07. proses elektrokimia dan diagram potensial - ph

07. Proses Elektrokimia dan Diagram Potensial - pH

Zulfiadi Zulhan

Teknik MetalurgiFakultas Teknik Pertambangan dan PerminyakanInstitut Teknologi BandungINDONESIA

Termodinamika Metalurgi (MG-2112)

201 Zulfiadi Zulhan MG2112 Termodinamika Metalurgi 2021

Jangan Mengunggah

Materi Kuliah ini di

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1. Pendahuluan, istilah-istilah dan notasi

2. Hukum I Termodinamika

3. Hukum II Termodinamika

4. Hubungan Besaran-Besaran Termodinamika

5. Kesetimbangan

6. Kesetimbangan Kimia dan Diagram Ellingham

7. Proses Elektrokimia dan Diagram Potensial - pH (Pourbaix)

8. Ujian Tengah Semester

9. Aktivitas Ion

10. Termodinamika Larutan

11. Penggunaan Persamaan Gibbs - Duhem

12. Penggunaan Metoda Elektrokimia untuk menentukan Sifat-Sifat / Besaran-Besaran

Termodinamika

13. Keadaan Standar Alternatif

14. Koefisien Aktivitas dalam Larutan Encer Multi-Komponen

15. Diagram Fasa

16. Ujian Akhir Semester

Materi Perkuliahan


Electrochemical Cell

For reaction of hydrogen and oxygen to produce water, electrons are transfered from the hydrogen

atoms to the oxygen atoms.

H2 + ½ O2 = (2H+ . O2-) or H2O

H2 O2

e

H2 → 2H+ + 2e ½ O2 + 2H+ + 2e → H2O

Electrolite must be

able to conduct ions,

but it should not

conduct electrons

Anode : electrode at which oxidation

takes place (increase of oxidation number)

Cathode : electrode at which reduction

takes place (reduction in oxidation state)

In case of reaction of hydrogen and oxygen to form

water:

Anode : H2 = 2H+ + 2e

Cathode : ½ O2 + 2H+ + 2e = H2O

This process is called fuel cell.

At hydrogen pressure and oxygen pressure are

each equal to one atmosphere, circuit voltage will be

1.229 V

Anode Cathode


Water electrolysis to produce hydrogen and oxygen

anodecathode

H2O

Hydrogen

is evolved

Oxygen is

evolved

Electrodes are inert (Pt)

DC Power

Supply

+-

e e

H+H2

H+

In case of water electrolysis to produce

hydrogen and oxygen (hydrogen – oxygen

generator)

Anode : H2O = ½ O2 + 2H+ + 2e

Cathode : 2H+ + 2e = H2

Electrical potential greater than circuit (1.229

V) is applied to the cell, the polarity is opposite

to natural cell potential.

In alkaline or neutral solution, following

reactions take place:

Anode : 2 OH- = ½ O2 + H2O + 2e

Cathode : 2H2O + 2e = H2 + 2OH-


Calculation of Cell Voltage

H2O2e

H2 → 2H+ + 2e ½ O2 + 2H+ + 2e → H2O

Anode Cathode

Consider the fuel cell of reaction between hydrogen

and oxygen to produce water as a system.

If the apparatus is considered to be isothermal and at

steady state, then Wrev = G

Suppose that the electrodes 1 and 2 at

different electrical potential levels, 1 and

2, then:

dW1-2 = -(2 - 1) dQ (Volt * Coulomb)

(1 Ampere = 1 Coulomb /second)

W (Joule) = V I t (Watt.second)

https://en.wikipedia.org/



H2O2e

H2 → 2H+ + 2e ½ O2 + 2H+ + 2e → H2O

Anode Cathode

Suppose that the electrodes 1 and 2 at different electrical

potential levels, 1 and 2, then:

dW1-2 = -(2 - 1) dQ (Volt * Coulomb)

(1 Ampere = 1 Coulomb /second)

Integrating the equation above:

W1-2 = -(2 - 1) Q (Volt * Ampere * second)

(2 - 1) is called E, electromotive force (EMF) of the cell

(voltage)

Q is moles of electron flowed in the process

W (Joule) = V I t (Watt.second)



H2O2e

H2 → 2H+ + 2e ½ O2 + 2H+ + 2e → H2O

Anode Cathode

Q = n F

F = NA e = 6.022 x 1023 (electrons / mole) x 1.6 x 10-19

(C/electron)

F = 96 480 (C per mole of electrons)

F is Faraday constant, electrical charge on one mole of

electrons.

The work done is: W = -E n F (Volt * mole * Ampere *

second / mole)

Q is moles of electron

flowed in the process

1 mole of electrons is a quantity equal to Avogadro’s number (6.022 x 1023)

Q = ne e

Q = n NA e

Where ne is number of electrons, n is mole of electron, and NA is Avogadro’s

number.

Wrev = G G = - E n F

Italian scientist

(1776–1856)

en.wikipedia.org

W1-2 = -(2 - 1) Q



H2O2e

H2 → 2H+ + 2e ½ O2 + 2H+ + 2e → H2O

Anode Cathode

If each of the reactants and products in an electrochemical cell is in its standard state:

Go = - Eo n F

Example: at 298 K, Gibbs free energy of formation of water

from hydrogen and oxygen, each in its standard state, is

–237 191 J/mol. Number of electrons involves is 2. Calculate

standard potential of hydrogen – oxygen fuel cell!

Go = - Eo n F

-237 191 = -Eo x 2 x 96480

Eo = +1.229 V

Wrev = G G = - E n F

At hydrogen pressure and oxygen pressure are

each equal to one atmosphere, circuit voltage will be

1.229 V


Direction of Reaction

H2O2e

H2 → 2H+ + 2e ½ O2 + 2H+ + 2e → H2O

Anode Cathode

If electrohemical potential of the cell is greater than zero (positive), then the change in Gibbs free energy

(G) is negative, the reaction proceeds spontaneously.

Anode : H2 = 2H+ + 2e

Cathode : ½ O2 + 2H+ + 2e = H2O In this mode, the cell operate as a generator or electrical

current.

The reaction can be pushed in the opposite direction by

applying a voltage in opposite direction greater than open

circuit voltage of the cell.


Reference point: Standard Hydrogen Electrode.

Hydrogen gas and hydrogen ions are in their standard states

(pH2 = 1 atm, [H+]=1M, one mole per liter)

Half Cell Reaction

Standard Potential (voltage) of a cell can be calculated from the

standard Gibbs free energy change of a reaction as follows:

Go = - Eo n F

In the table, reactions at anode and at cathode are written

separately (half cell reactions)

Standard half cell potention for the reduction of oxygen to form water

at 298 K is 1.229 V

O2 + 4H+ + 4e = 2H2O Eo = 1.229 V

Standard potential for hydrogen oxidation is zero (reference)

H2 = 2H+ + 2e Eo = 0 V


Potensial Reference (Conversion)

-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0

V (SHE)

Hg/H2SO4,K2SO4 (sat)

+0.64

Cu/CuSO4

+0.316

SCE

+0.241

Ag/AgCl, KCl(sat)

+0.197

Zn/seawater

-0.8

https://www.castle-electrodes.com/ https://www.fishersci.co.uk/

https://intl.hannainst.com/

https://products.corrosionservice.com/

https://www.buch-holm.com/


Nernst Equation: Variation of Voltage

with Concentration

Consider following chemical equation:

bB + cC = dD +eE

Gibbs free energy change for this reaction can be written as follows:

K ln RT G a a

a a ln RT G ΔG

c

C

b

B

e

E

d

D +=+=

F n E ΔG −=

K ln F n

T RE

a a

a a ln

F n

T RE E

c

C

b

B

e

E

d

D −=−=

At T = 298 K: K log n

05916.0E E −=

en.wikipedia.org


Pourbaix Diagram (E-pH)

services.jacobi.net

Marcel Pourbaix was a Belgian chemist and pianist.

He performed his most well known research at

the University of Brussels, studying corrosion. His

biggest achievement is the derivation of potential-pH,

better known as “Pourbaix Diagrams”

en.wikipedia.org

Pourbaix diagrams display the stability of

compounds in electrical and chemical terms at a

given temperature.

Ordinate : electrical potential

Abscissa : pH

Standard state for ions in aqueous solutions is at

a concentration of one mole per liter (1M).

Assuming linearity of activity with concentration,

the activity of H+ is equal to its concentration in

moles per liter:

pH = -log (aH+) = - log [H+]

NOTE:

In Pourbaix diagram, the reaction is REDUCTION (not OXIDATION)

https://en.wikipedia.org/wiki/Pianist

https://en.wikipedia.org/wiki/Free_University_of_Brussels_(1834%E2%80%931969)

https://en.wikipedia.org/wiki/Corrosion

https://en.wikipedia.org/wiki/Pourbaix_diagram

https://en.wikipedia.org/wiki/Pourbaix_diagram


O2

Pourbaix Diagram: O2, H2O, H2

In a cell containing pure water at 298K, reaction at oxygen

electrode:

O2 + 4H+ + 4e = 2H2O Eo = 1.229 V

2

2

O

4

H

2

OH

p a

a log

4

05916.0-1.229K log

n

05916.0E E

+

=−=

Activity of water ~ 1, then:

( )2O

4

Hp a log

4

05916.01.229 E ++=

At a pressure of oxygen = 1 atm:

pH 05916.01.229 E −=

At pH = 0, E = 1.229 V

• Water is stable at E<1.229 V

• O2 is evolved at E > 1.229 V

E = 1.229 − 0.05916 pH

→ pH

→ E

(V

olt

)

H2O

H2

a



In a cell containing pure water at 298K, reaction at hydrogen

electrode:

2H+ + 2e = H2 Eo = 0 V

2

H

H

a

p log

2

05916.0-0K log

n

05916.0E E 2

+

=−=

At a pressure of hydrogen = 1 atm:

pH 05916.0 E −=

At pH = 0, E = 0 V

• Water is stable at E > 0 V

• H2 is evolved at E < 0 V

→ pH

→ E

(V

olt

)

O2

H2O

H2

a

b



→ pH

→ E

(V

olt

)

O2

H2O

H2

a

b


→ pH

→ E

(V

olt

)

O2

H2O

H2

a

b

Evolution of Hydrogen

and Oxygen

anodecathode

H2O

Hydrogen is

evolved if

electrode

interface

potential vs

SHE is

lower than

line a on

pourbaix

diagram

Oxygen is evolved if electrode interface

potential vs SHE is more than line a on

pourbaix diagram

Electrodes are inert (Pt)

DC Power

Supply

+-

e e

H+H2

H+


Pourbaix Diagram for Aluminum

The species participating in the various chemical and electrochemical equilibria are solids Al and Al2O3, and ions

Al3+ and AlO2-

Standard Gibbs free energies of formation:

Reaction Go298K, J

Al 0

½ H2 = H+ + e 0

2 Al + 3/2 O2 = Al2O3 - 1 608 900

Al = Al3+ + 3e - 481 200

Al + O2 + e = AlO2- - 839 800

H2 + ½ O2 = H2O - 237 200

Al3+



1). Equilibrium between Al3+ and AlO2-

Procedure for deriving the expression for the equilibrium is as

follows:

a. Balance the oxygen with H2O

Al3+ + 2H2O AlO2-

b. Balance the hydrogen with H+

Al3+ + 2H2O AlO2- + 4H+

c. Balance the charge with electron, (if necessary)

in this case, this step is not necessary (equilibrium is not

electrochemical)

Al3+ + 2H2O AlO2- + 4H+

Go = (-839 800) – (2x -237 200) – (-481 200)

= 115 800 J

][Al

][AlO][H log 2.303 x 298 x 8.3144 800 115

3

2

4

+

−+

−=

][Al

][AlO log ][H log 4 20.29-

3

2

+

−

+ +=

Hp 4 20.29 ][Al

][AlO log

3

2 +−=+

−

At pH = 5.07, [Al3+] = [AlO2-]

At pH > 5.07 then [AlO2-] > [Al3+]

At pH < 5.07, [Al3+] > [AlO2-]



1). Equilibrium between Al3+ and AlO2-

][Al

][AlO][H log 2.303 x 298 x 8.3144 800 115

3

2

4

+

−+

−=

][Al

][AlO log ][H log 4 20.29-

3

2

+

−

+ +=

Hp 4 20.29 ][Al

][AlO log

3

2 +−=+

−

At pH = 5.07, [Al3+] = [AlO2-]

At pH > 5.07 then [AlO2-] > [Al3+]

At pH < 5.07, [Al3+] > [AlO2-]

1



2). Equilibrium between Al and Al2O3

Procedure for deriving the expression for the

equilibrium is as follows:

1. Balance the oxygen with H2O

Al2O3 2Al + 3H2O

2. Balance the hydrogen with H+

Al2O3 + 6H+ 2Al + 3H2O

3. Balance the charge with electron, (if necessary)

Al2O3 + 6H+ + 6e 2Al + 3H2O

Go = (3 x -237 200) – (- 1 608 900)

= 897 300 J = - 6 x 96 480 Eo

Eo = -1.55 V

6

][HOAl

2

AlOHo

a a

a a ln

F n

T R - E E

32

2

+

=

Hp 0591.055.1E −−=

With aAl = aAl2O3 = aH2O =

1, then:

1

2



1

2

3). Al3+ + 3e Al

Eo = -481 200 / (3 x 96 480)

= -1.66 V

][Al log 10 x 971.166.1E 3-2 ++−=

4). Al2O3 + 6H+ 2Al3+ +3H2O

pH 3 - 5.70 ][Al log 3 =+

5). AlO2- + 4H+ + 3e Al + 2H2O

][AlO log 0.0198 pH 0.0789 - 1.26- E 2

−+=

14.59 - pH ][AlO log 2 =−

6). H2O + Al2O3 2AlO2- + 2H+

3

4

5

6


Eh - pH


Chromium

M. Pourbaix, 1974


Battery


Eh - pH


Daniell Cell

Zn2+ Cu2+

V

Zn Cu

Half cell reaction:

Zn → Zn2+ + 2e Eo = + 0.763 V

Cu → Cu2+ + 2e Eo = - 0.337 V

Cell reaction:

A: Zn → Zn2+ + 2e Eo = + 0.763 V

C: Cu2+ + 2e → Cu Eo = + 0.337 V

+

Zn + Cu2+ → Cu + Zn2+ Eo = + 1.10 V

ZnSO4 CuSO4

Zn

Cu

eq

Zinc

electrolite

interface

Copper electrolite

interface

Potential,

Cu - Zn

en.wikipedia.org


Concentration Cell

Cu2+ Cu2+

V

Cu Cu

1 2

Cell 1: Cu → Cu2+ [I] + 2e

Cell 2: Cu2+ [II] + 2e → Cu

+

Cu2+ [II] → Cu2+ [I]

[II]Cu

[I]Cu

2

2

a

a ln

F z

T R- E

+

+

=

If activity of copper ions (related to concentration) on one side is known,

then a measurement of voltage would determine the copper ion activity

(concentration) on the other side.

Can be used to determine activity of copper in an alloy copper.


Determine of Activity

The activity of a component A in an alloy can be determined by measuring the reversible emf of the following cell:

Cell reaction can be represented as follows:

A (pure metal) = A (in alloy)

a

a ln TR G G

(pure)A

(alloy)A o

+=

Pure Metal Aelectrolyte containing ions

of metal A (valency nA)Metal A in an alloy

a

a ln

nF

TR - E -E

(pure)A

(alloy)A o

=

In case of both electrodes are pure A, then Eo = 0

a ln nF

TR - E (alloy)A =

The Gibbs free energy change of the reaction:


Problems:

1. A thermodynamic study of molten lead-bismuth alloys was carried out by the electromotive force (emf) method

using the reversible concentration cell:

Pb(l) / PbCl2 – LiCl(l) / Pb-Bi(l)Find the activity and activity coefficient of lead in the Pb-Bi melt (lead mole fraction is 0.60) at 900 C using the

experimental emf value E=0.0193 V, R = 9.314 J.K-1 mol-1, F = 96480 C

2. In the commercial production of aluminum, alumina is decomposed by electrolysis in the presence of carbon,

according to the reaction:

½ Al2O3 (s) + ¾ C (s) = Al (l) + ¾ CO2 (g) G = 132,230 + 1.88 T log T – 46.26 T calorie

Assuming that the condensed phases are pure and CO2 is formed at 1 atm pressure, determine the theoretical

minimum voltage required to effect the decomposition if the cell is operated at 950 C.


Temperature Dependence of Voltage

The temperature dependence of the cell voltage can be determined by

knowing the temperature dependence of the Gibbs free energy function:

F n E - G =( ) dE F n - dT S - Gd ==

Temperature coefficient of a cell dE/dT is

F n

S

dT

dE =

F n

S

dT

dE oo =

S is the entropy change for the overall

chemical reaction of the cell

The heat effect or enthalpy change of the electrochemical reaction can be calculated by knowing the cell

voltage and the temperature coefficient of cell voltage as follows:

S T - H G = dT

dE F n T - H F n E =−

dT

dE T - E F n - H

=

dT

dE T - E F n - H

ooo

=

At P constant:

dU = T dS - P dV

dH = T dS + V dP

dF = -S dT - P dV

dG = -S dT + V dP


Problems:

1. A galvanic cell is set up with electrodes of solid aluminium and solid aluminium-zinc

alloy, and an electrolyte of a fused AlCl3-NaCl mixture. When the mole fraction of

aluminium in the alloy electrode is 0.38, the emf of the cell is 7.43 milivolt at 380oC and

the temperature coefficient of the emf is 2.9 x 10-5 volt/deg. Calculate the activity, and

the partial molar free energy and enthalpy of mixing of aluminium in the alloy at 380oC.


Problems:

2. Emf between pure magnesium and magnesium-zinc alloy containing 63.5 atom%

magnesium in a fused KCl-LiCl-MgCl2 electrolyte may be represented by:

E = 16.08 x 10-3 + 1.02 x 10-5 T

Where E and T in Volt and Kelvin respectively. Calculate the activity coefficient and

excess partial molar free energy of mixing of magnesium in the above alloy at 727oC.


Oxygen Pressure Determination

Principle of concentration cell can be used to measure differences in

thermodynamic activity (pressure) of gases. An example is to measure

oxygen activity of oxygen pressure.

Electrolyte for this cell could be liquid, but in practice solid electrolytes are

used. (Electrolyte is an ionic conductor, not an electonic conductor)

Ceramics solid solutions of zirconia and yitria (Zr2O3 and Y2O3) are ion

conductors at high temperatures and do not conduct electrons.

Voltage of the cell:

ref,O

O

2

2

p

p ln

F z

T R- E =

If one electrode is as reference pressure, e.g. oxygen pressure is 0.21 atm, then

oxygen pressure in on the other side can be determined by measuring the voltage.

This technique is used to measure oxygen pressure (activity) in exhaust of internal

combustion engines to control air fuel mixture.Measure of oxygen activity in molten

metal baths

Probe to measure oxygen

activity in metal baths


Determination of Oxygen Content


Terima kasih!Zulfiadi Zulhan

Program Studi Teknik Metalurgi

Fakultas Teknik Pertambangan dan Perminyakan

Institut Teknologi Bandung

Jl. Ganesa No. 10

Bandung 40132

INDONESIA

www.metallurgy.itb.ac.id

[email protected]

07. proses elektrokimia dan diagram potensial - ph

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