GCC CHM 130LL: Reactivity of Metals page 1 of 6 CHM 130LL: Reactivity of Metals Introduction In a single-replacement reaction, a solid metal “replaces” the cation of another metal in solution. For example, in the following reaction, Cd (s) + CuCl2(aq) → CdCl2(aq) + Cu (s) the Cd “replaces” the Cu+2 in the solution, so copper comes out in its natural form, Cu (s), and Cd becomes Cd+2 in solution with Cl–. To understand single-replacement reactions, you must make a careful distinction between the free metal and the metal cation. The free metal still has all of its original electrons; it has no charge because the number of protons is equal to the number of electrons. The free metal is never part of a compound. It exists as a metallic-looking solid or a coating (if it plated out). When the free metal loses electrons, it becomes positively charged because the number of protons (which does not change) is greater than the number or electrons (since some electrons are lost). The positively charged metal cation is shown as being part of an ionic compound in solution. In the following single-replacement reaction, 2 Al (s) + 3 CdCl2 (aq) → 2 AlCl3 (aq) + 3 Cd (s) the free metals are Al on the reactant side and Cd on the product side. The metal ions are Cd+2 in the CdCl2 compound on the reactant side and Al+3 in the AlCl3 compound on the product side. Since the ionic compounds in these reactions are all soluble, the physical state is shown as aqueous, (aq). Metals differ in their tendency to release valence electrons. Some metals give their electrons a forceful “push” while other metals can only manage a weaker “push.” The chemical name for this tendency to release electrons is “activity.” The activity series is the arrangement of elements according to their ability to release electrons. The activity series is shown below: Li > K > Ba > Sr > Ca > Na > Mg > Al > Mn > Zn > Fe > Cd > Co > Ni > Sn > Pb > H2 > Cu > Ag > Au For a single-replacement reaction to occur, the free metal must be more active—i.e. be higher on the activity series—than the metal cation in solution. Note that in the following reaction, 2 Al (s) + 3 CdCl2 (aq) → 2 AlCl3 (aq) + 3 Cd (s) Al is higher than Cd on the activity series. Thus, it can force its electrons onto the Cd+2 ion to make it plate out as Cd (s) while the Al atoms go into solution as Al+3 cations. (Note: The formula for aluminum chloride, AlCl3, is based on the charges for aluminum ion, Al+3, and chloride ion, Cl–.) If the free metal is less active than the metal cation in solution, then it is not strongGCC CHM 130LL: Reactivity of Metals page 2 of 6 enough to replace the cation in solution, and no reaction occurs. For example, consider what happens when a piece of silver, Ag (s), is placed in a CdCl2 solution. Since Ag is lower on the activity series than Cd, it is not strong enough to replace Cd+2 in solution. Consequently, no reaction occurs, as shown below: Ag (s) + CdCl2 (aq) → NR where NR means “no reaction.” Even though hydrogen is a nonmetal, it can also behave like a metal under certain conditions. For example, like a metal, it can lose an electron to form H+. Compounds of H+ in solution are called acids—e.g. HCl(aq) is hydrochloric acid. Metals that are higher than hydrogen on the activity series react with acids to replace hydrogen in solution. Hydrogen is released in its natural form, H2(g). Consider the reaction between Cd metal and hydrochoric acid, HCl(aq): Cd(s) + 2 HCl(aq) → CdCl2(aq) + H2 (g). The free metal is Cd on the reactant side, but the product is hydrogen gas, H2(g). When the hydrogen gas is produced in this type of single-replacement reaction, bubbles are observed instead of a metal plating out. Note that all the metals above hydrogen on the activity series will react with any acid in a similar manner. In this experiment, • You will observe how each free metal reacts with metal solutions and an acid solution. • You will rank each metal based on its reactivity to develop an activity series. • You will identify the four unknown metals by comparing the experimentally determined activity series with the accepted Activity Series. You will be given four metals labeled A, B, C, and D, as well as four solutions of each metal cation also labeled A, B, C, and D. You will mix each solid metal with the metal cation or acid solutions and observe the results. You will record the results of each test in your Data Table. Using your results, you will make an abbreviated “activity series” indicating the free metal (A, B, C, or D) with the most reactions as the most active, then the next most active metal, etc., with the free metal having the least number of reactions as the least active. Use your activity series to match up the letters (A, B, C, or D) with the element symbols in the activity series provided to identify each metal. Next, include hydrogen in your experimentally determined activity series. Be sure to place hydrogen below all the free metals that react with the acid solution but above any metals that do not react with the acid solution. Afterwards, you will write balanced chemical equations for all the single-replacement reactions that did occur.GCC CHM 130LL: Reactivity of Metals page 3 of 6 Laboratory Technique Safety Information Procedure The experiment will be carried out in microplates. Follow the directions below: 1. To use the metals, hold the small metal strip and sandpaper both sides of it thoroughly. Then cut it into small squares as needed. 2. Add a few drops of liquid—enough to cover the metal. You need not completely fill the well on the microplates. (Note: You do not need to test the metal with its corresponding solution—i.e. you do not need to test free metal A in the metal cation A solution.) Allow each metal to react with each solution for at least 3-5 minutes. 3. If necessary, observe reactions through magnifying glasses. You will be looking for a plating reaction—a darkening or coating of the metal’s surface. In the case of the metals in HCl(aq) solution, you will be looking for bubbles of hydrogen gas. Note that HCl is the only hydrogen-producing solution. Bubbles in any other solution are due