A Cycle of Copper Reactions

Transcription

1 EXPERIMENT 5 A Cycle of Copper Reactions PURPOSE To observe a sequence of reactions of copper that form a cycle, along with the color and physical property changes that indicate those reactions. To gain skill in recording observations and interpreting them in terms of chemical equations. To use a simple classification scheme for grouping chemical reactions by reaction type. To practice quantitative laboratory techniques by determining the percentage of the initial sample of copper that is recovered. PRE-LAB PREPARATION To a beginning student of chemistry, one of the most fascinating aspects of the laboratory is the dazzling array of sights, sounds, odors, and textures that are encountered there. Among other things, we believe that this experiment will provide an interesting aesthetic experience. You will be asked to carry out a series of reactions involving the element copper and to carefully observe and record your observations. The sequence begins and ends with copper metal, so it is called a cycle of copper reactions. Because no copper is added or removed between the initial and final steps, and because each reaction goes to completion, you should be able to quantitatively recover all of the copper you started with if you are careful and skillful. This diagram shows in an abbreviated form the reactions of the cycle of copper: HNO 3 Cu B Cu(NO 3 ) 2 B Cu(OH) (1) (2) 2 B Zn, HCl (5) CuSO 4 H 2 SO 4 b (4) NaOH CuO heat (3) Like any good chemist, you will probably be curious to know the identity of each reaction product and the stoichiometry of the chemical reactions for each step of the cycle. Here they are, numbered to correspond to the steps shown in the chemical equation. 1 8HNO 3 (aq) 3Cu(s) O 2 (g) B 3Cu(NO 3 ) 2 (aq) 4H 2 O(l) 2NO 2 (g) (1) Cu(NO 3 ) 2 (aq) 2NaOH(aq) B Cu(OH) 2 (s) 2NaNO 3 (aq) (2) Cu(OH) 2 (s) B CuO(s) H 2 O(l) (3) CuO(s) H 2 SO 4 (aq) B CuSO 4 (aq) H 2 O(l) (4) CuSO 4 (aq) Zn(s) B ZnSO 4 (aq) Cu(s) (5) These equations summarize the results of a large number of experiments, but it s easy to lose sight of this if you just look at equations written on paper. You can easily be overwhelmed by the vast amount of information found in this lab manual and in chemistry textbooks. It is in fact a formidable task to attempt to learn or memorize isolated bits of information that are not reinforced by your personal experience. This is one reason why it is important to have a laboratory experience. Chemistry is preeminently an experimental science. As you perform this and any other experiment, watch closely and record what you see. Each observation should be a little hook in your mind on which you can hang a more abstract bit of information, such as the chemical formula for the compound you are observing. 1 Labels specify the states of the reactants and products: (s) means a solid, (l) means a liquid, (g) means a gas, and (aq) means an aqueous (water) solution. 71

2 72 A CYCLE OF COPPER REACTIONS It is also easier to remember information that is organized by some conceptual framework. Observations and facts that have not been assimilated into some coherent scheme of interpretation are relatively useless. It would be like memorizing the daily weather reports when you have no knowledge of or interest in meteorology. Chemists look for relationships, trends, or patterns of regularity when they organize their observations of chemical reactions. The periodic table, which groups the elements into chemical families, is a product of this kind of thinking. Each element bears a strong resemblance to other members of the same chemical family but also has its own unique identity and chemistry. In a similar fashion, it is useful to classify reactions into different types. Because no one scheme is able to accommodate all known reactions, several different kinds of classification schemes exist. A simple classification scheme we will use at the beginning is one based on ideas of precipitation (ion combination), acid-base (proton transfer), and redox (electron transfer). Here we present an outline and some examples of this kind of classification: A SIMPLE SCHEME FOR CLASSIFYING CHEMICAL REACTIONS 1. PRECIPITATION REACTIONS (THE COMBINA- TION OF POSITIVELY CHARGED IONS WITH NEGATIVELY CHARGED IONS TO FORM AN INSOLUBLE NEUTRAL COMPOUND THAT PRE- CIPITATES FROM SOLUTION). If we add a solution of sodium chloride (NaCl) to a solution containing silver nitrate (AgNO 3 ), an insoluble white solid forms. The solid, called a precipitate, is silver chloride (AgCl), and we may write the following chemical equation to describe the reaction, using symbols to represent the substances in solution. Na (aq) Cl (aq) Ag (aq) NO 3 (aq) B AgCl(s) Na (aq) NO 3 (aq) First we eliminate the ions called spectator ions, which appear on both sides of the equation but do not participate in the precipitation reaction. What is left is the net ionic equation: Ag (aq) Cl (aq) B AgCl(s) solid AgCl precipitate is easily separated from the solution containing the soluble sodium nitrate salt. If we desired, we could also recover the sodium nitrate by evaporating the water from the solution. 2. ACID-BASE (PROTON TRANSFER REACTIONS). An acid is a substance that reacts with water to form hydronium ions (H 3 O ) by transfering a proton to a water molecule. (a strong acid) HCl(g) H 2 O B H 3 O (aq) Cl (aq) (a weak acid) CH 3 COOH(aq) H 2 O j H 3 O (aq) CH 3 COO (aq) HCl is a strong acid, completely dissociated in aqueous solution, while acetic acid (CH 3 COOH) is a weak acid that only partially dissociates into hydronium ion and acetate ion. A base is a substance that forms hydroxide ions when dissolved in water: (a strong base) NaOH(s) B Na (aq) OH (aq) (a weak base) NH 3 (g) H 2 O j NH 4 (aq) OH (aq) NaOH is a strong base, completely dissociated in aqueous solution, while NH 3 is a weak base, only partially dissociating into ammonium ion and hydroxide ion. Acids react with bases to form salts and (usually) water. Both are neutral compounds, being neither strongly acidic nor strongly basic. So acid-base reactions are also called neutralization reactions. Two examples follow: (a strong acid a strong base) H 3 O (aq) Cl (aq) Na (aq) OH (aq) B 2 H 2 O Na (aq) Cl (aq) (the net ionic equation) H 3 O (aq) OH (aq) B 2 H 2 O (a weak acid a weak base) CH 3 COOH(aq) NH 3 (aq) B NH 4 (aq) CH 3 COO (aq) Note that in these two examples of acid-base reactions a proton is transferred from the acid to the base. The net ionic equation concisely summarizes the net result of mixing the two solutions: the formation of an insoluble precipitate when a positively charged silver ion is combined with a negatively charged chloride ion. The 3. REDOX REACTIONS (ELECTRON TRANSFER REACTIONS). Oxidation-reduction reactions, called redox reactions, are reactions that involve the shift or transfer of electrons from one kind of atom to another. In

3 A CYCLE OF COPPER REACTIONS 73 some reactions, the transfer is obvious, as in this reaction: Mg(s) 2H 3 O (aq) 2Cl (aq) B H 2 (g) Mg 2 (aq) 2Cl (aq) Here, each magnesium atom is giving up two electrons to two hydrogen ions, forming magnesium ion and hydrogen gas. Sometimes, the transfer of electrons between atoms is less obvious, as in the following reaction: 2SO 2 (g) O 2 (g) B 2SO 3 (g) Here, the reactants and products are all gases, and no ions are formed. In classifying this reaction as an oxidationreduction reaction, we use the concept of assigning oxidation numbers (also called oxidation states) to each atom in the compound. A simple set of rules defines the procedure for assigning the oxidation number. For a simple binary compound (a compound composed of two different elements), we imagine that all of the electrons in the chemical bonds are assigned to the atoms with the greatest affinity for electrons. The ability of an atom to attract electrons to itself is called electronegativity, and the atoms having the greatest electronegativity are found in the upper right hand corner of the periodic table, with fluorine having the most. In sulfur dioxide, SO 2, and sulfur trioxide, SO 3,we imagine that all of the electrons in the S O bonds are assigned to the O atoms, giving each oxygen atom a full valence shell; this formally gives each oxygen atom a net charge of 2. So if oxygen is assigned oxidation number 2, the sulfur in SO 2 must have oxidation number 4 and the oxidation number of S in SO 3 would be 6, since the sum of the oxidation numbers on all the atoms must add up to the net charge on the molecule (zero, in this case). The oxidation number of atoms in their elemental form is always assigned zero. This seems reasonable for O 2 because the oxygen atoms are equivalent so that there would be no tendency for one oxygen atom to transfer electrons to its partner in the O 2 molecule. Once we have assigned oxidation numbers to each element in the chemical reaction, we will see that in this particular reaction the oxidation number of the sulfur atoms increases from 4 to 6 while the oxidation number of the oxygen atoms decreases from zero (in O 2 ) to 2 (in SO 3 ). From this viewpoint, the change in oxidation number is formally equivalent to transferring electrons from sulfur to oxygen. We say that the sulfur atoms have been oxidized (because their oxidation number increases), while the oxygen atoms in O 2 have been reduced (because the oxidation number of oxygen atoms decreases from zero in O 2 to 2 in SO 3 ). We must be careful to note, however, that the oxidation numbers we assign do not necessarily represent the real distribution of electronic charge in the molecule. By assigning the oxidation number according to fixed rules, we have artificially assigned integer changes in oxidation number to particular atoms (sulfur and oxygen in this case), but the changes in the electron density on the sulfur and oxygen atoms may not be as large as implied by the assigned oxidation numbers. Nevertheless, it is reasonable to suppose that the sulfur atom in SO 3 is more positive than the sulfur atom in SO 2 because the added oxygen atom would tend to draw electrons away from the sulfur atom. 4. DECOMPOSITION REACTIONS (A SUBSTANCE BREAKING DOWN INTO SIMPLER SUBSTANCES UNDER THE INFLUENCE OF HEAT). Although many chemical reactions can be classified into one of the three reaction types we described earlier, it is possible to find examples of chemical reactions that do not neatly fit into this scheme. For example, when calcium carbonate is heated, it breaks down into simpler substances: CaCO 3 (s) B CaO(s) CO 2 (g) This reaction might be called a decomposition or dissociation reaction. Any substance heated to a sufficiently high temperature will begin to decompose into simpler substances, so this kind of reaction is common. DOES A COMPREHENSIVE CLASSIFICATION SCHEME EXIST? If we searched we would find examples of other reactions that do not fit into these four categories. Indeed, it is probably fair to say that there is no completely comprehensive classification scheme that would accommodate all known chemical reactions. However, many, if not most, of the chemical reactions described in your general chemistry text will fit into this simple scheme. As you carry out each step of the cycle of copper reactions, think about what is happening in each reaction, and try to fit it into one of the four categories we described. EXPERIMENTAL PROCEDURE Special Supplies: Infrared lamps or steam baths, porcelain evaporating dish. Chemicals: 18- to 20-gauge copper wire, concentrated (16 M) HNO 3, 3 M NaOH, 6 M H 2 SO 4, 30-mesh zinc metal, 6 M HCl, methanol.! SAFETY PRECAUTIONS: Concentrated nitric acid, HNO 3, is hazardous. It produces severe burns on the skin, and the vapor is a lung irritant. When you handle it, you should use a fume hood while wearing safety glasses (as always) and rubber or polyvinyl chloride gloves. A polyethylene

4 74 A CYCLE OF COPPER REACTIONS FIGURE 5-1 If a fume hood is not available, substitute this apparatus. A waste container should also be provided for the methanol used to dry the product in Step Cu TO Cu(NO 3 ) 2. Cut a length of pure copper wire that weighs about 0.5 g (about a 10-cm length of 20- gauge copper wire). If it is not bright and shiny, clean it with steel wool, rinse it with water, and dry it with a tissue. Weigh it to the nearest milligram, recording the weight in your laboratory book. Coil the wire into a flat spiral, place it in the bottom of a 250-mL beaker, and in the fume hood add 4.0 ml of concentrated (16 M) nitric acid, HNO 3. (If a fume hood is not available, use the apparatus shown in Figure 5-1.) Record in your notebook a description of what you see. Swirl the solution around in the beaker until the copper has completely dissolved. What is in the solution when the reaction is complete? After the copper has dissolved, add deionized water until the beaker is about half full. Steps 2 through 4 can be conducted at your lab bench. 2. Cu(NO 3 ) 2 TO Cu(OH) 2. While stirring the solution with a glass rod, add 30 ml of 3.0 M NaOH to precipitate Cu(OH) 2. What is formed in the solution besides Cu(OH) 2? Record your observations in your lab book. 3. Cu(OH) 2 TO CuO. Stirring gently with a glass rod to prevent bumping (a phenomenon caused by the formation of a large steam bubble in a locally overheated area), heat the solution just barely to the boiling point over a burner, using the apparatus shown in Figure 5-2. If the solution bumps you may lose some CuO, so don t neglect the stirring. Record your observations. When the transformation is complete, remove the burner, continue squeeze pipet can be useful for transferring the HNO 3 from a small beaker to your 10-mL graduated cylinder. Rinse your hands with tap water after handling HNO 3. The dissolution of the copper wire with concentrated HNO 3 should be carried out in a fume hood. If no hood is available, construct the apparatus shown in Figure 5-1 to substitute for the fume hood. The brown NO 2 gas that is evolved is toxic and must be avoided. NaOH solutions are corrosive to the skin and especially dangerous if splashed into the eyes wear your safety glasses. Methanol and acetone are flammable and their vapors are toxic. Use them in the hood to avoid breathing the vapor, and keep them away from all open flames. WASTE COLLECTION: The supernatant solution that is decanted in Step 5 contains zinc sulfate and zinc chloride and should be collected. Warm gently, only as necessary FIGURE 5-2 Setup for heating Cu(OH) 2 to convert it to CuO. CuO suspension

5 A CYCLE OF COPPER REACTIONS 75 stirring for a minute or so, then allow the CuO to settle. Then decant (pour off) the supernatant liquid, being careful not to lose any CuO. Add about 200 ml of hot deionized water, allow to settle again, and decant once more. What is removed by this washing and decantation process? 4. CuO TO CuSO 4. Add 15 ml of 6.0 M H 2 SO 4, while stirring. Record your observations. What is in solution now? Now transfer operations back to the fume hood. 5. CuSO 4 TO Cu. In the fume hood, add all at once 2.0 g of 30-mesh zinc metal, stirring until the supernatant liquid is colorless. What happens? What is the gas produced? When the evolution of gas has become very slow, decant the supernatant liquid, and pour it into the waste container provided. If you can see any silvery grains of unreacted zinc, add 10 ml of 6 M HCl and warm, but do not boil, the solution. When no hydrogen evolution can be detected by eye, decant the supernatant liquid, and transfer the copper to a porcelain dish. A spatula or rubber policeman is helpful for making the transfer. Wash the product with about 5 ml of deionized water, allow it to settle, and decant the wash water. Repeat the washing and decantation at least two more times. Move to the hood, away from all flames. Wash with about 5 ml of methanol, allow to settle, and decant. Dispose of the methanol in the proper receptacle. Place the porcelain dish under an infrared lamp or on a steam bath or hot plate, and dry the copper metal. What color is it? Using a spatula, transfer the dried copper metal to a preweighed 100-mL beaker and weigh to the nearest milligram. Calculate the mass of copper you recovered by subtracting the weight of the empty beaker from the weight of the beaker plus the copper metal. CALCULATION OF PERCENTAGE RECOVERY. Express the percentage of copper recovered as percentage recovery mass of recovered copper 100% initial mass of copper wire If you are careful at every step, you will recover nearly 100 percent of the copper you started with. Consider This If you used a penny as the source of your original copper in this experiment, would it matter if you used a pre penny (essentially pure copper) or a post-1982 penny (copper cladding over a zinc core)? Describe what would happen in each step if you used a post-1982 penny. Test your predictions. Why would it be hard to perform a cycle of oxygen or a cycle of hydrogen experiment similar to this cycle of copper exercise? Can you design an experimental apparatus for a cycle of oxygen lab? BIBLIOGRAPHY Bailar, Jr., J.C. A Further Improvement on the Copper Cycle Experiment, J. Chem. Educ., 1983, 60: 583. Condike, G. F. Near 100% Yields With the Cycle of Copper Reactions Experiment, J. Chem. Educ., 1975, 52: 615. Todd, D. and Hobey, W.D. An Improvement in the Classical Copper Cycle Experiment, J. Chem. Educ., 1985, 62: 177. Umans, T. and de Vos, W. An Improved Copper Cycle Experiment, J. Chem. Educ., 1982, 59: 52.

6

7 REPORT 5 A Cycle of Copper Reactions NAME SECTION INSTRUCTOR LOCKER DATE DATA AND CALCULATIONS 1. Initial mass of copper wire g 2. Mass of beaker plus dry copper g 3. Mass of empty beaker g 4. Mass of recovered copper g 5. Percentage recovery percentage recovery mass of recovered copper 100% initial mass of copper wire % Show your calculation. EQUATIONS AND OBSERVATIONS For each step of the cycle, write the products of the reaction and balance the chemical equation(s). Using the classification scheme presented in the Pre-Lab Preparation, write the reaction type (combination, decomposition, or single/double replacement). Also record your observations of what happens at each step, and answer the questions posed earlier in the Experimental Procedure. 1. Cu HNO 3 O 2 B What is in the solution after reaction is complete? Observations (Be sure to include color and texture changes that occur in each step.) 77

8 2. Cu(NO 3 ) 2 NaOH B HNO 3 NaOH B What is formed in the solution besides Cu(OH) 2? Observations: 3. Cu(OH) 2 heat ßßßßB What is removed by the washing and decantation process at the end of Step 3? (Consider the products of the reaction as well as reagents from previous steps.) Observations: 4. CuO H 2 SO 4 B What is in the solution at the end of Step 4? Observations: 78

9 REPORT 5 Sheet 2 NAME 5. Zn CuSO 4 B Zn H 2 SO 4 B Zn HCl B What happens when the zinc is added? What is the gas produced in the reaction? What is removed by the washing and decantation near the end of Step 5? What color is the recovered copper? Observations: 79

10 QUESTIONS 1. Describe whether the error introduced by each of the following problems would result in a high or a low value for the Cu recovery, or would not affect the results. (a) Some of the copper nitrate solution is splashed out of the beaker in Step 1. (b) Insufficient NaOH is added in Step 2. (c) The solution bumps in Step 3, splashing out some CuO. (d) Some solid is lost in the decanting process. (e) An excess of H 2 SO 4 is added in Step 4. (f) Some unreacted zinc remains with the product at the end of the experiment. 80

11 REPORT 5 Sheet 3 NAME (g) The washings in Step 5 are insufficient to remove all of the solution residues from the copper. (h) The copper is not completely dried. 2. How many milliliters of 3.0 M NaOH are required to react with 4.0 ml of 16 M HNO 3? 3. How many milliliters of 3.0 M NaOH are required to react with 0.5 g of Cu 2 to form Cu(OH) 2? 4. (a) Add together the results you calculated for Questions 2 and 3 and compare the sum with the milliliters of 3.0 M NaOH added in Step 2. (b) Is there an excess of NaOH above that required to react with the HNO 3 and Cu 2? (c) What would be the effect on the percentage recovery if not enough NaOH were added in Step 2 to react with both the unreacted HNO 3 and the Cu 2 present at the end of Step 1? 81

12 Consider This If you used a penny as the source of your original copper in this experiment, would it matter if you used a pre-1982 penny (essentially pure copper) or a post-1982 penny (copper cladding over a zinc core)? Describe what would happen in each step if you used a post-1982 penny. Test your predictions. Why would it be hard to perform a cycle of oxygen or a cycle of hydrogen experiment similar to this cycle of copper exercise? Can you design an experimental apparatus for a cycle of oxygen lab? 82