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06/11/22 1 CHAPTER 2 CHAPTER 2 CORROSION PRINCIPLES CORROSION PRINCIPLES Chapter Outlines Chapter Outlines 2.1 Oxidation and Reduction Reactions 2.2 Standard Electrode Half- Cell Potent 2.3 Standard EMF Series 2.4 Galvanic Cells With 1 Molar Electrol 2.5 Galvanic Cells Not 1Molar Electrolyt

CHAPTER 2 CORROSION PRINCIPLES

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CHAPTER 2 CORROSION PRINCIPLES. Chapter Outlines 2.1 Oxidation and Reduction Reactions 2.2 Standard Electrode Half- Cell Potentials 2.3 Standard EMF Series 2.4 Galvanic Cells With 1 Molar Electrolytes 2.5 Galvanic Cells Not 1Molar Electrolytes. 2.1 Oxidation and Reduction Reactions. - PowerPoint PPT Presentation

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Page 1: CHAPTER 2 CORROSION PRINCIPLES

04/20/23 1

CHAPTER 2CHAPTER 2

CORROSION PRINCIPLESCORROSION PRINCIPLES

Chapter OutlinesChapter Outlines

2.1 Oxidation and Reduction Reactions 2.2 Standard Electrode Half- Cell Potentials2.3 Standard EMF Series2.4 Galvanic Cells With 1 Molar Electrolytes2.5 Galvanic Cells Not 1Molar Electrolytes

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In metal, corrosion process is normally electrochemical @ electrochemistry

(a chemical reaction in which there is transfer of electrons from one chemical species to another)

2 reactions that occur during corrosion process:

i. Oxidation reaction

ii. Reduction reaction

2.1 2.1 Oxidation and Reduction ReactionsOxidation and Reduction Reactions

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i. Oxidation reaction @ anodic reaction

Definition:the removal of one or more electrons from an atom, ion or molecule

Equation:

(in which M becomes an n+ positively charged ion and in the process loses its n valence electrons; e- is used to symbolize an electron)

Example:

Anode is the side at which oxidation takes place.

M Mn+ + ne-

Zn Zn2 2e

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ii. Reduction reaction

Definition:

the addition of one or more electrons to an atom, ion or molecule (because the electrons generated from each metal atom that is oxidized must be transferred to and become a part of another chemical species = reduction reaction)

Equation:

(some metals undergo corrosion in acid solutions, which have a high concentration of hydrogen (H+) and hydrogen gas (H2) is evolved)

Cathode is the side at which reduction occurs

M+ + e- M(n-1)+

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There are 3 possibilities of reaction that can occur at cathode (reduction):

First possibilities

Cathodic half- cell reaction : Condition : if the electrolyte is an acid

solutionReaction : hydrogen ions in the acid solution

will be reduced to hydrogen atom to form diatomic hydrogen gas

Second possibilities

Cathodic half- cell reaction :Condition : if the electrolyte also contain

oxidizing agentReaction : oxygen will combine with hydrogen

ions to form water molecules

O2 4H 4e 2H2O

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Third possibilities

Cathodic half- cell reaction :

Condition : if the electrolyte is basic or neutral and oxygen

is present

Reaction : oxygen and water molecules will react to form hydroxyl

ions

O2 2H2O 4e 4(OH)

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iii. Overall Electrochemical Reaction

Consist of at least one oxidation (half reaction) and one reduction (half reaction), and will be the sum of them

Example:

(Zinc metal immersed in an acid solution)

Zinc

oxidation reactionZn Zn2+

2e-Acid solution

reduction reaction

H+H+

H2(gas)

H+

H+

H+

H+

H+

flow of e- in the metal

At some regions on the metal surface, zinc will experience oxidation or corrosion

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Oxidation half reaction:

Since Zn is a metal and good electrical conductor, these electrons

may be transferred to an adjacent region at which the H+ ions are

reduced.

Reduction half reaction:

Total electrochemical reaction:

Zn + 2H+ Zn2+ + H2 (gas)

Zn Zn2 2e

2H 2e H2(gas)

Zn Zn2 2e

2H 2e H2(gas)

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04/20/23 Asyadi 9Fig. Reaction of hydrochloric acid with zinc to produce hydrogen gas

Chemical reaction:Zn + 2HCl ZnCl2 + H2

Ionic form:Zn + 2H+ Zn2+ + H2

Half- cell reaction:Zn Zn2+ + 2e- (oxidation)2H+ + 2e- H2 (reduction)

Zinc metal

hydrochloric acid

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Every metal has a different tendency to corrode in a particular environment

E.g. ‘zinc is chemically attacked or corroded by dilute hydrochloric acid, whereas gold is not’

Method for comparing the tendency for metals to form ions in aqueous solution is to compare their half- cell oxidation or reduction potentials (voltages) to a standard hydrogen- hydrogen ion half- cell potential.

2.2 2.2 Standard Electrode Half- Cell PotentialsStandard Electrode Half- Cell Potentials

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Experimental Setup for the Determination of Half- cell Standard Electrode Potentials

Experimental setup for the determination of the standard emf of zinc. In a beaker a Zn Electrode is placed in a solution of 1 M Zn2+ ions. In the other there is a

standard hydrogen reference electrode consisting of a platinum electrode immersed in a solution of 1 MH+ ions which contains H2 gas at 1 atm.

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Standard Hydrogen Electrode

Represent only differences in electrical potential and thus it is convenient to establish a reference point/ reference cell to which other cell halves may be compared.

It consist of an inert platinum electrode in a 1M solution of H+ ions, saturated with hydrogen gas that is bubbled through the solution at a pressure of 1 atm and temperature of 25°C.

The platinum itself does not take part in the electrochemical reaction: it acts only as a surface on which hydrogen atoms may be oxidized or hydrogen ions may be reduced.

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Electromotive force (EMF) series:

is generated by coupling to the standard hydrogen electrode, std half- cells for various metals and ranking them according to measured voltage.

Table 17.1- show the list of the standard half- cell potentials of some selected metals which represents the corrosion tendencies for the several metals

2.2 2.2 Standard EMF SeriesStandard EMF Series

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Increasingly inert(cathodic)

Increasingly active(anodic)

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those metals at the top (Au & Pt) --- are noble or chemically inert

Moving down --- metals become increasingly more active (more susceptible to oxidation) (sodium & potassium)

The voltages --- are for the half- reactions

oxidation reaction: electron on the right hand side

reduction reaction: electron on the left hand side (sign of the voltage changed)

M1 Mn+ + ne-

M+ + e- M(n-1)+

V1º

V2º

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Overall cell potential ΔV° is:

ΔVcell° = V° 1 + V° 2

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Galvanic coupleGalvanic couple: :

Two metals electrically connected in a liquid Two metals electrically connected in a liquid

electrolyte wherein one metal becomes electrolyte wherein one metal becomes anode and anode and

corrodes, while the other acts as a cathodecorrodes, while the other acts as a cathode

GALVANIC CELLSGALVANIC CELLS

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Can be constructed with two dissimilar metal electrodes each immersed in a 1M solution of their own ions

The two solutions are separated by a porous wall to prevent their mechanical mixing, and an external wire in series with a switch and a voltmeter connects the two electrodes

E.g.:

zinc electrode immersed in a 1 M solution of Zn2+ ions and another of copper immersed in a 1 M solution of Cu2+ ions with the solutions at 25°C

2.4 Galvanic Cells With 1 Molar ElectrolytesGalvanic Cells With 1 Molar Electrolytes

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A macroscopic galvanic cell with zinc and copper electrodes. When the switch is closed and the electrons flow, the voltage difference between the zinc and copper electrodes

is -1.10V. The zinc electrode is the anode of the cell and corrodes.

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Calculation of electrochemical potential of Zn- Cu galvanic cell

From the Standard emf Series: Zn Zn2+ + 2e- E° = -0.763 V

Cu Cu2+ + 2e- E° = +0.340 V

Oxidation half- cell reaction: (ANODE)

Zn Zn2+ + 2e- E° = -0.763 V°1

Reduction half- cell reaction: (CATHODE)

Cu2+ + 2e- Cu E° = -0.340 V°2

Overall reaction (by adding):

Zn + Cu2+ Zn2+ + Cu E°cell = V°1 + V°2

= -0.763 + (-0.340)

= -1.103 V

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Problem 1:

A galvanic cell consist of an electrode of zinc in a 1M ZnSO4

solution and another of nickel in a 1 M NiSO4 solution. The two

electrodes are separated by a porous wall so that mixing of the

solutions is prevented. An external wire with a switch connects the

two electrodes. When the switch is just closed:

(a) At which electrode does oxidation occur

(b) Which electrode is the anode of the cell?

(c) Which electrode corrodes?

(d) Write the equation for the half- cell reaction at the anode?

(e) Write the equation for the half- cell reaction at the cathode?

(f) What is the emf of this galvanic cell when the switch is just closed?

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Answer:

(a) Oxidation occurs at the zinc electrode since the zinc half- cell reaction has a more negative E° potential of -0.763 V as compared to -0.250 V for the nickel half- cell reaction.

(b) The zinc electrode is the anode since oxidation occurs at the anode

(c) The zinc electrode corrodes since the anode in a galvanic cell corrodes.

(d) Zn Zn2+ + 2e- E° = -0.763V(e) Ni2+ + 2e- Ni E° =+0.250V(f) The emf of the cell is obtained by adding the half- cell reactions

together:

Anode reaction: Zn Zn2+ + 2e- E° = -0.763 VCathode reaction: Ni2+ + 2e- Ni E° = +0.250 V

Overall reaction:Zn + Ni2+ Zn2+ + Ni E°cell = -0.513 V

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Most electrolytes for real corrosion galvanic cells are not 1 M, but are usually dilute solutions that are much lower than 1 M.

If the concentration of the ions in an electrolyte surrounding an anodic electrode is less than 1 M, the driving force for the reaction to dissolve or corrode the anode is greater since there is a lower concentration of ions to cause the reverse reaction

2.5 Galvanic Cells Not 1 Molar ElectrolytesGalvanic Cells Not 1 Molar Electrolytes

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Nernst equation:

E = E° + 0.0592 log Cion

n

Where: E = new emf of half- cell

E° = standard emf of half- cell

n = number of electrons transferred (for example, M Mn+ + ne-)

Cion = molar concentration of ions

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Problem 2:

A galvanic cell at 25ºC consist of an electrode of zinc in a 0.10 M

ZnSO4 solution and another of nickel in a 0.05 M NiSO4 solution.

The two electrodes are separated by a porous wall and connected

by an external wire. What is the emf of the cell when a switch

between the two electrodes is just closed?

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Answer:

Half cell reactions:

Zn Zn2+ + 2e- E° = -0.763V (ANODE)

Ni Ni2+ + 2e- E° = -0.250V (CATHODE)

Apply Nernst Equation:

Ecell = E° + 0.0592 log Cion

n

Anode reaction: EA = -0.763 V + 0.0592 log 0.10 = -0.793 V

2

Cathode reaction: Ec = - (- 0.250 V + 0.0592 log 0.05) = +0.289 V

2

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Emf of the cell (Ecell) = EA + EC

= -0.793V + 0.289 V

= -0.505 V