Electrochemical cells Introduction: Sudha Madhugiri D.Chem. Collin College Department of Chemistry Have you used a battery before for some purpose? I bet you have. The type of chemistry that is used in batteries is always happening around us. Tarnishing of silver and restoring the shine is another example of the same phenomenon. The chemistry we are talking about here is the process of oxidationreduction (also known as redox). Loss of electrons is oxidation, and gain of electrons is reduction. Though these reactions always occur together, they can be physically separated and studied. The substance that undergoes oxidation is called a reducing agent and the substance that undergoes reduction is called an oxidizing agent. Redox Reactions and Voltaic Cells: In a redox reaction, the electrons lost by the reducing agent are gained by the oxidizing agent. If these two reactions are made to occur in separate regions connected by a metal wire that allows the flow of electrons between the two, it forces the electrons to travel through the external wire from the oxidizing agent to the reducing agent compartments, or half-cells. The natural tendency for the electrons to flow through the external wire is called the electromotive force (EMF), or potential. This flow creates electricity. Electricity is typically viewed as the flow of electrons or charge through a medium. This principle is used in building an apparatus called a voltaic cell. In a voltaic cell, a spontaneous reaction (negative G o ) is used to create electricity. The magnitude of the potential is a direct measure of the thermodynamic driving force of the reaction. If a voltmeter is connected between the two electrodes (the conducting metal or substance) immersed in each of the oxidation and reduction reaction chambers, the voltage registered is the sum of oxidation and reduction potentials. E o cell = E o oxidation + E o reduction Standard conditions for redox reactions are 1 M concentration for solutions and 1 atm pressure for gases and a temperature of 298 K. Although not always the case, redox reactions are typically carried out at room temperature, 298K. By convention, all the half cell potentials are tabulated as standard reduction potentials. It is for you to determine which electrode undergoes oxidation and which undergoes reduction. The electrode with more positive reduction potential undergoes reduction and the one with less positive reduction potential undergoes oxidation. That means for oxidation, the half reaction is flipped, which also means the standard reduction potential carries an opposite sign for oxidation potential. As a voltaic cell is spontaneous, it always has a positive cell potential (E o cell). In general, in a voltaic cell there are two half cell compartments, an anode (assigned the - sign) and a cathode (assigned the + sign) compartment. In a half cell, an electrode is dipped in an electrolytic solution, which is typically a salt solution of the metal electrode. If the electrode is used to make the connection and if it is also part of the reaction, it is known as active electrode. However, if the reactant is an aqueous solution, or a liquid or a gas, an electrode made of either a graphite rod or a platinum wire is used to make the connection. These are called inactive electrodes. Oxidation (loss of electrons) always occurs at the anode and reduction (gain of electrons) always occurs at the cathode. A shorthand form of representing a cell, called line notation, is shown below: 1
The electron flow which occurs when the anode and cathode are connected produces electricity in a spontaneous cell. As the cell is discharging (the redox reaction is producing electron flow), the half cells end up electrically charged as the anode compartment loses electrons and becomes more positive while the cathode compartment gains electrons to become more negatively charged. This electrical charge hinders the flow of electrons in the redox reaction. To overcome this obstacle, one can use either a porous cup for one of the compartments or a salt bridge. Both work on the same principle. The salt bridge contains a saturated solution of an inert salt. As the charge begins to build up in the compartments, the ions present in the salt bridge will naturally flow from the salt bridge into each compartment to make it electrically neutral. The porous cup allows ions to flow through the pores to accomplish electrical neutrality in both compartments. Electrolytic Cell: Contrary to a voltaic cell that produces electricity from a spontaneous chemical reaction, an electrolytic cell uses electricity to drive a non-spontaneous chemical reaction (positive G o ). A power source, usually a battery, causes oxidation, still at the anode (now assigned a + sign), and reduction, still at the cathode (now assigned a - sign), where as the overall reaction is non-spontaneous. Note the reversal in the signs for the electrodes as compared to voltaic cells. Electrolysis (electro means electricity, lysis means breaking up) has great industrial importance as this is the main process used in recovering metals from their ores. Electrolysis of brine (NaCl solution) is extensively used to produce chlorine gas, sodium hydroxide and hydrogen gas. In electrolysis of aqueous salt solutions, two oxidation reactions and two reduction reactions are possible. At the cathode (-): The metal cation itself can be reduced or water can be reduced based on the following reduction half reactions. M + (aq) + e - M (s) 2H 2 O (l) + 2e - H 2 (g) + 2 OH - (aq) E = -0.83 V In general, water gets reduced at the cathode. At the anode (+): The non-metal anion can be oxidized or water can be oxidized based on the following two half reactions: X - (aq) X (s) + e - 2H 2 O (l) O 2 (g) + 4H + (aq) + 4 e - E = -1.23 V The non-metal normally gets oxidized at the anode. However, these results depend on the individual half cell potentials. The reaction with less negative reduction potential will be reduced (for the most part it is water) at the cathode and the reaction with less positive reduction potential will be oxidized (which is normally the metal) at the anode (M. Silberberg 2007). 2
From the reactions at the anode and cathode, it can be deduced that more the # of moles of electrons transferred, greater will be the product formed at a given electrode. This relationship was first put forth by Michael Faraday and is known as Faraday s law of electrolysis. This law states that "the amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell. 1 The charge in coulombs (C) can be calculated by: Charge (C) = Current (amps) * time (seconds) Using Faraday s constant, it takes 96,500 C of charge to transport 1 mole of electrons. Using Faraday s law, the current transferred in amperes (amps) can also be calculated: 1 amp = 1 coulomb/second This simple relationship combined with a balanced half reaction can be used to calculate the required parameters involving stoichiometry of electrolysis. Electroplating: Electrolysis has a variety of applications as mentioned earlier and one is electroplating. Electroplating is a technique that uses electric current to coat a layer of a desired metal by reducing cations of that metal from a solution. This process is performed to bestow properties that are not intrinsic to the material being coated. For example, chrome plating is done on a variety of materials including parts of automobiles to provide corrosion resistance, ease of cleaning, and aesthetic appeal. Gold or silver plating can be done on pieces of jewelry and to keep expensive electronic material from tarnishing. The material that needs to be coated can be made a cathode and using an electroplating solution, the cathode can be coated with the desired metal using electrochemical reduction. PART I: Voltaic Cell You are provided with the following materials: Strips of zinc, copper, iron and 1M aqueous solutions of ZnSO 4, CuSO 4, FeSO 4, and KBr A porous container, graphite rods, a voltmeter, and a few wires with alligator clips. 3
Objective: You are trying to design a voltaic cell with the maximum voltage to be used as a power source for your experiment. Use Figure 1 as a guide to construct 4 different voltaic cells using the above materials. Cathode (+) Anode (-) Porous cup acting as a cathode compartment Anode compartment Figure 1. Schematic of a voltaic cell NOTE: If you are using a salt solution for more than one cell, you can reuse it before carefully disposing of it in the designated waste container. Make sure you don t contaminate your solutions through the trials as this will affect your results. If your half cell uses an active electrode, be sure to place the metal strip (which functions as an active electrode) in its respective salt solution (i.e., copper rod needs to be placed in CuSO 4 solution). If your half cell uses an inactive electrode, use a graphite rod as your inactive electrode to make the connection. Use a 100 ml beaker for one half cell and the porous cup for the other half cell. Connect the electrode you decide to be the anode to the negative voltmeter terminal and the cathode to the positive voltmeter terminal and record the voltage. While using KBr solution, use a graphite rod as an inactive electrode to make the contact. Be sure to discuss and decide the possible anode and cathode for each cell among your group members before you put the voltaic cells together. 4
Prelab Questions for Part I: 1. A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr 3+ and a gold electrode in a solution that is 1.0 M in Au 3+. a) Provide the half cell reactions for each of the electrodes. b) Which is the anode and which is the cathode? How did you decide? c) Give an overall balanced equation for the cell. d) Calculate the E o cell at 25 o C. 2. Is it possible to create a cell that would follow the following cell reaction? Fe 2+ (aq) + Ni(s) Fe(s) + Ni 2+ (aq) If yes, calculate E o cell at 25 o C. If it is not possible, explain why. Critical Data/Discussion to Include in Your Lab Report for Part I: For each cell, which electrode was the anode? Cathode? The anode and cathode half reactions Balanced equation for each of the four cells you successfully constructed Both calculated and measured E o cell Line notation for the four cells Discussion of the four cells; which gave the most voltage? The least? Do your observations agree with the calculated values? If not, discuss why not? What is the use of porous cup in the set up? References cited in the format specified by your instructor Part II: Electrolytic Cell Objective: Construct an electrolytic cell based on the schematic provided in Figure 2 and coat a piece of copper strip with nickel metal to understand the process of electroplating. Using the data collected by electroplating nickel, calculate the current supplied in amperes (amps) in this process. Current (amps) and time (s) given Calculate charge Calculate Calculate (coulombs) moles of moles of electrons product Mass of product (g) 5
Battery Anode (+) Cathode (-) Electroplating Solution Figure 2. Schematic of the cell for electroplating Set up of the electrolytic cell: Attach a brightly sanded, weighed, copper strip to the negative pole of a 6 V battery source. The positive pole of the battery source should be connected to a graphite rod using a lead wire. Place the electrodes in a 50 ml beaker containing 40 ml of nickel plating solution. You should be noticing a layer of shiny metal getting deposited on the copper strip. Allow this process to continue for 10 minutes. Disassemble the electrolytic cell, pat dry and weigh the copper strip and dispose the chemicals in appropriate waste containers. Remove the nickel coating on the copper strip using sand paper before returning and perform the required calculations. Prelab Questions for Part II: 1. Calculate the amount of copper produced in 1.0 h when aqueous CuBr 2 solution was electrolyzed by using a current of 4.50 A. 2. In another electroplating experiment, if electric current was passed for 3 hours and 2.00 g of silver was deposited from a AgNO 3 solution, what was the current used in amperes? Critical Data/Discussion to Include in Your Lab Report for Part II: 1. Construct a table using Microsoft Word containing the following data/information Process (oxidation or reduction) occurring at the cathode The half reaction for the reaction at cathode Observations at the graphite electrode The exact time for which the current was passed through solution Mass of the copper strip before electroplating Mass of the copper strip after electroplating Mass of the metal deposited 2. Calculations for the charge (in coulombs) passed through the solution in this process 3. Calculations for the current supplied (in amps) 4. Discussion of results, things learned and ways to improve the experiment 6
5. References cited in the format specified by your instructor 6. If you were to plate silver onto an object you would need to use either AgNO 3 in a basic medium along with ammonium hydroxide or silver cyanide (AgCN). This is matter of concern with respect to safety and the cost involved in electroplating silver onto an object. Perform a cost analysis on using AgNO 3 if you were to use 1M AgNO 3 solution in one electroplating experiment and 1M CuSO 4 5H 2 O solution for copper plating in an another experiment. (500 g of CuSO 4 5H 2 O costs $19 and 500 g of AgNO 3 costs $540). Show your calculations. Bibliography: Silberberg, Martin. (2007). Principles of general chemistry. 2007. 7