MODULE 1_UNIT 1_ENGGCHEM (1) PDF

ENGG...

ENGG CHEM Department of Chemical Engineering Prepared by: NLEscalante 1 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  ENGGCHEM COURSE LEARNING OUTCOMES At the end of the module, you should be able to: CLO 1: Express proficiency in General Inorganic Chemistry. CLO 2: Execute proper calculations involving General Inorganic Chemistry, Thermochemistry, Electrochemistry, Nuclear Reactions, and Chemistry of the Environment. CLO 3: Apply appropriate operating variables in Chemistry. CLO 4: Evaluate results of calculations made on different topics if Chemistry. CHEMISTRY FOR ENGINEERS (LEC) Prepared by: Engr. N. L. Escalante 0 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  COURSE INTRODUCTION This 3-unit course in the Chemical Engineering curriculum provides students with core concepts of chemistry that are important in the practice of engineering profession. It covers the Principles of General Inorganic Chemistry including Thermochemistry: energy flow and chemical change; Electrochemistry: chemical change and electrical work; Nuclear Reactions and their applications; Engineering Materials; The Chemistry of the Environment: water, atmosphere, and soil; and Chemical Safety. Each unit in this course is designed using the 5E constructivist model of learning, developed by Rodger Bybee, that encourages students to engage, explore, explain, elaborate, and evaluate their knowledge of topics covered therein. It means that at the end of each unit, each module, and the course as a whole, you will be assessed on your progress in attaining the course learning outcomes. Outcomes-based education dictates that only when you can demonstrate the course learning outcomes by the end of this course, can you be given a passing mark. Prepared by: Engr. N. L. Escalante 1 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  MODULE 1 ENERGY At the end of the module, you should be able to: TLO 1: Express high competence in the understanding of Electrochemical Energy TLO 2: Demonstrate appropriate concepts of Nuclear Chemistry and Energy. TLO 3: Demonstrate appropriate calculations in stoichiometry particularly in the combustion of fuels. Introduction What is Energy? Energy is one of the most fundamental and universal concepts of physical science, but one that is remarkably difficult to define in a way that is meaningful to most people. This perhaps reflects the fact that energy is not a “thing” that exists by itself, but is rather an attribute of matter (and also of electromagnetic radiation) that can manifest itself in different ways. It can be observed and measured only indirectly through its effects on matter that acquires, loses, or possesses it. The concept that we call energy was very slow to develop; it took more than a hundred years just to get people to agree on the definitions of many of the terms we use to describe energy and the interconversion between its various forms. But even now, most people have some difficulty in explaining what it is; somehow, the definition we all learned in elementary science ("the capacity to do work") seems less than adequate to convey its meaning. Although the term "energy" was not used in science prior to 1802, it had long been suggested that certain properties related to the motions of objects exhibit an endurance which is incorporated into the modern concept of "conservation of energy". In the 17th Century, the great mathematician Gottfried Leibnitz (1646-1716) suggested the distinction between vis viva ("live force") and vis mortua ("dead force"), which later became known as kinetic energy (1829) and potential energy (1853). “Chemical Energy” Electrostatic potential energy plays a major role in chemistry; the potential energies of electrons in the force field created by atomic nuclei lie at the heart of the chemical behavior of atoms and molecules. "Chemical energy" usually refers to the energy that is stored in the chemical bonds of molecules. These bonds form when electrons are able to respond to the force fields created by two or more atomic nuclei, so they can be regarded as manifestations of electrostatic potential energy. In an exothermic chemical reaction, the electrons and nuclei within the reactants undergo rearrangement into products possessing lower energies, and the difference is released to the environment in the form of heat. Prepared by: Engr. N. L. Escalante 1 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Unit 1 Electrochemical Energy UNIT LEARNING OUTCOMES TLO 1: Express high competence in the understanding of Electrochemical Energy ENGAGE All chemical reactions are fundamentally electrical in nature since electrons are involved (in various ways) in all types of chemical bonding. Electrochemistry is a branch of chemistry that deals with the relationship between electricity and chemical reactions. The relations between chemical change end electrical energy have theoretical as well as practical importance. Chemical reactions can be used to produce electrical energy (i.e., in cells are called voltaic or galvanic cells). Electrical energy can be used to bring about chemical transformations (i.e., in electrolytic cells). The study of electrochemical processes leads to an understanding as well as to systematization of oxidation-reduction phenomena that take place outside cells.  To have a quick overview of what we’re going to tackle on this module, let us watch the video entitled “ Electrochemistry: Crash Course Chemistry #36 ” using the link https://www.youtube.com/watch?v=IV4IUsholjg. EXPLORE Electrochemistry ELECTRICITY ↔ CHEMISTRY The study of electricity and how it relates to chemical reactions ▪ Certain chemical reactions can produce electricity. ▪ Electricity can be used to make certain chemical reactions. Electricity - the flow or movement of electrons in a conductor. Chemical Reactions: Oxidation and Reduction Reactions - electrons move between atoms ▪ Certain chemical reactions can produce electricity. (SPONTANEOUS REACTION) ▪ Electricity can be used to make certain chemical reactions. (REACTION DO NOT NORMALLY HAPPEN) Oxidation-Reduction Terminology and Concepts ▪ Oxidation occurs when a species loses electrons and increases its oxidation number (ON ↑). An equation that shows this loss is called an oxidation half-reaction. In this half-reaction, the electrons are in the product side. ▪ Reduction occurs when a species gains electrons and decreases its oxidation number (ON ↓). An equation that shows this gain is called a reduction half-reaction. In this half-reaction, the electrons are in the reactant side. Prepared by: Engr. N. L. Escalante 2 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  ▪ Oxidation and reduction occur together in the same reaction. There is no net change in the number of electrons, just an exchange of electrons. ▪ Oxidation Number – set of numbers obtained by arbitrary rules which are related to the combining ratios of elements. ▪ Ion – an atom or group of atoms carrying a + or a – electric charge. REDOX (Reduction-Oxidation) Reactions  OXIDATION IS LOSING of e- REDUCING AGENT (OILRA)  REDUCTION IS GAINING of e- OXIDIZING AGENT (RIGOA) Remember: electrons have NEGATIVE CHARGE  Need to know the oxidation numbers of each atom in a chemical reaction. How to determine the oxidation state? 1. The oxidation state of the uncombined element is zero. Examples: Be0, N20, Cs0 2. The sum of the oxidation state of a neutral compound is always equal to zero. Examples: CaO Ca+2 O-2 1(+2) + 1(-2) = 0 MgBr2 Mg+2 Br-1 1(+2) + 2(-1) = 0 Li2CO3 Li+1 (CO3)-2 2(+1) + 1(-2) = 0  The Group A elements number indicates it oxidation state Examples: Elements Group I II III VI VII Oxidation State +1 +2 +3 -2 -1 3. The oxidation number of polyatomic ion is equal to its charge. Oxygen is -2 except in peroxides. Examples: C in CO3-2 oxidation of C + oxidation of O = -2 C + 3(-2) = -2 C + (-6) = -2 C = -2 + 6 C = +4 Cr in KCrO4-2 Oxidation of K + oxidation of Cr + oxidation of O = -2 1(+1) + Cr + 4(-2) = -2 1 + Cr + (-8) = -2 Cr = +5 Prepared by: Engr. N. L. Escalante 3 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  4. Hydrogen + non-metal, the oxidation state of hydrogen is +1. Examples: HI H+1 I-1 H2SO4 H +1 (SO4)-2 5. Metal + hydrogen, the oxidation state of hydrogen is -1. Example: LiH Li+1 H-1  To better understand the determination of the oxidation state of an element, you may want to watch the video entitled “ Assigning Oxidation Numbers - Chemistry Tutorial ” using the link https://www.youtube.com/watch?v=w0RfMRDy34w. Activity 1: Determination of Oxidation State Self-Assessment No. 1 Determine the oxidation state of the following: 1. O2 2. Cr in KCr2O7 3. S in MgSO4 4. P in Zn3PO4 5. Group VIA elements To be submitted in Google classroom on: Determining the oxidizing agent and the reducing agent. ▪ Oxidizing Agent (OA) - or oxidant, gains electrons and is reduced in a chemical reaction. Also known as the electron acceptor, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced. Examples of oxidizing agents include halogens, potassium nitrate, and nitric acid. ▪ Reducing Agent (RA) - or reductant, loses electrons and is oxidized in a chemical reaction. A reducing agent is typically in one of its lower possible oxidation states, and is known as the electron donor. A reducing agent is oxidized, because it loses electrons in the redox reaction. Examples of reducing agents include the earth metals, formic acid, and sulfite compounds. Examples: 2 Mg (s) + O2 (g) → 2 MgO (s) Mg0 O20 Mg+2 O-2 Mg 0 → +2 Loss 2 e- (Reducing Agent; Oxidized) O2 0 → -2 Gains 2 e- (Oxidizing Agent; Reduced) Zn + 2 H+ → Zn+2 + H2 Zn0 H+ Zn+2 H20 Prepared by: Engr. N. L. Escalante 4 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Zn 0 → +2 Loss 2 e- (Reducing Agent; Oxidized) H2 +1 → 0 Gains 1 e- (Oxidizing Agent; Reduced) Activity 2: Identifying Oxidizing and Reducing Agents Self-Assessment No. 2 Identify the oxidizing agent and reducing agent in each of the following: 1. 2 H2 (g) + O2 (g) → 2 H2O (g) 2. Cu (s) + 4 HNO3 (aq) → Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O (l) To be submitted in Google classroom on: Balancing REDOX Reactions When balancing redox reactions, make sure that the number of electrons lost by the reducing agent equals the number of electrons gained by the oxidizing agent. Two methods can be used: ▪ Oxidation Number Method ▪ Half-reaction Method Method 1: Oxidation Number Method 1. Assign oxidation numbers to all elements in the reaction. 2. From the changes in oxidation number, identify the oxidized and reduced species. 3. Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes. 4. Multiply one or both these numbers by appropriate factors to make the electrons lost equal the electrons gained, use the factors as balancing coefficients. 5. Complete the balancing by inspection, adding states of matter. Examples: Method A: Al (s) + H2SO4 (aq) → Al2(SO4)3 (aq) + H2 (g) Step 1: Assign oxidation numbers to all elements 0 +1 +6 -2 +3 +6 -2 0 Al (s) + H2SO4 (aq) → Al2(SO4)3 (aq) + H2 (g) Step 2: Identify oxidized and reduced species Al was oxidized (oxidation number of Al: 0 → +3) H2SO4 was reduced (oxidation number of H: +1 → 0) Step 3: Compute e- lost and e- gained In the oxidation: 3 e- were lost from Al In the reduction: 1 e- was gained by H Prepared by: Engr. N. L. Escalante 5 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Step 4: Multiply by factors to make e - lost equal to e- gained, and use the factors as coefficients Al lost 3 e-, so the 1 e- gained by H should be multiplied by 3. Put the coefficient 3 before H2SO4 and H2. Al (s) + 3 H2SO4 (aq) → Al2(SO4)3 (aq) + 3 H2 (g) Step 5: Complete the balancing by inspection. 2 Al (s) + 3 H2SO4 (aq) → Al2(SO4)3 (aq) + 3 H2 (g) Method B: PbS (s) + O2 (g) → PbO (s) + SO2 (g) Step 1: Assign oxidation numbers to all elements +2 -2 0 +2 -2 +4 -2 PbS (s) + O2 (g) → PbO (s) + SO2 (g) Step 2: Identify oxidized and reduced species PbS was oxidized (oxidation number of S: -2 → +4) O2 was reduced (oxidation number of O: 0 → -2) Step 3: Compute e- lost and e- gained In the oxidation: 6 e- were lost from S In the reduction: 2 e- were gained by each O Step 4: Multiply by factors to make e - lost equal to e- gained, and use the factors as coefficients S lost 6 e-, O gained 4 e- (2 e- each O). Thus, put the coefficient 3/2 before O2. PbS (s) + 3/2 O2 (g) → PbO (s) + SO2 (g) Step 5: Complete the balancing by inspection 2 PbS (s) + 3/2 O2 (g) → 2 PbO (s) + 2 SO2 (g) Method 2: Half-reaction Method (ACIDIC) 1. Divide the skeleton reaction into two half-reactions, each of which contains the oxidized and reduced forms of one of each of the species. 2. Balance the atoms and charges in each half-reaction. ▪ Atoms are balanced in order: atoms other than O and H ▪ Add H2O to balance O and H+ to balance H ▪ Charge is balanced by adding electrons - To the left in reduction half-reactions - To the right in oxidation half-reactions 3. If necessary, multiply one or both half-reactions by an integer to make the number of e- gained equal to the number of e- lost 4. Add the balanced half-reactions, and include states of matter. 5. Check the atoms and charges are balanced. Prepared by: Engr. N. L. Escalante 6 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Examples: Method A: ACIDIC SOLUTION ClO3- (aq) + I- (aq) → I2 (s) + Cl- (aq) [acidic] Step 1: Divide the reaction into half-reactions ClO3- (aq) → Cl- (aq) I- (aq) → I2 (s) Step 2: Balance atoms and charges in each half-reaction - Atoms other than O and H ClO3- (aq) → Cl- (aq) Cl is balanced 2 I - (aq) → I2 (s) I is now balanced - Balance O atoms by adding H2O molecules ClO3- (aq) → Cl- (aq) + 3 H2O Add 3 H2O 2 I (aq) - → I2 (s) No change - Balance H atoms by adding H+ ions ClO3- (aq) + 6 H+ → Cl- (aq) + 3 H2O Add 6 H + 2 I - (aq) → I2 (s) No change - Balance the charge by adding electrons ClO3- (aq) + 6 H+ + 6 e- → Cl- (aq) + 3 H2O Add 6 e- 2 I- (aq) → I2 (s) + 2 e- Add 2 e- Step 3: Multiply each half-reaction by an integer to equalize number of electrons ClO3- (aq) + 6 H+ + 6 e- → Cl- (aq) + 3 H2O x1 2 I (aq) - → I2 (s) + 2 e- x3 Step 4: Add the half-reactions together ClO3- (aq) + 6 H+ + 6 e- → Cl- (aq) + 3 H2O 6 I- (aq) → 3 I2 (s) + 6 e- ClO3 (aq) - + 6H + + 6 I (aq) → - Cl- (aq) + 3 H2O + 3 I2 (s) Step 5: Check that atoms and charges balance OA: ClO3- RA: I - Method B: BASIC SOLUTION ▪ The only difference in balancing a redox equation that takes place in basic solution is in Step 4 ▪ At this point, we add one OH- ion to both sides of the equation for every H + ion present ▪ The H+ ions on one side are combined with the added OH - ions to form H2O and OH- ions appear on the other side of the equation ▪ Subtract H2O from both side of the equation if possible ▪ Do a final check and make sure atoms and charges are balanced. Prepared by: Engr. N. L. Escalante 7 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Fe(OH)2 (s) + Pb(OH)3- (aq) → Fe(OH)3 (s) + PbS (s) [basic] Step 1: Divide the reaction into half-reactions Pb(OH)3- (aq) → PbS (s) Fe(OH)2 (s) → Fe(OH)3 (s) Step 2: Balance atoms and charges in each half-reaction - Atoms other than O and H Pb(OH)3- (aq) → PbS (s) Pb is balanced Fe(OH)2 (s) → Fe(OH)3 (s) Fe is balanced - Balance O atoms by adding H2O molecules Pb(OH)3- (aq) → PbS (s) + 3 H2O Add 3 H2O Fe(OH)2 (s) + H2O → Fe(OH)3 (s) Add H2O - Balance H atoms by adding H+ ions Pb(OH)3- (aq) + 3 H+ → PbS (s) + 3 H2O Add 3 H + Fe(OH)2 (s) + H2O → Fe(OH)3 (s) + H+ Add H + - Balance charge by adding electrons Pb(OH)3- (aq) + 3 H+ + 2 e- → PbS (s) + 3 H2O Fe(OH)2 (s) + H2O → Fe(OH)3 (s) + H+ + e- Step 3: Multiple each half-reaction by an integer to equalize number of electrons Pb(OH)3- (aq) + 3 H+ + 2 e- → PbS (s) + 3 H2O x1 Fe(OH)2 (s) + H2O → Fe(OH)3 (s) + H+ + e- x2 Step 4: Add the half-reactions together Pb(OH)3- (aq) + 3 H+ + 2 e- → PbS (s) + 3 H2O 2 Fe(OH)2 (s) + 2 H2O → 2 Fe(OH)3 (s) + 2 H+ + 2 e- Pb(OH)3- (aq) + H+ + 2 Fe(OH)2 (s) → PbS (s) + H2O + 2 Fe(OH)3 (s) Step 4 (basic): Add OH- Pb(OH)3- (aq) + H+ + 2 Fe(OH)2 (s) → PbS (s) + H2O + 2 Fe(OH)3 (s) O OH- → OH- Pb(OH)3- (aq) + 2 Fe(OH)2 (s) → PbS (s) + 2 Fe(OH)3 (s) + OH- Step 5: Check OA: Pb(OH)3- (aq) RA: Fe(OH)2 (s)  For more examples on balancing redox equations, you may want to watch the videos entitled “ Oxidation Number Method ” using the link https://www.youtube.com/watch?v=nSQl6qI_JWk and “ Half Reaction Method ” using the link https://www.youtube.com/watch?v=j-xcAvrv9ow. Prepared by: Engr. N. L. Escalante 8 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Activity 3: Balancing REDOX Reactions Self-Assessment No. 3 Method 1: Fe2O3 + CO → Fe + CO2 Method 2: Zn + HNO3 ↔ Zn(NO3)2 + NH4NO3 To be submitted in Google classroom on: Metallic Conduction ▪ Conduction of electricity through metal by electron displacement ▪ An electric current is the flow of electric charge. In metals this charge is carried by electrons. Electrical Units I - electric current - rate of flow of an electric charge - unit: ampere (A) Q - quantity of an electric charge measured in coulombs (C) - quantity of electricity carried past a point in one second by a current of 1 A 1 A = 1 C/s - from a point of view of fundamental particles: 1 proton = 1 electron = 1.602 x 10-19 C V - potential difference - force through a circuit measured in volts 1 V = 1 J/C OHM’S LAW - expressed the quantitative relation between voltage and current V=IxR R = resistance in ohms (Ω) P=IxV P = power in watts (W) 1 W = 1 J/s Sample Problems: 1. A lamp draws a current of 2.0 A. Find the charge in coulombs used by the lamp in 30 seconds. Given: I = 2A Req’d: Q =? t = 30 sec Solution: Q = It ( ) = 60 C Prepared by: Engr. N. L. Escalante 9 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  2. Compute the time required to pass 36,000 C through an electroplating bath using a current of 5 A. Solution: 3. A dynamo delivers 15 A at 120 V. a.) Compute the power in KW supplied by the dynamo. Solution: ( ) b.) How much electric energy, in KW-hr, is supplied by the dynamo in 2 hours? Solution: c.) What is the cost of this energy at Php 120/KW-hr. Solution: 4. How many electrons per second pass through a cross section of copper wire carrying ? Solution: ( )= Electrolytic Conduction ▪ Conduction of electricity by the movement of ions through a solution or a molten salt. A sustained current requires that chemical changes at the electrodes also occur. ▪ Cations move towards the cathode and anions move towards the anode FARADAY’S LAWS OF ELECTROLYSIS - the mass of any substance liberated or deposited at an electrode is proportional to the electrical charge. mαQ or mαIxt where: m = mass I = current t = time - the mass of different substances liberated or deposited by the same quantity of electricity are proportional to the equivalent weights of various substance. Prepared by: Engr. N. L. Escalante 10 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  1.602x1019 C 6.023x1023 electrons 1F  x electron mol 1F  9.56x10 C / mol 4 Sample Problems: 1. Exactly 0.2000 mol of electrons is passed through 3 electrolytic cells in series. One contains silver ions, one zinc ion and one ferric ion. Assume that the only cathode reaction in each cell is the reduction of the ion to the metal. How many grams each metal will be deposited? Atomic Weights (g/mol): Ag = 107.9 Zn = 65.39 Fe = 55.85 Solution: 1 mol of electron liberates 1 g-eq wt of an element. Equivalent-weights: is = 32.69 is = 107.9 =( )( ) =( )( ) ( )( ) 2, A current of 5.00 A flowing for exactly 30.0 minutes deposits 3.048 g of zinc at the cathode. Calculate the equivalent weight of zinc from this information. ( )( ) ( ) = 32.7 3. Find the charge in coulomb on 1 g-ion of N3-. Solution: = 4. How much charge is required to reduce (a) 1 mole of to Al and (b) 1 mole of to ? Solution: (a) The reduction reaction is +3 Thus, 3 mole of electrons are needed to reduce 1 mole of Al 3+ Q = 3 x F = 3 x 96500 = 289500 coulomb Prepared by: Engr. N. L. Escalante 11 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  (b) The reduction reaction is 1mole 5mole Q = 5 x F = 5 x 96500 = 48500 coulomb Electrolysis ▪ Electrolysis is a process by which electric current is passed through a substance to effect a chemical change. ▪ The chemical change is one in which the substance loses or gains an electron (oxidation or reduction). ▪ Electrolysis is used extensively in metallurgical processes such as in extracting (electrowinning) or purification (electrorefining) of metals from ores or compounds and in deposition of metals from solution (electroplating). ▪ It is the use of electric current to bring about oxidation-reduction (REDOX) reaction. ▪ Electric charge in electrolytic conduction is carried by CATIONS (+) moving toward the cathode and anion Na+ Cl- Cl- Na+Na+ Cl2 For a complete circuit, electrode reactions must accompany the movement of ions. At the cathode, come chemical species (not necessarily the charge carrier) must accept electrons and can be reduced. At the anode, electrons must be removed from some chemical species which as a consequence is oxidized. In the given diagram, Na+ ions are reduced at the cathode: Na+ + e- → Na0 And Cl ions are oxidized at the anode: - Cl- → Cl2 + 2 e- Proper addition of these partial equations gives the reaction for the entire cell: 2 NaCl (l) → 2 Na + Cl (g)  Here’s a video entitled “ What Is Electrolysis | Reactions | Chemistry | FuseSchool ” explaining what electrolysis is for a better understanding. Use the link https://www.youtube.com/watch?v=7uIIq_Ofzgw. Voltaic Cell or Galvanic Cell ▪ A cell that is used as a source of electrical energy. It is also called a GALVANIC CELL after ALESSANDRO VOLTA or LUIGI GALVANI, who first experimented with conversion of chemical energy or electrical energy. ▪ A device in which spontaneous oxidation-reduction occurs in the passage of electron through an external circuit. Prepared by: Engr. N. L. Escalante 12 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  ▪ In voltaic cells, half reactions are made to occur at different electrodes so that the transfer of electrons takes place through the external circuit rather than directly between the metal electrodes and the ions. Oxidation - a process in which a substance loses electrons. It occurs at the anode. Reduction - a process in which a substance gains electrons. It occurs at the cathode. The cell diagram in the figure is designed to make use of this reaction to produce an electric current. The cell on the LEFT contains a zinc metal electrode and ZnSO4 solution. The cell on the RIGHT consists of a copper metal electrode in a solution of CuSO4.  To understand better how galvanic cells work, watch the video entitled “ Introduction to galvanic/voltaic cells | Chemistry | Khan Academy ” using the link https://www.youtube.com/watch?v=9Xncz_mMc5g. Cell Notation - a useful notation that describes the components of a cell - expresses a certain reaction in an electrochemical cell Rules in writing cell notations ▪ solid | aqueous || aqueous | solid ▪ Anode on the left (oxidation) || cathode on the right (reduction) ▪ Single line; different phases ▪ Double line; porous disk or salt bridge ▪ If all the substances on one side are aqueous, a platinum electrode is indicated. ▪ If non standard concentration of the aqueous solution was used, indicate the concentration enclosed with an open and close parenthesis. Example: Zn (s) + Cu+2 (aq) → Zn+2 (aq) + Cu (s) Half Reactions: Anode (oxidation): Zn (s) → Zn+2 (aq) + 2 e- Cathode (reduction): Cu+2 (aq) + 2 e- → Cu (s) Cell Notation: Zn|Zn+2||Cu+2|Cu Prepared by: Engr. N. L. Escalante 13 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  Standard Half Cell Potentials Cell Potential - difference between two electrode potentials, one associated with the cathode and the other associated with the anode. A galvanic cell have a reduction reaction (in cathode half-cell) and an oxidation reaction (in anode half-cell). Electrons flows through a wire from the anode half-cell to the cathode half-cell. The driving force that allows force that allows electrons to flow is called the electromotive force (emf) or the cell potential (Ecell).  Determining Standard State Cell Potentials A cell’s standard state potential is the potential of the cell underd standard state conditions, which is approximated with concentrations of 1 mole per lite (1 M) and pressure of 1 atmosphere at 25 oC. To calculate the standard cell potential for a reaction: ▪ Write the oxidation and reduction half-reactions for the cell. ▪ Look up the reduction potential, Ereduction o , for the reduction half-reaction in a table of reduction potentials. See Appendix for the Table of Standard Electrode Potentials in Aqueous Solution at 25 oC ▪ Look up for the potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For oxidation half-reaction, Eoxidation  Ereduction o o ▪ Add the potentials of the half-cells to get the overall standard cell potential. o Ecell  Ereduction o  Eoxidation o >0 Spontaneous 0 Ecell Li(s) -3.04 K+(aq) + e- -> K(s) -2.92 Ca2+(aq) + 2e- -> Ca(s) -2.76 Na+(aq) + e- -> Na(s) -2.71 Mg2+(aq) + 2e- -> Mg(s) -2.38 Al3+(aq) + 3e- -> Al(s) -1.66 2H2O(l) + 2e- -> H2(g) + 2OH-(aq) -0.83 Zn2+(aq) + 2e- -> Zn(s) -0.76 Cr3+(aq) + 3e- -> Cr(s) -0.74 Fe2+(aq) + 2e- -> Fe(s) -0.41 Cd2+(aq) + 2e- -> Cd(s) -0.40 Ni2+(aq) + 2e- -> Ni(s) -0.23 Sn2+(aq) + 2e- -> Sn(s) -0.14 Pb2+(aq) + 2e- -> Pb(s) -0.13 Fe3+(aq) + 3e- -> Fe(s) -0.04 2H+(aq) + 2e- -> H2(g) 0.00 Sn4+(aq) + 2e- -> Sn2+(aq) 0.15 Cu2+(aq) + e- -> Cu+(aq) 0.16 ClO4-(aq) + H2O(l) + 2e- -> ClO3-(aq) + 2OH-(aq) 0.17 Prepared by: Engr. N. L. Escalante 19 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  AgCl(s) + e- -> Ag(s) + Cl-(aq) 0.22 Cu2+(aq) + 2e- -> Cu(s) 0.34 ClO3-(aq) + H2O(l) + 2e- -> ClO2-(aq) + 2OH-(aq) 0.35 IO-(aq) + H2O(l) + 2e- -> I -(aq) + 2OH-(aq) 0.49 Cu+(aq) + e- -> Cu(s) 0.52 I2(s) + 2e- -> 2I-(aq) 0.54 ClO2-(aq) + H2O(l) + 2e- -> ClO-(aq) + 2OH-(aq) 0.59 Fe3+(aq) + e- -> Fe2+(aq) 0.77 Hg22+(aq) + 2e- -> 2Hg(l) 0.80 Ag+(aq) + e- -> Ag(s) 0.80 Hg2+(aq) + 2e- -> Hg(l) 0.85 ClO-(aq) + H2O(l) + 2e- -> Cl-(aq) + 2OH-(aq) 0.90 2Hg2+(aq) + 2e- -> Hg22+(aq) 0.90 NO3-(aq) + 4H +(aq) + 3e- -> NO(g) + 2H2O(l) 0.96 Br2(l) + 2e- -> 2Br-(aq) 1.07 O2(g) + 4H +(aq) + 4e- -> 2H2O(l) 1.23 Cr2O72-(aq) + 14H +(aq) + 6e- -> 2Cr3+(aq) + 7H2O(l) 1.33 Cl2(g) + 2e- -> 2Cl-(aq) 1.36 Ce4+(aq) + e- -> Ce3+(aq) 1.44 MnO4-(aq) + 8H +(aq) + 5e- -> Mn2+(aq) + 4H2O(l) 1.49 H2O2(aq) + 2H +(aq) + 2e- -> 2H2O(l) 1.78 Co3+(aq) + e- -> Co2+(aq) 1.82 Prepared by: Engr. N. L. Escalante 20 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.  S2O82-(aq) + 2e- -> 2SO42-(aq) 2.01 O3(g) + 2H+(aq) + 2e- -> O2(g) + H2O(l) 2.07 F2(g) + 2e- -> 2F-(aq) 2.87 Prepared by: Engr. N. L. Escalante 21 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited.