1

Chapter 19

Applications of

Standard Electrode Potentials

• 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 cathode(+, right) anode

(1) For half cell:

0.0592ln lg (@ 298 )

(2) For electrochemical cell:

= -

(3) At chemical equilibrium:

= -

OX OX

Red Red

a aRTE E E K

nF a n a

E E E

E E E

(-,left)

0 0

0

When the reaction reaches equalibrium

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

nF E n EK K

RT

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

2

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

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 /Fe Fe /Fe

4+ 3+0 0

3+ 2+Ce

e

/Ce

q

eq

eq Fe /Fe

[Ce ]0.0592 lg (1)

[Ce ]

[Fe ]0.0592 lg (2)

[Fe ]

Adding eqs. (1) and (2)

[Ce ] [Fe ]2 ) 0.059

and

( 2 lg{ }[Ce ] [Fe ]

F

=

=

=

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+

eq

e

0 0

Ce /Ce Fe /Fe

0 0

Ce /Ce Fe /Feq

Fe + Ce For at equilibrium

[ ]=[ ], [ ]=[ ]

2 ) 0.0

e + Ce

F

592 lg{1}

2

e Ce F

(

e Ce

=

=

E E

E

E

EE

2+ 4+ 4+ 3+2

4+ 4+ 2+ 3+ +2 2

2 4

2+ + 4+ 0

0 0'

UO / U Ce /Cee

2 2

4 + 0'

q

+ 3

For system:

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

in 1.0 M H SO

UO + 4H +2e U + 2H O = 0.334 V

Ce + e Ce = 1.44 V

2 0.0592= l

3 3

E

E

E EE

4

eq

g[H ]

is pH dependent.E

3

Generally,

1

2

1 1 1 2 2 2

1 2

1 Ox 1

2 Ox 2

1 2 1 2

Ox Ox / Red Ox Ox / Redeq

Ox Ox

For Ox + n e Red

Ox + n e Red

Ox Red Red Ox

n n

n n

E EE

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.

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.1000 M Ce4+ in a medium that is1.0 M in H2SO4.

Fe2+ + Ce4+ Ce3+ + Fe3+

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

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

4

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.1000 M Ce4+ in a medium that is1.0 M in H2SO4.

Fe2+ + Ce4+ Ce3+ + Fe3+

(2) At 15.00 mL of Ce4+ added, VFeMFe > VCeMCe

= 2.308x10-2 M

VCe MCe (15.00 mL)(0.1000 M)[Fe3+] = --------------- = ----------------------------

VFe + VCe (50.00 + 15.00)mL

At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe

[Fe3+] = 2.308 x 10-2 MVFe MFe - VCe MCe

[Fe2+] = ------------------------VFe + VCe

(50.00 mL)(0.0500 M) - (15.00 mL)(0.1000 M)

[Fe2+] = --------------------------------------------------------(50.00 + 15.00)mL

= 1.54 x 10-2 M

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.1000 M Ce4+ in a medium that is 1.0 M in H2SO4.

Fe2+ + Ce4+ Ce3+ + Fe3+

At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe

[Fe3+] = 2.308 x 10-2 M(50.00 mL)(0.0500 M) - (15.00 mL)(0.1000 M)

[Fe2+] = --------------------------------------------------------(50.00 + 15.00)mL

= 1.54 x 10-2 M

5

Fe2+ + Ce4+ Ce3+ + Fe3+

At 15.00 mL of Ce4+ added, VFeMFe > VCeMCe

[Fe3+] = 2.308 x10-2 M [Fe2+] = 1.54 x 10-2 M

for Fe2+ + e = Fe3+ E0’ = 0.68 V

0.0592 [Fe2+]Esystem = E0’ - ----------- log ----------- = 0.69 V

1 [Fe3+]

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.1000 M Ce4+ in a medium that is 1.0 M in H2SO4.

Fe2+ + Ce4+ Ce3+ + Fe3+

(3) At 25.00 mL of Ce+4 added,VFeMFe = VCeMCe, equivalence point

nFe EFeo + nCe ECe

o 1(1.44) + 1(0.68)Esystem= ----------------------- = ------------------------ V

nFe + nCe 1 + 1

= 1.06 V

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.1000 M Ce4+ in a

medium that is 1.0 M in H2SO4.

Fe2+ + Ce4+ Ce3+ + Fe3+

At 26.00 mL of Ce4+ added, VCeMCe > VFeMFe

VFe MFe (50.00 mL)(0.0500 M)[Ce3+] = ---------------- = -------------------------------

VFe + VCe (50.00 + 26.00)mL= 3.29 x 10-2 M

6

Fe2+ + Ce4+ Ce3+ + Fe3+

At 26.00 mL of Ce4+ added, VCeMCe > VFeMFe

[Ce3+] = 3.29 x 10-2 M

VCe MCe - VFe MFe[Ce4+] = ---------------------------

VFe + VCe

(26.00 mL)(0.1000 M) - (50.00 mL)(0.0500 M)[Ce4+] = -------------------------------------------------------------

(50.00 + 26.00)mL= 1.32 x 10-3 M

EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe2+ with 0.1000 M Ce4+ in a

medium that is 1.0 M in H2SO4.

Fe2+ + Ce4+ Ce3+ + Fe3+

At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe

[Ce3+] = 3.29 x 10-2 M [Ce4+] = 1.32 x 10-3 M

for Ce4+ + e = Ce3+ E0’ = 1.44 V

0.0592 [Ce4+]Esystem = E0’ + ----------- log ----------= 1.39 V

n [Ce3+]

(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

7

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 (lgKeq=nE0/0.0592), the greater the 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• General redox indicators—substances that change

color on being oxidized or reduced, i.e., color changes for general redox indicators depend only on the potential of the system.

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

(thiocyanate) for Fe(III)

Starch + I3- Starch-I3

- complex

(colorless) (blue)

8

General Redox Indicators

ox red

ox red

0 OXox red In /In

Red

OX

Red

OX

Red

0In /In

[In ]0.0592In + ne In lg

[In ]

[In ] 1The color change is seen when

[In ]

overall change 100 tim

10

[In ] 10changes to ( )

[In ] 1

0.0592

0.118 / V is need

es

e

E En

E En

n

system

system

d for the change.

Many indicaors, 2, 59 mV change sufficient.

E

n E