ce Fe

Fe

; ( ) ( )

0.1000 (50.00 0.0500)

25.00 Ce

Ce

n n CV CV

V

V mL

0 0Fe Fe Ce Ce

Fe Ce

1 1.44 1 0.V

68

21.06 system

n E n EE

n n

(3) Before the eq point, e.g., at 15.00 mL of Ce4+

added:Fe2+ + Ce4+ Ce3+ + Fe3+

I 50.00×0.0500 0 0 0C -15.00×0.1000 +15.00×0.1000 1.50 mmol 1.50 mmolE 1.00 mmol 0 1.50 mmol 1.50 mmol Vtotal 50.00 + 15.00 = 65.00 mLConc. 1.00/65.00 0 1.50/65.00 1.50/65.00

3+ 2+

3+0

2+Fe /Fe

[Fe ] 1.500.0592log 0.68 0 V.0592log

[F 10.6

e ] .0090 systemE E

16

17

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7

(4) After eq. point, e.g., at 26.00 mL of Ce4+

added,

2+ 4+ 3+ 3+ Fe + Ce Fe + Ce

I 50.00 0.0500

C -50.00 0.0500 +26.00 0.1000 +50.00 0.0500 +50.00 0.0500

E ~0 0.10 mmol 2.

4+ 3+

4+0

system 3+Ce /Ce

50 mmol 2.50 mmol

[Ce ] 0.10 = + 0.0592log =1.44 +0.0592log

[Ce ] 2.50

= 1.357 V

E E

(A)50.00 mL of 0.05000 M Fe2+ in 1.0 M H2SO4

(B) 50.00 mL of 0.02500 M U4+ in 1.0 M H2SO4

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20

21

8

Effect of Variables on Redox Titration Curves

• Esystem is usually independent of dilution Redox titration curves are usually independent of analyte and titrant concentrations--Different from neutralization, precipitation and complexation titrations.

• The larger the equilibrium constants or the standard E0 (logKeq=nE0/0.0592), the greater the change in equivalence-point region—Similar to other types of titrations.

Effect of titrant electrode potential on reaction completeness.

E0A—Analyte E0, 0.200 V

E0T—Titrant E0.

Oxidation/Reduction Indicators

• Specific indicators: (1) Starch for I3-, (2) SCN-

(thiocyanate) for Fe(III)

-3 3

- 3 III 2+ - 2

(blue)

O

S

( ) (colorle

ltarch + I Starch-I -comp ex Starch + 3I

colorless ?

SCN + Fe Fe (SC

s

ss)

( )

N) SCN + Fe

colorle s ? (

e

e

)r ) (cg oe lR od ra l n e ess/

22

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9

• General redox indicators—substances change color upon redox reaction, which is dependent only on the potential of the titration system.

General Redox Indicators

ox red

ox red

0 OXox red In /In

Red

OX

Red

OX

Red

0In /In

[In ]0.0592In + ne In log

[In

d

o

g

v

e

s

r

s

a

e

l

l

h

n

c

[

h

n

a

a

n

[

g

n

e

]

I ] 1The color

ch n e i e n w e I ] 10

[In ] 10changes to ( )

[In ]

.

100 t1

0 0592

0.118 / V is n

e

ee

im s

E En

E En

n

system

system

ed for the change.

Many indicaors, 2, 59 mV change sufficient.

E

n E

25

26

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10

The Strong Oxidants - Potassium Permanganate and Cerium (IV)

+ 2+ 02

-

4

4

4+ 3+ 02 4

-

+ 8H + 5e Mn +4H O = 1.51 V

Ce + e Ce ' = 1.44 V (1 M H SO )

Reactions are in strong acidic media.

MnO (purple) is a self-indicator.

Can be standardized w

MnO (purple) E

E

- + 2+4 2 2 4 2 2

4+ 3+ +2 2 4 2

ith oxalate:

2MnO + 5H C O + 6H 2Mn 10CO ( ) 8H O

Ce + H C O 2Ce + 2CO ( ) + 2H

g

g

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11

Strong Reducing Agents—Fe(II) and thiosulfate

2+ 3+

2- 2-2 3 4 6

- - +3 2 2

2- 2-2 2 3 4 6

(1) Fe Fe + e (freshly prepared from stable agent)

(2) 2S O (thiosulfate) S O (tetrathionate) + 2e

Standardized with pottasium iodate:

IO + 5I (excess) + 6H 3I +2H O

I 2S O S O +

-

- 2-3 2 2 3

2

2- 2- -2 2 3 4 6

I

1 mol IO 3 mol I 6 mol S O

Indicator: I -starch (blue)

I -starch 2S O ( ) S O + I

blue colorless

excess starch

Calculation Examples1. Aqueous solutions containing approximately 3% (w/w) H2O2 aresold in drug stores as a disinfectant. A method for determining theperoxide content involves the use of 0.01145 M KMnO4. Assumethat you wish to use between 30 and 45 mL of the above KMnO4

reagent for a titration. Calculate the amount of H2O2 needs to take.

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32

33

12

2. - + 2+

4 2 2 4 2 22MnO + 5H C O + 6H 2Mn 10CO ( ) 8H Og

Na2C2O4

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