NAME PER DATE DUE ACTIVE LEARNING IN CHEMISTRY EDUCATION CHAPTER 24 OXIDATION AND REDUCTION (Part 2) Corrosion Electrochemical Cells 24-1 1997, A.J. Girondi
NOTICE OF RIGHTS All rights reserved. No part of this document may be reproduced or transmitted in any form by any means, electronic, mechanical, photocopying, or otherwise, without the prior written permission of the author. Copies of this document may be made free of charge for use in public or nonprofit private educational institutions provided that permission is obtained from the author. Please indicate the name and address of the institution where use is anticipated. 1997 A.J. Girondi, Ph.D. 505 Latshmere Drive Harrisburg, PA 17109 alicechem@geocities.com Website: www.geocities.com/athens/oracle/2041 24-2 1997, A.J. Girondi
ACTIVITY 24.1 Redox Reactions in a Petri Dish Corrosion is a type of oxidation in which an uncombined metal reacts with an oxidizing agent to form compounds. Corrosion involves an oxidation reduction reaction. Rusting is a very common type of corrosion. In rusting, iron atoms lose electrons (usually 3) to become Fe 3+ ions. These iron (ferric) ions then react with oxygen and water to form iron (III) hydroxide, Fe(OH)3, which eventually reacts with more oxygen in the air to form the reddish brown compound we call rust (Fe2O3). Oxygen molecules are reduced (gain electrons). The half equations for these reactions are written below. oxidation: Fe ---> Fe 3+ + 3e - reduction: O2 + 2 H2O + 4e - ---> 4 OH 1- You can see, then, that for iron to corrode, both water and an oxidizing agent like oxygen are needed. Theoretically, you could prevent this kind of corrosion of iron or steel objects by keeping them in an environment free of water or oxidizing agents. In this activity you will observe the corrosion (oxidation) of some metals. The metals will be placed into a "gel" made by mixing agar with water. The oxidizing agent will be the dissolved oxygen (O2) contained in the water in the gel. Procedure: (Watch your time. About 40 minutes may be needed.) 1. Get 2 petri dishes and 4 iron nails. Obtain a strip of bare zinc wire or ribbon 4 to 5 cm in length, and a 5 cm piece of bare copper wire. 2. Prepare 100 ml of agar-agar solution by obtaining 1.0 g of powdered agar-agar. Heat 100 ml of distilled water to a boiling in a 250 ml beaker. Stop heating, and then slowly add the agar-agar to the hot water with constant stirring. Next, add 10 drops of 0.1 M K3Fe(CN)6 and 10 drops of phenolphthalein solution to the agar solution and stir again. 3. While the agar solution is cooling, set aside the nails for dish 1 which include straight nail "A" and a nail "B" bent to a very sharp 90 degree angle. Use pliers to bend nail B, sharply. 4. Next, prepare the nails for dish 2. Wrap nail "C" with the zinc metal wire and nail "D" with the copper wire as shown in Figure 24.1. The wire coils should not touch each other, and should extend the entire length of the nails. plain nails nail wrapped with copper wire Agar nail wrapped with zinc wire Figure 24.1 Redox in Petri Dishes 24-3 1997, A.J. Girondi
5. Using forceps or tongs, dip the 4 nails in a beaker of 1,1,1- trichloroethane (or another degreasing agent) to remove any oil or grease. Dry the nails with a towel and try not to touch them more than necessary after the degreasing. 6. Place nails A and B in dish 1, and nails C and D in dish 2. Do not allow the nails to touch each other or the edge of the dishes. 7. While keeping the nails well separated, slowly pour the agar-agar solution into the dishes to cover the nails. Do not cover or disturb the dishes until the solution forms a solid gel. 8. After the agar has solidified, cover the dishes and turn them upside down. If time allows, you can watch for any changes which may be occurring in the dishes. The results will be observed after 24 hours have passed. Before leaving class, place the dishes (upside down) in your lab drawer. 9. After the 24 hours have passed, observe the dishes and answer the questions below. Empty the contents of the dishes into a paper towel, and clean and dry the dishes. Return them to the materials shelf. Discussion: The chemicals you added to the gel will cause the color pink to appear wherever reduction has occurred and blue or green to appear wherever oxidation (corrosion) of iron has occurred. White areas indicate sites when zinc has been oxidized. Metals will be oxidized, while the dissolved oxygen, O2, in the water in the agar solution will be reduced. Questions: 1. List the locations on the nails where oxidation occurred. When nails are produced, the metal is drawn out into a wire. The wire is coated with a substance which inhibits oxidation. The wire is then cut into nails. Why do you think the oxidation occurred at the ends of the nails? 2. List the locations in the dishes where reduction occurred. Why do you think it occurred in those areas? 3. Any white areas around the zinc wire represent substances formed by the oxidation of zinc. Explain why there was no oxidation of the iron in the nail which was wrapped with zinc, while there was oxidation of the iron nail which was wrapped with copper wire. 24-4 1997, A.J. Girondi
4. Why do you think that zinc blocks are attached to the iron hulls of naval vessels? 5. Steel which is given a coating of zinc is said to be galvanized. List 3 or more possible uses for galvanized steel, since it does not corrode. SECTION 24.2 Introduction to Electrochemistry In the remainder of this chapter, we will be studying the role of oxidizing and reducing agents in a field of study known as electrochemistry. Electrochemistry is the study of chemical reactions that involve electric currents. Some redox reactions can be made to produce an electric current. Other redox reactions will not occur unless an electric current is passed through the reactants. When a chemical reaction occurs, there is usually a net increase or decrease in potential energy. In most cases, this energy takes the form of heat given off or absorbed from the surroundings. Occasionally, however, the change in potential energy appears as electrical energy. It is possible to produce an electric current from an oxidation-reduction reaction if it occurs spontaneously. In the previous chapter, you carried out an experiment with copper and zinc. This redox reaction involved energy changes; but, the experiment did not require that you measure or use any of this energy. By changing the set-up of the experiment, we can use the energy difference between the reactants and the products to produce an electric current. The secret is to allow the oxidation and reduction halfreactions to occur in containers that are connected in such a way as to complete an electrical circuit. This can be done by organizing your equipment into an arrangement called a voltaic cell. A spontaneous redox reaction produces an electric current when it occurs in a voltaic cell. Now let's take a detailed look at a typical voltaic cell. Just as redox reactions have two parts (two half reactions), a voltaic cell is also made of two parts (two half cells). Oxidation occurs in one half cell, and reduction occurs in the other half-cell. Look at the diagram of the voltaic cell in Figure 24.2 below. Porous Cup Cu Zn CuSO4(aq) Cu ZnSO4(aq) VOLTMETER Cathode Anode Figure 24.2 A Voltaic Cell 24-5 1997, A.J. Girondi
At first glance, a voltaic cell it may seem complicated; but, if you study it carefully, you will see that it is not. There are several ways to organize a voltaic cell. The one pictured above is only one example. First, look at the basic parts of the cell. Every voltaic cell must have two electrodes. An electrode is a surface, usually (but not always) made of a metal where oxidation or reduction occurs. The electrode where oxidation takes place is called the anode. The electrode where reduction takes place is called the cathode. In Figure 24.2, a piece of zinc metal serves as the anode, while the cathode is a copper strip. Notice the use of the word "where" in this paragraph. Sometimes it is useful to think of the anode and cathode as places rather than as things. For example, the surface of the anode is the place where oxidation occurs. Depending on the reaction, the anode material itself may or may not be involved in the reaction. In the cell pictured in Figure 24.2, the zinc anode actually is a reactant in the reaction, while the copper cathode is just the place where reduction of Cu 2+ ions in the solution will occur. The two half reactions in this cell are: Zn ---> Zn 2+ + 2e - Cu 2+ + 2e - ---> Cu oxidation reduction Notice that the beaker contains a solution of ZnSO4 which would be in the form of dissociated Zn 2+ and SO4 2- ions. The porous cup contains a solution of CuSO4 in the form of dissociated Cu 2+ and SO4 2- ions. The electrodes are attached to a voltmeter which measures the force with which electrons are moved through a circuit. They could just as well have been attached to a motor, a light bulb, or some other device that makes use of electricity. In a setup like this one, we are not really putting the electricity that is produced to a useful purpose. We are simply measuring the voltage produced. Everything in the cell has now been located except the electrons. The wires connected to the electrodes allow the electrons to move around the system, so we will have to follow them. Follow the path of the electrons through the voltaic cell as you read. Pathway of electrons: 1. Begin on the surface of the zinc electrode. The oxidation-reaction, Zn ---> Zn 2+ + 2e -, occurs here. A solid zinc atom loses two electrons and becomes a Zn 2+ ion which becomes part of the solution in the beaker. If you allow this oxidation to continue for a time, what would happen to the mass of the zinc electrode? {1} 2. The electrons move up the wire and pass through the voltmeter. You know that electrons are passing through the voltmeter if some voltage registers on the meter. 3. The electrons leave the voltmeter and continue through the wire and into the copper electrode. They move to the surface of the copper electrode where they meet the Cu2+ ions in the solution in the cup. A Cu 2+ ion will pick up two electrons to become a copper atom (the reduction reaction): Cu 2+ + 2e - ---> Cu. The copper atom that forms stays on the surface of the copper electrode as a "plating." This process happens every time a zinc atom loses electrons which pass through the circuit. If this reaction continues for a time, what will happen to the mass of the copper electrode? {2} Theoretically, the oxidation and reduction reactions should continue for as long as the zinc keeps giving up its electrons to the Cu 2+ ions. The reaction would stop when the half-cells have consumed either all of the zinc electrode or all of the Cu 2+ ions. 24-6 1997, A.J. Girondi
You may be wondering why the porous cup is needed. Porous means that certain things can pass through the walls of the cup. In this case, the SO4 2- ions can seep through its walls. If this did not happen in the cell, there would be an accumulation of positive charges in the zinc half cell as a result of zinc ions being produced and added to the solution in the beaker: Zn ---> Zn 2+ + 2e - In addition, there would be an accumulation of negative charges in the copper half cell as a result of Cu 2+ ions being reduced and removed from the copper solution in the porous cup: Cu 2+ + 2e - ---> Cu The movement of the SO4 2- ions from the copper solution to the zinc solution through the walls of the cup manages to keep both solutions electrically neutral. Electricity is nothing more than the movement of charge through a conductor such as a wire or a solution. In a wire the charge is carried by a flow of electrons, while charge is carried by moving ions in solutions. The redox reaction is producing electricity which we can measure and use. In physics we learn that a force is needed to move something or to do work. The force behind the flow of the electrons (which the voltmeter measures) can be used by us to do useful work. Actually this same redox reaction would occur if you simply drop a piece of zinc metal into a solution of Cu 2+ ions. However, the electrons would flow from the zinc to the Cu 2+ ions inside the solution. When that happens, we cannot "tap into the flow" of the electrons so that we can make use of the force. A voltaic cell arrangement causes the electrons to flow outside the cell where we can harness some of their energy. ACTIVITY 24.3 Construction and Testing of Voltaic Cells 1. Look back at Figure 24.2 and assemble the materials needed to construct a voltaic cell. 2. Clean one metal strip of copper and one of zinc, if needed, with a piece of steel wool or fine sandpaper. 3. Place the Cu electrode in the porous cup which should be about 2 / 3 full of CuSO4 solution. The zinc electrode should be placed in a 250 ml beaker which should be about 1 / 2 to 2 / 3 full of ZnSO4 solution. 4. Connect the electrodes to a voltmeter so that you get a positive voltage. If you are using a multimeter which has numerous scales, check with your teacher for the correct setting. If the meter needle goes the wrong way, or if you get a negative reading with a digital meter, reverse the wires. You should read the voltage immediately after you connect the meter. What reading do you get from the voltmeter? V The half reaction occurring at the anode is: Zn ----> Zn 2+ + 2e - Is this an oxidation or a reduction? {3}. The half reaction occurring at the cathode is: Cu 2+ + 2e - ----> Cu Is this an oxidation or a reduction? {4}. 24-7 1997, A.J. Girondi
The net reaction in this voltaic cell is: Zn + Cu 2+ ----> Zn 2+ + Cu Are the electrons moving from the zinc metal to the copper ions or from the copper ions to the zinc metal? {5}. What is the reducing agent in your cell? {6}. What is the oxidizing agent in your cell? {7}. Does the number of SO4 2- ions in the entire cell increase, decrease, or stay the same during the reaction? {8} Why? {9}. 5. While you still have all of your materials, measure the voltage produced in a cell where a different redox reaction is occurring. Pour the ZnSO4 solution out of the cell into the original container, because it can be reused. 6. Rinse the cell with water, and replace the zinc solution with a solution of lead nitrate, Pb(NO3)2. 7. Replace the zinc electrode with a lead strip (clean it if needed). 8. Keep the Cu electrode and CuSO4 solution the same. Connect the voltmeter and read the voltage.. The measured voltage is: V 9. The Pb(NO3)2 solution can also be reused, so pour it back into the original container. 10. The solution in the porous cup can be discarded. 11. Rinse the porous cup with water, and return all materials to the materials shelf. The half-equation for the reaction occurring at the anode is: Pb ----> Pb 2+ + 2e - Is this an oxidation or a reduction? {10}. The half-equation for the reaction occurring at the cathode is: Cu 2+ + 2e - ----> Cu Is this an oxidation or a reduction? {11}. The overall equation for the redox reaction in this voltaic cell is: Pb + Cu 2+ ----> Pb 2+ + Cu Are the electrons moving from the lead metal to the copper ions or from the copper ions to the lead metal? {12}. What is the reducing agent in your cell? {13}. What is the oxidizing agent in your cell? {14}. 24-8 1997, A.J. Girondi
SECTION 24.4 Predicting Voltages Of Voltaic Cells It is possible to perform a large number of experiments in which many different electrodes and solutions are used. Some combinations of electrodes and solutions produce large voltages, while other combinations produce no voltage at all. Measurements such as these have already been done by chemists. They have measured the contribution of each half reaction to the voltage of the cell. An arbitrary value of zero has been assigned to the hydrogen reduction half reaction: 2 H 1+ + 2e - ---> H2 Half-Cell Voltage (E o ) = 0.00 Scientists have combined many other half-reactions, one at a time, with hydrogen and measured the cell voltage. These measured values are called half cell potentials and are represented by the symbol E o. The potential of a half-cell representing a reduction is called a reduction potential and is given the symbol, E o red. The potential of a half-reaction representing an oxidation, is called an oxidation potential and is given the symbol, E o ox. The overall cell potential or cell voltage is equal to the sum of E o red and E o ox, and is given the symbol E o cell or simply E o. E o red + E o ox = E o A listing of many of these values is shown in Table 24.1. They are all listed as reduction potentials. The greater the value of a reduction potential, the greater the ability of a substance to be reduced (gain electrons). Table 24.1 is extremely useful. It allows you to predict the amount of voltage which will be produced by any combination of half-reactions. Let's say you wish to determine the voltage produced by the combination of magnesium and cesium in a voltaic cell. Begin by writing the two half-equations and their E o red values, taken from Table 24.1. Cs 1+ + 1e - ----> Cs E o red = 2.92 volts Mg 2+ + 2e - ----> Mg E o red = 2.37 volts The half-equations are both written as reduction reactions. You must decide which reaction to write as an oxidation. The substance with the smallest E o red value always undergoes oxidation. The substance with the largest E o red value undergoes reduction. To convert the Cs half-reaction into oxidation, you merely reverse the reaction and the sign in front of the E o red value which converts it into an E o ox value. Obtain the net reaction in the same way you did earlier in this chapter: 2 Cs ----> 2 Cs 1+ + 2e - E o ox = +2.92 V oxidation Mg 2+ + 2e - ----> Mg E o red = 2.37 V reduction 2 Cs + Mg 2+ ---> 2 Cs 2+ + Mg E o = +0.55 V overall reaction Notice that the oxidation equation has been multiplied by two so that the number of electrons lost in the oxidation half equation equals the number of electrons gained in the reduction half equation. However, you see that the E o ox value for that reaction was NOT multiplied by two. 24-9 1997, A.J. Girondi
Table 24.1 Standard Half-Cell Reduction Potentials Reduction---> <---Oxidation E o red (Volts) Li 1+ + e - <====> Li -3.00 Rb 1+ + e - <====> Rb -2.92 K 1+ + e - <====> K -2.92 Cs 1+ + e - <====> Cs -2.92 Ba 2+ + 2e - <====> Ba -2.90 Sr 2+ + 2e - <====> Sr -2.89 Ca 2+ + 2e - <====> Ca -2.87 Na 1+ + e - <====> Na -2.71 Mg 2+ + 2e - <====> Mg -2.37 Al 3+ + 3e - <====> Al -1.66 Mn 2+ + 2e - <====> Mn -1.18 Zn 2+ + 2e - <====> Zn -0.76 Cr 3+ + 3e - <====> Cr -0.74 Cd 2+ + 2e - <====> Cd -0.40 Fe 2+ + 2e - <====> Fe -0.44 Cr 3+ + e - <====> Cr 2+ -0.41 Stronger Co 2+ + 2e - <====> Co -0.28 Reducing Ni 2+ + 2e - <====> Ni -0.25 Agents Sn 2+ + 2e - <====> Sn -0.14 Pb 2+ + 2e - <====> Pb -0.13 ------------- 2 H 1+ + 2e - <====> H2(g) 0.00 ------------- Sn 4+ + 2e - <====> Sn 2+ +0.15 Cu 2+ + e - <====> Cu 1+ +0.15 Stronger Cu 2+ + 2e - <====> Cu +0.34 Oxidizing Cu 1+ + e - <====> Cu +0.52 Agents I2 + 2e- <====> 2 I 1- +0.53 Fe 3+ + e - <====> Fe 2+ +0.77 Hg 2+ + 2e - <====> Hg(l) +0.78 Hg2 2+ + 2e - <====> 2 Hg(l) +0.79 Ag 1+ + e - <====> Ag +0.80 Br2(l) + 2e - <====> 2 Br 1- +1.06 Cl2(g) + 2e - <====> 2 Cl 1- +1.36 Au 3+ + 3e - <====> Au +1.50 F2(g) + 2e - <====> 2 F 1- +2.87 Oxidizing Agents Reducing Agent Use the same procedure to write the overall reaction, and calculate the cell voltages for the pairs of elements listed in the problems which follow. Show the proper half-equations and E o values. 24-10 1997, A.J. Girondi
Problem 1. Ni and Al oxidation: E o ox = V reduction: E o red = V overall: E o = V Problem 2. Zn and Cl2 oxidation: E o ox = V reduction: E o red = V overall: E o = V Problem 3. Co and K oxidation: E o ox = V reduction: E o red = V overall: E o = V Problem 4. Mn and Al oxidation: E o ox = V reduction: E o red = V overall: E o = V Problem 5. Li and Au oxidation: E o ox = V reduction: E o red = V overall: E o = V Problem 6. F2 and Ag oxidation: E o ox = V reduction: E o red = V overall: E o = V 24-11 1997, A.J. Girondi
Problem 7. Mg and Au oxidation: E o ox = V reduction: E o red = V overall: E o = V ACTIVITY 24.5 BATTERIES As was mentioned earlier, there are a number of types of voltaic cells. These include the hydrogen/oxygen fuel cells used in space technology, and dry cells used in devices such as flashlights. We often misuse the term battery. A battery is actually two or more voltaic cells joined together. For example, the lead battery in your car most likely contains six cells linked together in an arrangement known as a "series." Each cell produces two volts, giving the battery a total of twelve volts. A standard nine volt battery such as you might use in a radio or a smoke detector is actually six 1.5 volt cells joined together (6 X 1.5 = 9). So, the nine volt battery really is a battery, but a 1.5 V dry cell is not. Dry cells get their name from the fact that they do not contain solutions like the voltaic cell did that you put together. Instead, they contain a moist paste. A typical carbon zinc dry cell contains a carbon rod in the center which is surrounded by a dark paste. This whole arrangement is enclosed in a zinc metal case. The zinc metal is the anode and the carbon rod is the cathode. The zinc gets oxidized, while manganese dioxide, MnO2, in the paste get reduced at the surface of the carbon rod. The half- reactions are: Zn(s) ----> Zn 2+ (aq) + 2e - E o ox = +0.76 V 2 MnO2(s) + 2 NH4 1+ (aq) + 2e - ---> Mn2O3(s) + 2 NH3(aq) + H2O(l) E o red =??? On the line below, write the overall redox equation that occurs in a dry cell: {15} E o = +1.50 V Problem 8. Since you know that a carbon-zinc dry cell produces 1.50 V, use the information given to calculate the half-cell potential for the reduction half-reaction shown above. Show the calculation. E o red = V Your teacher has some different types of dry cells which have been cut in half for you to examine. In the space below, sketch cross sections of a 1.5 V carbon-zinc cell and a 9 V carbon zinc battery. Label the anodes and cathodes. 1.5 V Cell 9 V Battery 24-12 1997, A.J. Girondi
SECTION 24.6 Spontaneous and Nonspontaneous Redox Reactions You have learned how to calculate the standard voltage (E o ) of a number of redox reactions. By "standard" we mean that any solutions in the cell are 1 M and any gases involved would be at 1 atm of pressure. If you go on to study more advanced chemistry, you will learn that changes in these conditions can change the voltage of a cell. Most of the reactions we have worked with so far have resulted in positive cell voltages. In general, when the standard voltage is calculated for a spontaneous redox reaction, the cell voltage will be a positive value. A spontaneous reaction happens on its own without us having to do anything to make it happen. The opposite is also true. If we calculate a negative standard voltage, then the redox reaction will be nonspontaneous. Reactions that are nonspontaneous require additional energy to proceed. If E o is positive, the reaction is spontaneous. If E o is negative, the reaction is nonspontaneous. You may recall an activity you performed in Chapter 23. You placed a piece of copper metal in an acid (H 1+ ) solution and nothing happened: Cu(s) + H 1+ (aq) ----> NO REACTION If something would have happened, it would have had to have been: Cu(s) + 2 H 1+ (aq) ----> Cu 2+ (aq) + H2(g) However, this reaction does not occur. Let's figure out why it doesn't. Problem 9. The half-equations for this redox reaction are given below. Look up the half cell reduction potential for the first one using Table 24.1 or Reference Table 11, and enter the data in the blank to the right. Since the half reaction involving Cu is an oxidation, use the same table to find the oxidation potential for this half reaction. Enter the value in the blank to the right. Add the two half cell potentials to find E o for the reaction, and enter the result in the proper blank below. 2 H 1+ + 2e - ----> H2 E o red = V Cu ----> Cu 2+ + 2e - E o ox = V Cu + 2 H 1+ ----> Cu 2+ + H2 E o = V What evidence exists to support the conclusion that the reaction represented by the equation above, is nonspontaneous: {16} Problem 10. Indicate whether the reactions below would be spontaneous or nonspontaneous. a. Fe + 2 H 1+ ----> Fe 2+ + H2 spontaneous nonspontaneous b. Ba + Pb 2+ ----> Ba 2+ + Pb spontaneous nonspontaneous c. 2 Ag + Cu 2+ ----> 2 Ag 1+ + Cu spontaneous nonspontaneous 24-13 1997, A.J. Girondi
Problem 11. An interesting nonspontaneous reaction involves the decomposition of water into hydrogen and oxygen gases. We can make this reaction happen by adding energy in the form of electricity. The half equations are listed on the next page. It is interesting because water is the reactant in both of the half reactions. In other words, some water molecules get oxidized and some get reduced as the reaction proceeds. The half cell potentials are given. Calculate the cell voltage, and enter the value in the space provided. oxidation: 2 H2O ----> O2 + 4 H 1+ + 4 e - E o ox = 1.23 V reduction: 2(2 H2O + 2 e - ----> H2 + 2 OH 1- ) E o red = 0.83 V sum: 6 H2O + 4e - ----> O2 + 2 H2 + 4 H2O + 4e - If we subtract some H2O and some electrons from both sides we get: net: 2 H2O ----> O2 + 2 H2 E o = V Referring to the E o value for the reaction, explain why energy is needed to decompose water: {17} ACTIVITY 24.7 The Electrolysis of Water The teacher will have set up equipment to demonstrate the decomposition of water using electricity. Fill in the blanks below as the teacher operates the equipment for you. 1. Turn on the power supply and slowly increase the voltage until you see gases being given off at the electrodes. Note the voltage as soon as evidence of a reaction is seen. Repeat this procedure several times, by turning the power down so that the voltage returns to zero, and then turning it back up again. When electricity is used to decompose molecules in this way, the process is known as electrolysis. Although your result will be just a rough estimate, you should note a relationship between the cell voltage you calculated in Problem 11 above, and the amount of voltage needed to cause this reaction to occur. In order to make this reaction happen, electricity must be supplied in excess of how many volts? {18} V The equipment pictured In Figure 24.3 is an example of an electrolytic cell. The reaction in an electrolytic cell is nonspontaneous, so we must add energy to make the cell function. Electrolytic cells have (positive/negative) {19} cell voltages (potentials). 2. Increase the voltage just high enough to allow the reaction to proceed at a good rate. Allow it to run long enough to permit you to determine the ratio of the volumes of the two gases being produced. What does the ratio appear to be? to. Based on the balanced equation for this reaction, does the observed ratio appear to be in agreement with the balanced equation for this reaction?. Explain: {20}. 24-14 1997, A.J. Girondi
water + acid gas gas POWER SUPPLY VOLTMETER Figure 24.3 Electrolysis Apparatus ACTIVITY 24.8 Electroplating Electrolytic cells are used to electroplate metals. For example, copper plating is done using a solution which contains Cu 2+ ions. By running a current through the solution the Cu 2+ ions are reduced to Cu and the Cu metal "sticks" to the object being plated which is one of the electrodes in the cell. Should the object being plated be the anode in the cell or the cathode? {21} In this activity you will be electroplating a coin with a layer of copper metal. Nickels, dimes or quarters seem to produce the best results. Procedure: 1. Have your teacher advise and assist you as you complete this activity. Obtain a coin and clean it well by washing it with soap and water. Place a few ml of 1 M HCl into a small beaker or flask and rinse the coin in it for a few minutes. Power Supply 2. Decant the HCl into the sink without dumping the coin. Rinse the coin with water. Decant the water. 3. Obtain a battery or power supply and two wire leads with alligator clips. Attach one clip to a strip of copper metal and the other clip to your coin. Try to cover as little of the coin as possible when you attach the clip. Solution of Cu 2+ ions Copper anode Object to be plated Cu 2+ Figure 24.4 Copper Plating 24-15 1997, A.J. Girondi
4. Place the copper strip and the coin into a small (100 or 150 ml beaker) as shown in figure 24.4 which illustrates silver plating. 5. Obtain the bottle of copper sulfate electroplating solution from the materials shelf, and pour enough of the solution into the beaker to completely submerse the coin. Handle the solution which also contains sulfuric acid with care. 6. Connect the battery or power supply to the wire leads making sure that the coin is connected to the cathode (the negative terminal). If you are using a power supply, connect the leads in the same manner and adjust the voltage to between 1 and 3 V. There may be a mark on the dial to help you with this. If not, use a voltmeter. Your instructor will assist you. DO NOT allow the copper strip to touch the coin. This will cause a short circuit. 7. Allow the current to flow for a few minutes. Pull the coin out of the solution every minute or so to check on the progress of the plating. Change the position of the clip and immerse the coin again. After a few minutes, the coin should be covered with a thin layer of copper. 8. Remove the coin and rinse with water. If polishing compound is available, put some on the coin and rub it between your fingers or with a soft moist cloth. Buff the coin to a nice shine with a dry rag. 9. Pour the plating solution back into the bottle from which it came. It can be reused many times. Rinse the copper strip and the beaker with water. Return all equipment to the materials shelf. The electrons gained by the copper ions in the solution turned them into copper atoms. This reaction occurred at the surface of the coin. These electrons came from the copper metal in the copper strip. Reaction at the coin (reduction): Cu 2+ (aq) + 2e - ----> Cu(s) Reaction at the copper strip (oxidation): Cu(s) ----> Cu 2+ (aq) + 2e - Even though the plating process removes Cu 2+ ions from the solution, the concentration of Cu 2+ ions in the plating solution remains constant. How is this possible? {22} What do you think is happening to the mass of the coin as it is being plated? {23} What do you think is happening to the mass of the copper strip as the plating of the coin goes on? {24} Why? {25}. SECTION 24.9 Learning Outcomes This concludes the discussion of redox processes. As you have seen, redox reactions are very important, indeed. Review the learning outcomes below. When you feel you have mastered them, arrange to take any quizzes or exams on Chapter 24, and move on to Chapter 25. 1. Explain how redox reactions and "cells" can be used to produce electrical currents. 2. Trace the pathway of electrons through a voltaic cell. 3. Explain the difference between voltaic and electrolytic cells. 4. Use a table of oxidation potentials to determine the overall voltage of a redox reaction. 24-16 1997, A.J. Girondi
SECTION 24.10 Answers to Questions and Problems Questions: {1} It would decrease; {2} It will increase; {3} oxidation; {4} reduction; {5} from zinc metal to copper ions; {6} zinc metal; {7} copper ions, Cu 2+ ; {8} Stays the same; {9} they are not involved in the reaction; {10} oxidation; {11} reduction; {12} from lead metal to copper ions; {13} lead metal; {14} copper ions; {15} Zn(s) + 2 MnO2(s) + 2 NH4 1+ (aq) ----> Zn 2+ (aq) + Mn2O3(s) + 2 NH3(aq) + H2O(l) {16} The cell voltage has a negative value; {17} Energy is needed because the reaction is nonspontaneous (has a negative cell voltage); {18} +2.06 V; {19} negative; {20} The ratio of H2 to O2 in the equation is 2:1, the same as the volume ratio of H2 to O2 when water is decomposed; {21} cathode; {22} The piece of Cu metal oxidizes to form Cu 2+ ions to replace those Cu 2+ ions which are plated out of the solution; {23} It is increasing in mass; {24} It is decreasing in mass; {25} Because it is being oxidized. Problems: 1. ox: 2(Al ----> Al 3+ + 3e - ) E o ox = +1.66 V red: 3(Ni 2+ + 2e - ----> Ni) E o red = 0.25 V overall: 3 Ni 2+ + 2 Al ----> 3 Ni + 2 Al 3+ E o = +1.41 V 2. ox: Zn ----> Zn 2+ + 2e - E o ox = +0.76 V red: Cl2 + 2e - ----> 2 Cl 1- E o red = +1.36 V overall: Cl2 + Zn ----> 2 Cl 1- + Zn 2+ E o = +2.12 V 3. ox: 2(K ----> K 1+ + e - ) E o ox = +2.92 V red: Co 2+ + 2e - ----> Co E o red = 0.28 V overall: Co 2+ + 2 K ----> Co + 2 K 1+ E o = +2.64 V 4. ox: 2(Al ----> Al 3+ + 3e - ) E o ox = +1.66 V red: 3(Mn 2+ + 2e - ----> Mn) E o red = 1.18 V overall: 2 Al + 3 Mn 2+ ----> 2 Al 3+ + 3 Mn E o = +0.48 V 5. ox: 3(Li ----> Li 1+ + e - ) E o ox = +3.00 V red: Au 3+ + 3e - ----> Au E o red = +1.50 V overall: 3 Li + Au 3+ ----> 3 Li 1+ + Au E o = +4.50 V 6. ox: 2(Ag ----> Ag 1+ + e - ) E o ox = 0.80 V red: F2 + 2e - ----> 2 F 1- E o red = +2.87 V overall: 2 Ag + F2 ----> 2 Ag 1+ + 2 F 1- E o = +2.07 V 7. ox: 3 (Mg ----> Mg 2+ + 2e - ) E o ox = +2.37 V red: 2(Au 3+ + 3e - ----> Au) E o red = +1.50 V overall: 3 Mg + 2 Au 3+ ----> 3 Mg 2+ + 2 Au E o = +3.87 V 24-17 1997, A.J. Girondi
8. 0.74 V carbon cathode Zn anode carbon cathodes 1.5 V Cell 9 V Battery zinc anodes 9. E o ox = 0.34 V E o red = 0 V E o = 0.34 V 10. a. spontaneous; b. spontaneous; c. nonspontaneous 11. 2.06 V 24-18 1997, A.J. Girondi