Lecture 1 (Electrochemistry & corrosion engineering).pptx

Electrochemistry

1. Applied Electrochemistry & Corrosion Engineering Dr. Marwa Abdeen Lecture 1

2. Course Title Applied Electrochemistry & Corrosion Engineering Course Code CHE 211 Semester Second term Academic Year 2024-2025 Course Hours Lec Tut Lab   2 --- 3   Grades Distribution Quizzes Mid-term Final Experimental Total   20 25 75 30 150   Course Staff Dr. Marwa Maher Abdeen

3. Course Objectives Understand the basic principles of electrochemical reactions, electrochemical cells, and electrode kinetics. Learn about thermodynamics and kinetics of electrochemical processes. Explore industrial applications of electrochemistry, including batteries, fuel cells, and electroplating. Analyze different types of corrosion, including uniform, galvanic, pitting, crevice, and intergranular corrosion. Understand the electrochemical basis of corrosion and how environmental factors influence corrosion rates. Learn about various corrosion protection methods, including cathodic and anodic protection, coatings, inhibitors, and material selection.

4. Reference Textbook:  P. W. Atkins, Atkins Physical Chemistry, Oxford University Press, 9th.Ed., 2006.  Fontana, M. G.& Greene, N. D., Corrosion Engineering, McGraw - Hill Int. Book Co, 1986. References:  Theodore L. Brown, et al, Chemistry the Central Science, Prentice Hall Int. ( PearsonInternational , 2010.  Zaki Ahmad, Principles of Corrosion Engineering and Corrosion Control, Butterworth -Heinemann, 1 st. Ed., 2006.

5. Contents   Chapter 1. Introduction Chapter 2. Redox Reactions   Chapter 3. Electro chemistry   Chapter 4. Batteries & Fuel cells   Chapter 5. Corrosion (types of corrosion)   Chapter 6. Corrosion Control  

6. Introduction

7. Oxidation-Reduction Oxidation: When a substances loses electrons. Reduction: When a substance gains electrons. Consider: Ca (s) + 2H + (aq)  Ca 2+ (aq) + H 2(g) . The neutral Ca (s) has lost two e - to 2 H + to become Ca 2+ . We say Ca has been oxidised to Ca 2+ At the same time 2 electrons are gained by 2 H + to form H 2 . We say H + is reduced to H 2 .

8. Redox Reaction with Air Consider the reaction of Ca with O 2 : 2Ca (s) O 2(g)  2CaO (s) Ca is easily oxidised in air. On the left we see freshly cut Ca. On the right we see Ca with a dull surface. On the surface, there is a coat of CaO. Again, Ca(s) gains electron and is oxidised to Ca +2 And the neutral O 2 has gained electrons from the Ca to become O 2- in CaO. We say O 2 has been reduced to O 2- .

9. In all reduction-oxidation (redox) reactions, one species is reduced at the same time as another is oxidised . Oxidizing Agent: The species which causes oxidation is called the oxidizing agent. The substance which is oxidised loses electrons to the other. The oxidizing agent is always reduced Reducing Agent: The species which causes reduction is called the reducing agent. The substance which is reduces gains electrons from the other. The Reducing agent is always oxidised

10. Oxidation of Metals with Acids It is common for metal to produce hydrogen gas when they react with acids. For example, the reaction between Mg and HCl: Mg (s) + 2HCI ( aq )  MgCl 2( aq ) + H 2(g) . Mg is oxidised and H in HCl is reduced. Note the change in oxidation state for these specie: Mg  Mg +2 in MgCl 2 & H + in HCl  H in H 2

11. It is possible for metals to be oxidised with salt: Fe (s) + Ni(N0 3 ) 2 ( aq )  Fe(N0 3 ) 2 ( aq ) + Ni (s) . Molecular Equation The overall ionic equation shows the redox chemistry : Fe (s) + Ni +2 ( aq )  Fe 2+ ( aq ) + Ni (s) Net ionic Equation In this reaction iron has been oxidised to Fe 2+ while the Ni +2 has been reduced to Ni . What determines whether the reaction occurs ?

12. The Activity Series Metals can be placed in order of their tendencies for losing electrons. This is called the activity series.

13. Competition For e - Transfer Consider: Na, Mg, Al, Metallic character decreases left to right. Metal tend to give up electrons. Now consider the reaction: Na + AlCl 3  ??? (NaCl + Al) To determine if the reaction occurs, the question is to determine which metal has a greater affinity for electrons (or which is willing to lose e - ). Na is more willing to lose e- than Al Al is more willing to accept e- (less metallic) Conclude: The reaction occurs. 3Na + AlCl 3  3NaCl + Al

14. Reading Activity Table A metal in the activity series can only be oxidised by a metal ion below it. In our example, Na is oxidised by Al. The metals at the top of the activity series are called active metals . The metals at the bottom of the activity series are called noble metals .

15. Example: Silver and Copper If we place Cu into a solution of Ag+ ions, will copper plate out of solution ? Cu (s) + 2AgNO 3( aq )  ? [Cu(NO 3 ) 2( aq ) + 2Ag (s) ] or Cu (s) + 2Ag+ ( aq )  ? [Cu 2+ ( aq ) + 2Ag (s) ] Which metal is active? Which is noble ? g Cu g Ag Therefore, Cu 2+ ions is be formed because Cu is above Ag in the activity series. Copper Cu g Cu 2+ + 2 e - Silver Ag g Ag + + e -

16. Example: Redox Reaction Based on the activity series, what is the outcome of the following reaction ? b) Ag (s) + PbNO 3 ( aq )  ? c) Cr (s) + NiSO 4 ( aq )  ? e) H 2 (g) + CuCl 2 ( aq )  ? f) Ba (s) + H 2 O (l)  ? b) Ag vs. Pb , Pb is more active, reaction not occurs c) Cr vs. Ni , Cr is more active, reaction occurs Cr (s) + NiSO 4 ( aq )  Ni (s) + CrSO 4 ( aq ) d) H 2 vs. Cu , H 2 is more active, reaction occurs H 2 (g) + CuCl 2 ( aq )  2HCl ( aq ) + Cu (s) e) Ba vs. H 2 , Ba is more active, reaction occurs Ba (s) + H 2 O (l)  H 2 (g) + Ba(OH) 2 ( aq )

17. Rules for determining oxidation numbers 1) The oxidation number of an uncombined atom is zero. This is true for elements that exist as polyatomic molecules such as O 2 , Cl 2 , H 2 , N 2 , S 8 . 2) The oxidation number of a monatomic ion is equal to the charge on the ion. For example, the oxidation number of a Ca2+ ion is +2, and the oxidation number of a Br— ion is —1. 3) The oxidation number of the more electronegative atom in a molecule or a complex ion is the same as the charge it would have if it were an ion. In ammonia (NH 3 ). (SiCl 4 ), 4) The most electronegative element, fluorine, always has an oxidation number of —1 when it is bonded to another element. 5) The oxidation number of oxygen in compounds is always —2, except in peroxides, such as hydrogen peroxide (H 2 O 2 ), where it is —1. When it is bonded to fluorine. 6) The oxidation number of hydrogen in most of its compounds is +1. 7) The metals of groups 1A and 2A and aluminum in group 3A form compounds in which the metal atom always has a positive oxidation number equal to the number of its valence electrons (+1, +2, and +3, respectively). 8) The sum of the oxidation numbers in a neutral compound is zero N=-3, Si=+4

18. The equation is separated into two half-equations, one for oxidation, and one for reduction. The equation is balanced by adjusting coefficients and adding H 2 O, H + , and e -  in this order: Balance the atoms in the equation, apart from O and H. To balance the Oxygen atoms, add the appropriate number of water (H 2 O) molecules to the other side. To balance the Hydrogen atoms (including those added in step 2), add H +  ions. Add up the charges on each side. They must be made equal by adding enough electrons (e - ) to the more positive side. The e -  on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers to be made the same. The half-equations are added together, cancelling out the electrons to form one balanced equation. Cancel out as much as possible. (If the equation is being balanced in a basic solution, the appropriate number of OH -  must be added to turn the remaining H +  into water molecules) The equation can now be checked to make sure it is balanced. Balancing Redox Equations by Half- Reactions

19. Balance this equations