Electrochemistry Lecture PDF

ELECTRO CHEMISTRY Manuel O. Bibit  AGENDA Learning Objectives Definition of Terminology Concept and Importance Lesson Proper(Lecture & Class Discussions) Problem Solving Process Summary LEARNING OBJECTIVES TO UNDERSTAND WHAT IS AN ELECTROCHEMISTRY TO REVIEW SOME PREVIOUS REL...

ELECTRO CHEMISTRY Manuel O. Bibit  AGENDA Learning Objectives Definition of Terminology Concept and Importance Lesson Proper(Lecture & Class Discussions) Problem Solving Process Summary LEARNING OBJECTIVES TO UNDERSTAND WHAT IS AN ELECTROCHEMISTRY TO REVIEW SOME PREVIOUS RELATED TOPICS ABOUT REDOX DIFFERENTIATE THE FOLLOWING TERMS: A. OXIDATION AND REDUCTION B. OXIDIZING AGENT AND REDUCING AGENT C. USING HALF- REACTIONS IN REDOX DESCRIBE THE PROCESS IN ELECTROLYSIS  DEFINITIONS OF 4 TERMINOLOGY ELECTROCHEMISTRY - is the study of chemical processes that cause electrons to move. This movement of electrons is called electricity, which can be generated by movements of electrons from one element to another in a reaction known as an oxidation-reduction ("redox") reaction. Oxidation is the loss of electrons. Reduction refers to the acquisition of electrons. The species being oxidized is also known as the reducing agent, and the species being reduced is called the oxidizing agent,  CONCEPTS OF 5 ELECTROCHEMISTRY The relation between electricity and chemical change. Spontaneously occurring chemical reactions that liberate electrical energy. Ex. Electrochemical Cell- also called Voltaic Cells or Galvanic Cell Non-spontaneous use of electrochemistry in batteries and fuel cells to produce electric power. An electrochemical cell is a device that:Generates electrical energy from chemical reactions123. Converts chemical energy into electrical energy, or vice versa1.  GENERAL RULES FOR 6 ASSIGNING OXIDATION NUMBER For an atom in its elemental form(Na, O₂, Cl₂, etc.) : O.N. = 0 For a monatomic ion. Naᶧ, Alᶟᶧ, Ca²ᶧ: O.N. = ion charge The sum of O.N. values for the atoms in a molecule or formula unit of a compound equals zero. MgCl₂ , let x= O.N. for Mg, Cl = -1 x + 2 (-1) = 0 , x = +2 O.N. of Mg = +2 The sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge. Ex. Nitrate: NO₃̄ , Carbonate: C0₃² ̄  ROLES FOR SPECIFIC ATOMS 7 OR PERIODIC TABLE GROUPS For Group 1A(1): O.N. = +1 in all compounds For Group 2A(2): O.N. =+2 in all compounds For Hydrogen: O.N. = +1 in compd w/ nonmetals = -1 in combination with metals and boron For Fluorine: O.N. = -1 in all compound For Oxygen: O.N. = -1 in peroxides O.N. = -2 in all other compd (except with F) For Group 7A(17): = O.N.=-1 in combination with metals, nonmetals(except O) and other halogens lower in the group  STEPS TO IDENTIFY 8 OXIDIZING & REDUCING AGENT First assign an oxidizing number (O.N.) to each atom (or ion) based on the rules stated in previous slides. The reactants is the reducing agent if it contains an atom that is oxidized (O.N. increased from left side to right side of the equation). The reactant is the oxidizing agent if it contains an atom that is reduced (O.N. decreased) MATHEMATICAL EXPRESSION REDOX 9 IN REDOX REACTION, THE SUBSTANCE THAT GAINS ELECTRONS UNDERGOES REDUCTION. SINCE IT TAKES ELECTRON OF THE OTHER SUBSTANCE IT IS CALLED THE OXIDIZING AGENT. THE SUBSTANCE THAT LOSES ELECTRONS UNDERGOES OXIDATION. SINCE IT GIVES UP ELECTRON, THE SUBSTANCE IS KNOWN AS THE REDUCING AGENT. HOW DO WE KNOW THAT THE REACTION INVOLVES OXIDATION-REDUCTION: * By identifying the presence of a strong oxidizing or reducing agents as reactants. * By recognizing a change in oxidation number - this means we need to determine the oxidation number of each element as it appears in a reactant and product. * By recognizing the presence of uncombined element as a reactant or product. An uncombined elements is always assigned as oxidation numbers of 0. Example: Consider the displacement reaction between magnesium, a reactive metal, and hydrochloric acid. 0 +1 -1 +2 -1 0 H is reduced - HCl is oxidizing agent Mg₍𐞥₎ + 2HCl₍ₐ𐞥₎ MgCl₂₍ₐ𐞥₎ + H₂₍ϧ₎ Mg is oxidized - Mg metal is reducing agent Mg(s) + 2 HCl(aq) --> MgCl 2(aq) M + H 2(g) Mg(s) + 2 HClMg(aq) --> MgCl 2(aq) +  10 PRACTICE PROBLEM GIVE THE OXIDATION NUMBER OF EACH ATOM AND IDENTIFY THE OXIDIZING AND REDUCING AGENT IN THESE BALANCE CHEMICAL EQUATIONS, 1. 2 Fe(s) + 3Cl₂(g) 2 FeCl₃(s) 2. 2 H₂ (g) + O₂ (g) 2 H₂O(g) 3. C (s) + O₂ (g) CO₂ (g)  11 USING HALF-REACTIONS TO UNDERSTAND REDOX REACTIONS A STRIP OF ZINC HAS BEEN PLACED IN A SOLUTION OF COPPER(II) SULFATE. AS TIME PASSES, THE ZINC METAL REACTS WITH THE Cu ²ᶧ IONS TO PRODUCE COPPER METAL AND Zn ²ᶧ IONS IN SOLUTION. reduction Zn(s) + Cu ²ᶧ (aq) Zn ²ᶧ (aq) + Cu(s) oxidation THE OVERALL REACTIONS AS THE RESULTS OF TWO SIMULTANEOUS HALF REACTIONS : ONE HALF REACTION FOR THE OXIDATION OF Zn AND ONE –HALF FOR THE REDUCTION OF Cu²ᶧ IONS. THE OXIDATION HALF REACTION: Zn (s) Zn²ᶧ(aq) + 2 e ̅ THE REDUCTION HALF REACTION : Cu ²ᶧ (aq) + 2 e ̅ Cu (s) NET REACTION : Zn(s) + Cu²ᶧ (aq) Zn ²ᶧ (aq) + Cu (s)  12 USING HALF-REACTIONS TO UNDERSTAND REDOX REACTIONS Example: Aluminum metal undergoes a redox reaction with Zn²ᶧ(aq) to produce Alᶟᶧ(aq) and zinc metal. (unbalanced equation) Al(s) + Zn²ᶧ Alᶟᶧ(ag) + Zn(s) Write the oxidation half-reactions, the reduction half reaction equations, and combine them to give the balanced equation for the net reaction. Classification Of Electrochemical Cells 1. Voltaic or Galvanic Cells electrochemical cells in which spontaneous oxidation–reduction reactions produce electrical energy. after the Italian scientists Luigi Galvani and Alessandro Volta, who constructed early versions of the device. Components of a Galvanic Cell the anode cathode is the is the electrode electrode at at which which oxidation occurs reduction occurs.  Electricity travels in a complete circuit An electric current flows from the anode to the cathode because there is a difference in electrical potential energy between the electrodes. A salt bridge serves three functions. 1. It allows electrical contact between the two solutions. 2. It prevents mixing of the electrode solutions. 3. It maintains the electrical neutrality in each half-cell as ions flow into and out of the salt bridge.  SUMMARY OF WORK FLOW FOR GALVANIC CELL Electrons move from the negative electrode (anode) to the positive electrode (cathode through an external circuit. cathode - electrode at which reduction occurs as electrons are gained by some species. anode -electrode at which oxidation occurs as electrons are lost by some species. The electrical circuit is completed in the solution by movement of ions, anion move from the salt bridge compartment into to the anode compartment and cations move from the salt bridge compartment to the cathode compartment. Components of a Electrolytic Cell the anode cathode is the electrode is the electrode at at which reduction which oxidation occurs. Pushes out occurs. Pulls in electron electron  SUMMARY OF WORK FLOW FOR ELECTROLYTIC CELL Electrical energy from the Battery initiates REDOX reactions that would not spontaneously occur. cathode – is the side of the reduction reaction occurs and is negatively charge. anode –is the side of the oxidation reaction and is positively charge. Cells contain two electrodes and electrolyte solutions that conducts electricity because it contains dissolved ions. THE PROCESS OF ELECTROLYSIS OF WATER.  PRESENTATION: SPONTANEOUS & NON SPONTANEOUS CHEMICAL 20 REACTIONS. Mg(s) + 2 HClMg(aq) --> MgCl 2(aq) +  Electrochemical cell (Galvanic Cell) Electrolytic cell 21 A Galvanic cell converts chemical energy into An electrolytic cell converts electrical energy electrical energy.. into chemical energy. The redox reaction is not spontaneous and Here, the redox reaction is spontaneous and is electrical energy has to be supplied to initiate responsible for the production of electrical energy. the reaction. Mg(s) + 2 HClMg(aq) --> MgCl 2(aq) + The two half-cells are set up in different Both the electrodes are placed in a same containers, being connected through the salt container in the solution of molten electrolyte. bridge or porous partition. Here the anode is negative and cathode is the Here, the anode is positive and cathode is the positive electrode. The reaction at the anode is negative electrode. The reaction at the anode is oxidation and that at the cathode is reduction. oxidation and that at the cathode is reduction. The electrons are supplied by the species getting The external battery supplies the electrons. They oxidized. They move from anode to the cathode in enter through the cathode and come out through the external circuit. the anode.  NOTIFICATION: 22 There is a shorthand notation for representing an electrochemical cell. For shown in Figure- presentation of spontaneous and non spontaneous reactions. Zn(s) + Cu²ᶧ (aq) Zn²ᶧ (aq) + Cu(s) the representation is Zn(s) I Zn²ᶧ(aq) II Cu²ᶧ (aq) I Cu(s) The anode half cell is represented on the left and the cathode half cell is represented on the right. The electrodes are written on the extreme left(anode, Zn) and extreme right( cathode, Cu) of the notation. The single vertical lines denote boundaries between phases and the double vertical lines denote the salt bridge and the separation between half cell.  The voltage across the electrodes of a galvanic cell is called the cell voltage, or cell potential (Ecell). Experimentally, this is measured by a voltmeter Another common term for the cell potential is the electromotive force or emf (E), which, is a measure of voltage, not force.  DEFINITION OF TERMINOLOGY: 24 Cell Potential (Ecell) – refers to the voltage of an electrochemical cell whose value can be affected by temperature, concentration and pressure. To calculate the cell potential, the electrical potential is taken from two electrodes to determine how much energy can be transferred. For electrons to be transferred from the anode to the cathode, there must be some sort of energy potential that makes this phenomenon favorable. The potential energy that drives the redox reactions involved in electrochemical cells is the potential for the anode to become oxidized and the potential for the cathode to become reduced. The electrons involved in these cells will fall from the anode, which has a higher potential to become oxidized to the cathode, which has a lower potential to become oxidized. This is analogous to a rock falling from a cliff in which the rock will fall from a higher potential energy to a lower potential energy. Note: The difference between the anode's potential to 25 become reduced and the cathode's potential to become reduced is the cell potential E°Cell = E° Red, Cathode − E° Red, Anode Both potentials used in this equation are Standard reduction potentials, which typically what found in Table However, the reaction of the anode is actually oxidation, we reversed to reduction reaction. This explain the minus sign. The superscript "o" in E^o indicates that these potentials are correct only when concentrations are 1 M and pressures are 1 bar. A correction called the "Nernst Equation" must be applied if conditions are different. If E°cell is positive, the reaction is spontaneous reaction If E°cell is negative, the reaction is nonspontaneous reaction If E°cell is zero , the reaction is in equilibrium. Standard Reduction Potentials at 25oC.  DETERMINE A HALF-CELL POTENTIAL 27 Example: The voltaic cell shown in the attached drawing generates a potential difference under standard condition of E°cell = 0.36 V at 25°C. The net cell reaction is Zn(s) + Cd²ᶧ(aq, 1M) Zn²ᶧ(aq), 1M) + Cd(s) Cd The standard half cell potential for Zn(s) I Zn²ᶧ(aq, 1M) is Zn Salt Bridge -0.76V. (a) Determine which electrode is the anode and which is the cathode. Cd²ᶧ (b) Show the direction of electron flow through the Zn²ᶧ circuit outside the cell, and complete the cell diagram (c) Calculate the standard half-cell potential for Cd²ᶧ(aq) + 2e ̄ Cd(s)  The Nernst Equation Electrical Energy(ΔG°) = - nFE°cell  Example Predict whether the following reaction would proceed spontaneously as written at 298 K: Co(s) + Fe+2 (aq) ---> Co+2 (aq) + Fe(s) given that [Co+2] =0.15 M Co+2 + 2e= -0.28V [Fe+2] =0.68 M Fe+2 +2e= -0.44V Anode: Co ---> Co+2 + 2e Cathode: Fe+2 + 2e---> Fe Eocell= Eo Fe + Eo Co = -0.44V + 0.28V = -0.16V  Example Eocell = -0.16V [Co+2] =0.15 M [Fe+2] =0.68 M E is negative, the reaction is non-spontaneous  Example Consider this electrochemical reaction: Zn(s) + Ni²ᶧ(aq) Zn²ᶧ + Ni(s) The standard cell potential E°cell = 0.51V. Calculate the cell potential if the Ni²ᶧ concentration is 5.0 and the Zn²ᶧ(aq) concentration is 0.050M. ELECTROPLATING - IS BASICALLY THE PROCESS 32 OF PLATING A METAL ONTO THE OTHER BY HYDROLYSIS MOSTLY TO PREVENT CORROSION OF METAL OR FOR DECORATIVE PURPOSES. The silver Mg(s) plating-->is + 2 HClMg(aq) connected MgCl 2(aq) + to the anode (+ve charged electrode) of the circuit and the spoon is kept at the cathode (-ve charged electrode). Both are kept immersed in a highly developed electrolytic bat (solution). At this stage, a DC current is supplied to the anode that oxidizes the silver atoms and dissolves them into the solution. The dissolved ions of silver are reduced at the cathode and plated on the spoon. Electroplating silver onto a spoon.  ELECTROPLATING 33 Example: The chemical equation to determine how many electrons are needed for 1 mole of the metal being electroplated. Using an example, if we take copper Cu as our metal with 25 amps, then each mole of copper Cu++ +will Mg(s) require 2e- electrons. 2 HClMg(aq) --> MgCl 2(aq) + Use the equation: Q = n(e)F Q=n(e)F, to solve for Q. Q is the amount of electricity or charge in coulombs C, n(e) is the number of moles of electrons and F is the Faraday constant 96,500 C mole-1. Using our example where we need 2e- for each mole of copper: Q=n(e)F=2 mol×96,500 C/mol=193,000 C Determine the time it will take to electroplate out one mole of the metal using the equation t = Q/I. Q is the amount of electricity in coulombs C, I is the current in amps A and t is the time in seconds. Using our example: t=Q/I=193,000C/25 A=7,720s=2.144 hours ELECTROPLATING - IS BASICALLY THE PROCESS 34 OF PLATING A METAL ONTO THE OTHER BY HYDROLYSIS MOSTLY TO PREVENT CORROSION OF METAL OR FOR DECORATIVE PURPOSES. Mg(s) + cell An electrolysis 2 HClMg(aq) -->gold that deposits MgCl 2(aq) (from + Auᶧ(aq) operates for 15.0 minutes at a current of 2.30 A. What mass of gold is deposited? Given : Gold (Auᶧ(aq) time (t) = 15.0 min. I = 2.30 A Required: masss of gold deposited Solution: First, write the balanced half reaction: Auᶧ(aq) + e ̄ Au(s) Next, calculate the mole of electron based on current and time: Q = I x t = (2.30 C/s ) ( 900 s) = 2.07 x 10ᶟ C (2.07 x 10ᶟ C) x ( 1 mol e ̄ ) = 2.15 x 10 ² mol e ̄ 96,485 C Note base on that the Auᶧ half reaction the mole ratio of gold is 1: 1, which means that we also have 2.15 x 10 ̄² mol of Au. Finally convert from to mass in grams. (2.15 x 10̄̄ ̄² mol Au) x 197 gm / mol) = 4.23 gm Au  Batteries A battery is a galvanic cell, or a series of combined galvanic cells, that can be used as a source of direct electric current at a constant voltage.  RESEARCH REPORT ABOUT BATTERY 36 The Research report must contain the following: 1. What are the three classification of a Battery. 2. Discuss each Battery, Give one example of each Battery & their uses. 3. Identify the major components of each example of the Battery. Describe each major components. 4. Determine which electrode is Anode and which is a Cathode 5. Provide the half reaction and the Overall Net (cell) reactions. Format should follow the Standard format in terms of Research reporting especially the margin, spacing, labeling & picture attachment. Reference must be attached. THANK YOU