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Oxidation / Reduction Experiment 9 Oxidation / Reduction Electron Transfer Reactions Expt 9 Redox.wpd INTENT In this experiment you will focus on the atomic model interpretation of the most easily recognized chemical reactions of common experience, namely the oxidation/reduction reactions. Key Terms and Concepts (familiarize yourself with them): C conjugate oxidizing agent & reducing agent relationship C oxidation & reduction (oxidation states) C oxidizing agent & reducing agent C metal & nonmetal activity series DISCUSSION Most people have no trouble recognizing the combustion of fuels, the rusting of iron, or the digestion of foods by an organism as examples of chemical reactions. These reactions are all examples of redox reactions. Reduction-oxidation (redox) reactions create new compounds by exchanging electrons and thereby fundamentally changing the reacting particles. These kinds of reactions will be examined in this experiment. Before we look at any specific reactions, let us define some of the basic terms of redox reactions. Originally the term oxidation described what happens to any substance that reacted with oxygen, O2, and in so doing lost electrons. The term has now been generalized so that: Oxidation is any process which increases the oxidation state of an atom due to the loss of electrons. Reduction is any process which reduces the oxidation state of an atom due to the gain of electrons. The actual species involved in these reactions are the oxidizing agents and the reducing agents. By taking on electrons the oxidizing agent causes something else to be oxidized (lose electrons). C The oxidizing agent is the electron acceptor and it is reduced. By giving up electrons, the reducing agent causes something else to be reduced (gain electrons). C The reducing agent is the electron donor and it is oxidized. These processes may seem very confusing at first. To avoid confusion we treat redox reactions as those in which electrons are transferred from the reducing agent to the oxidizing agent. 121 In order to understand the stoichiometry of redox reactions we will focus on the number of electrons gained by the oxidizing agent and lost by the reducing agent. We will need to identify how many moles of electrons are given up by each mole of reducing agent and how many moles of electrons are accepted by each mole of oxidizing agent. Since mass and energy are neither created nor destroyed in a chemical reaction, the number of electrons given up by the reducing agent must be the same as the number accepted by the oxidizing agent. That is, they must exchange an equal number of electrons. Thus their coefficients in the chemical equation must represent equivalent quantities of electrons. For example, consider the reaction of Zn and Cu2+ ion in solution: ! 2+ & oxidation: Zn (s) Zn (aq) + 2e & reduction: Cu2+ + 2e ! Cu (aq) (s) 2+ ! 2+ net reaction: Zn (s) + Cu (aq) Cu (s) + Zn (aq) In the example above the electrons lost in the oxidation equation equal the electrons gained in the reduction equation. Sometimes, in order for the number of electrons accepted to be equal to the number of electrons given up, the reactions must be multiplied by a whole number. For example, consider the reaction of F2 and H2O: & ! & oxidation: 2e + F2 2 F ! + & reduction: 6 H2O O2 + 4 H3O + 4e In order for the transfer to be equal, the first reaction must be multiplied by two before adding the two half-reactions to get the net equation. & ! & 2 (2 e + F2 2 F ) 6 H O ! O + 4 H O+ + 4e& 2 2 3 ! & + Net reaction: 2 F2 + 6 H2O 4 F + O2 + 4 H3O Note: 4 e&s absorbed = 4 e&s ejected Not every species has the same attraction for electrons. This is why electrons can be gained or lost. Even though we may sometimes refer to certain species as "wanting to give up electrons," substances do not give up electrons willingly. A species loses or gains electrons to achieve a lower energy state. Note these generalizations: The greater the tendency for a substance to take on electrons, the less its tendency, once having taken on the electrons, to give them back. That is, the stronger the oxidizing agent is, the weaker its conjugate reducing agent will be (and vice versa). The stronger oxidizing agent and stronger reducing agent will always react (eventually) to produce a weaker oxidizing agent and reducing agent. The weaker oxidizing agent and the weaker reducing agent will not react. & For the F2 and H2O reaction, F2 and O2 are the oxidizing agents; F and H2O are the reducing agents. For the reaction to proceed as written, F2 must be a stronger oxidizing agent than O2. 122 Since metals are good reducing agents, few metals are found pure in nature. So far you have learned that metals tend to lose electrons. The result of this is that very few metals are truly stable in our environment. What in the environment will take on these electrons? Some metals react with H2O to produce H2 gas. Those same metals and many additional ones may react with O2 and moisture in the air. Note that O2 in the presence of acid is a better oxidizing agent than just O2 and water. Consequently, it is easy to understand the concerns about acid emissions from combustion reactions that result in "acid rain." Rust formation on the surface of iron is actually the product of a reaction of iron with oxygen and moisture from the air. For this reason you would not want to make water pipes out of iron. You would want to make water pipes out of something which is less reactive under these conditions, like copper. The following equations illustrate the difference: ! 2 Fe (s) + 3 O2 (g) 2 Fe2O3 (s) But, ! Cu (s) + O2 (g) no appreciable reaction Therefore, iron is a more reactive metal, or is "more active," than copper. In other words, iron is a better reducing agent than copper. The reason that many metals even exist in our environment has to do with the rate at which they react. Many electron-transfer reactions are fast, but those which involve covalent bond making or breaking may be slow. (More about this in General Chemistry II.) The rate may also be affected by any oxide coating on the metal surface, which prevents further reaction. We will be looking at the relative abilities of substances to attract electrons. If we consider substances two at a time, we can establish a relative order of reactivity for them. From this approach, we can develop a complete order of reactivity, known as the "activity series." Those species which are higher on the list of reducing agent are "more active." How can we experimentally determine order of reactivity? In this experiment you will use observations to determine an order of activity, first for the metals, then the nonmetals. Some metals react with water or acid in water to produce H2. The evolution of hydrogen gas will prove which metal is more reactive. Before you start, give some thought to experimental procedure. Should you try to keep the size and shape of all the metals pieces approximately the same? Why? How will you distinguish between air bubbles on the metal surface and H2 gas being produced? Let' s inspect the reaction of K metal in water. The reaction is very exothermic. The H2 actually ignites when we toss a small piece of K in the water. ! + & 2 K (s) + 2 H2O (R) 2 K (aq) + 2 OH (aq) + H2 (g) reducing agent oxidizing agent conjugate oxidizing agent conjugate reducing agent Since both K and H2 are reducing agents and the reaction occurs as written (and not the other way around), we can state that K is a stronger reducing agent than H2. Potassium would be placed above hydrogen gas in an activity series. Note that if we try the reverse reaction, nothing will happen since hydrogen gas is less reactive. This conjugate relationship is very useful because it allows us to gain the 123 same information from the experiment no matter which reaction direction of the reaction we choose to investigate. Note that the reducing agents on both sides are being compared, not the reducing/oxidizing pair. As you saw in the example above, if one adds a very reactive metal to pure water, the metal reduces the & +n water to OH and H2 as the metal is oxidized to M . ! +n & M (s) + H2O M + OH (aq) + H2 (g) What happens to nonmetals, such as the halogens? If one adds Cl2, a very strong oxidizer, to pure water, + & the Cl2 oxidizes the water to H3O and O2 as it is reduced to Cl . So pure water can commonly function as either an oxidizer or a reducer. We will study several reductions of water. However, when we use chlorine water, the expected oxidation of water by Cl2 is slow (Cl-Cl bonds and O-H bonds are broken while O-O bonds are formed). The oxidation of water is not a factor in the fast reactions between & halogens (X2) and halide ions (X ). In this experiment you will discover the reactivity of halogens by reacting aqueous halogen solutions with halide solutions. For example, chlorine water will be mixed with bromide ions from a sodium bromide solution. Does the reaction written below occur? & ! & Cl2 (aq) + 2 Br (aq) 2 Cl (aq) + Br2 (aq) How will you know? You will again use visual clues. Unfortunately there is no bubbling in this reaction to help you out. So you will use the color of the halogens dissolved in hexane to tell you which halogen is present.
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