5 ELECTRON TRANSFER REACTIONS

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1 5 ELECTRON TRANSFER REACTIONS 5.1 TERMINOLOGY In a reaction between an acid and a base, a proton is transferred from acid to base. When an electron (or a number of electrons) is transferred from one species to another, the reaction is known as an oxidationreduction or redox reaction. Such reactions are very common, both in the laboratory for analysis of various species, or industrially, in the manufacture of chemicals, or economically, as in the undesirable process of corrosion. All redox reactions are, in theory, reversible, ie. they are equilibrium processes. The name oxidation is derived from the observations by chemists of centuries ago that certain materials gained oxygen during a reaction, e.g. rusting of iron to iron oxide. They didn't know anything about electrons, but could observe this common type of reaction. The opposite reaction, where species lost oxygen during a reaction, e.g. heating up copper oxide to form copper metal, they called reduction because the amount of oxygen in the substance had been reduced. Later studies more closely defined the two processes (in terms of electrons), and showed that there were plenty of redox reactions where there wasn't a single oxygen atom in sight. The species that gains the electron(s) is reduced, while the species that loses the electron(s) is oxidised. It should be emphasised that the processes are indivisible: the electrons lost in the oxidation must be gained by another species which is then reduced. A simple, nonanalytical example of oxidation and reduction is the reaction of metals with acids. When dropped into a HCl, zinc begins to dissolve, and hydrogen gas is given off. The zinc does not really dissolve, but in fact loses electrons (is oxidised) and becomes the ion, Zn 2+ (which is soluble), while the protons in solution gain those electrons (are reduced) and forms hydrogen gas. In acidbase reactions, the terms acid and base were used to describe the two species involved in the proton transfer. Whereas in redox, the terms you have met so far oxidation and reduction refer to the process. So what do we call the species involved? This is where it becomes a bit confusing! Referring back to our original example, the zinc, by losing electrons, causes the hydrogen ions to be reduced. Therefore, the zinc is called a reducing agent or reductant: a species that causes another to be reduced. On the other hand, the hydrogen ions, by gaining the electrons, cause the zinc to be oxidised. Thus, the hydrogen ion is called an oxidising agent or oxidant: a species that causes another to be oxidised. Why are these definitions possibly confusing? Consider what happens to the reducing agent it causes something else to be reduced, by getting itself oxidised. The reverse applies for the oxidising agents. 5.2 IDENTIFYING ELECTRON TRANSFER PROCESSES Recognising an oxidationreduction reaction is not necessarily as easy as one involving an acidbase reaction, but there are some general rules which help. The following changes definitely mean oxidation or reduction: elements changing into compounds or ions, or vice versa, e.g. Zn Zn 2+, H + H 2, C CO 2, or a change in valency by a monatomic species, e.g. Fe 2+ Fe 3+ But there are plenty of other reactions which are reduction or oxidation that don't fit these two rules, e.g. MnO 4 Mn 2+, and others, e.g. CaCO 3 CaCl 2, that might look candidates that aren't. So we need a better system. This is known as oxidation numbers. Chemistry 2

2 PRACTICE QUESTION 1. Which of the following processes are definitely redox reactions (without use of oxidation numbers)? (a) Sn Fe 2+ Sn Fe 3+ (b) HCl + NaOH H 2 O + NaCl (c) I + Fe 3+ I 2 + Fe 2+ (d) 2MnO 4 + 6H + + 5C 6 H 8 O 6 2Mn H 2 O + 5C 6 H 6 O 6 (e) ClO 4 + C 2 H 6 CO 2 + H 2 O + Cl (f) O 2 + H 2 O + Fe Fe(OH) 2 Calculating Oxidation Numbers The oxidation state of each element (a bit like the valency, but it applies to each element within a compound or ion) is the means of distinguishing redox reactions from others: if the oxidation state of an atom changes during a reaction, then it has been reduced or oxidised. The oxidation state is determined by the oxidation number system, which follows a set of simple rules, as below: The oxidation number of an atom when present in the elemental form is zero. The normal oxidation number of hydrogen in compounds is +1, except when it is present as hydride ion (H ), when it is 1. The normal oxidation number of oxygen in compounds is 2, except when it is present as peroxide ion (O 2 2 ), when it is 1. The sum of oxidation numbers of the atoms in a compound or ion is equal to the charge on the species. EXAMPLE Consider the following examples of a series of sulfur species: (a) sulfur (S 8 ), (b) sulfide, (c) sulfur dioxide, (d) sodium sulfate, (e) sulfuric acid, (f) thiosulfate (S 2 O 3 2 ). What is the oxidation number (ON) of sulfur in each of these species? (a) Sulfur is present as the elemental form, and therefore the oxidation number of sulfur is 0. (b) In sulfide, the one sulfur atom has a charge of 2. Therefore, the ON. is 2. (c) In sulfur dioxide, the overall charge is 0. Therefore, the sum of the ON of the one S and two O is zero. Since oxygen has an ON of 2 and there are two of them, the sulfur has an ON of +4. (d) In sulfate, the overall charge is 2. Therefore, the sum of the ON of sulfur and 4 oxygens (4 x 2) = 2. The ON of sulfur is +6. (e) In sulfuric acid, the sum is zero. Therefore, 2 x +1 + ON + 4 x 2 = 0. The ON of sulfur is + 6. (f) In thiosulfate, 2 x ON + 3 x 2 = 2. Therefore, ON of sulfur is +2. Sulfur is clearly capable of existing in many different oxidation states. The higher (more positive) oxidation number of an element, the more oxidised it is. Therefore, sulfur is most oxidised in sulfate and sulfuric acid (ON = +6), and most reduced in sulfide (ON = 2). How then do we use the oxidation number system in recognising redox reactions, and more importantly, what has been oxidised and reduced? The following simple rules help in the decision: If the ON of two atoms in a reaction change, then a redox reaction is indicated. If the ON of an atom within a species increases, the species has been oxidised. If the ON of an atom within a species decreases, the species has been reduced. Chemistry 2 5.2

3 PRACTICE QUESTIONS 2. Determine the oxidation number of the nonoxygen element in each of the following: (a) ClO 3, (b) P 2 O 7 2, (c) N 2 O 3, (d) MnO 4 2, (e)fe 2 O 3, (f)cu 2 O, (g)c 2 H 6 O. 3. Calculate the oxidation number of nitrogen in the following species, and determine which is the most oxidised and most reduced species: NO, NO 3, N 2, N 2 O, NO 2, N 2 O Determine the oxidation numbers of all elements in the reactions in Q1. Identify any other redox reactions, and the reducing and oxidising agents in each case. 5.3 REDOX EQUATIONS Clearly, a redox reaction occurs in two stages: the reductant gives up its electron and then the oxidant accepts it. These two halves of the reaction could be represented separately in equation, as below using the Zn/HCl example. Electrons become reactants or products in these equations Zn Zn e Oxidation = Electrons as products 2H + + 2e H 2 Reduction = Electrons as reactants Breaking up the reaction into its two halves is useful for a variety of reasons: firstly, we can see clearly what is happening, and secondly, we can use the information about what is happening when zinc is oxidised in any reaction where it is being oxidised. These equations are called halfequations, because they describe exactly that: either the reduction or oxidation part of the overall process. TABLE 5.1 Common half equations Ag + Cr 2 O 7 2 Cu 2+ + e Ag + 14H + + 6e 2Cr H 2 O + 2e Cu Fe e Fe Fe 3+ + e Fe 2+ 2H + + 2e H 2 H 2 O 2 + 2e + 2H + I 2 + 2e 2I MnO 4 MnO 4 2H 2 O + 4H + + 3e MnO 2 + 2H 2 O + 8H + + 5e Mn H 2 O O 2 + 2H 2 O + 2e 4OH Pb 2+ SO 4 2 S 4 O e Pb + 4H + + 2e SO 2 + 2H 2 O + 2e 2S 2 O 3 2 Sn e Sn 2+ Zn e Zn 2CO 2 + 2e C 2 O 4 2 Chemistry 2 5.3

4 You might notice that all these halfequations are written in the reduction form, i.e. with electrons on the lefthand side. However, this is simply a convention, and the doubleheaded arrows mean that any of these can be reversed. Obviously, when using these to describe an overall redox reaction, one of the two halfreactions must be an oxidation, i.e. reversed. Using Half Equations There is nothing special about a redox equation. It can be balanced in exactly the same way as any equation; it is just that some are more complicated than others. You wouldn't need half equations to write the balanced equation for the reaction of good old zinc and hydrochloric acid, but if balancing equations isn't your strong point, you might struggle to balance some of the more complex examples. The tables of half equations also mean that the other reactants and products involved don t have to be remembered. The steps to be followed in using half equations to produce a balanced overall equation are: 1. Identify the oxidant and reductant involved. 2. Locate half equations that have these species as reactants (on the left hand side). 3. Work out the correct half equation for each species, by knowledge of products and ensuring that one reduction and one oxidation reaction is involved (two reductions or two oxidations can t be right). 4. Add the two equations together, ensuring that no electrons remain in the overall equation. To achieve this, use the crossmultiplication method: multiply one equation completely by the number of electrons in the other half equation, and then do likewise for the opposite combination. 5. Cancel out common species on both sides. The reaction conditions (e.g. acidic, alkaline) can affect the reaction. For example, permanganate can form at least three species when it is reduced, depending on the solution. Remember that reactants have to end up on the left hand side of the final equation, but won't necessarily be there in the halfequations from the table, since they are all in reduction form. If this is the case, you must reverse the equation, taking all species on the left hand side and moving to the right hand side, and vice versa. EXAMPLE Write the balanced equation for the reaction of permanganate reacting with iron (II) in acidic solution. Looking at Table 5.1, there are two occurrences of permanganate and two of iron (II), as below. Fe e Fe Fe 2+ Fe 3+ + e (reversed to make iron (II) a reactant) MnO 4 MnO 4 + 8H + + 5e Mn H 2 O + 4H + + 3e MnO 2 + 2H 2 O In the case of permanganate, in acidic solution, it is more likely to form Mn 2+, not MnO 2. Therefore, the appropriate manganese halfequation is MnO 4 + 8H + + 5e Mn H 2 O. Which iron (II) equation is the right one? Reduction and oxidation go hand in hand. If permanganate is reduced, the iron (II) must be oxidised. Therefore, the correct half equation is Fe 2+ Fe 3+ + e. Chemistry 2 5.4

5 To produce the overall equation, cross multiply. x 1 MnO 4 + 8H + + 5e Mn H 2 O x 5 5Fe 2+ 5Fe e Overall MnO 4 + 8H + + 5Fe 2+ Mn H 2 O + 5Fe 3+ PRACTICE QUESTION 5. Write the balanced overall equations for the following reactions. (a) iodide and dichromate (b) oxalate and permanganate in acidic solution (c) hydrogen peroxide and iodide ion (d) thiosulfate and iodine (e) cerium (IV) ion and iron (II) (f) oxalate and sulfate 5.4 THE ACTIVITY SERIES OF METALS While redox processes are reversible, they only occur spontaneously in one direction. If an iron nail is dropped into a solution of copper ions, a copper coating appears instantly on the nail, and invisible to us, iron atoms are oxidised to ions and dissolve into the solution. if, however, a piece of copper is dropped into a solution of iron (II), nothing happens. Clearly, some metals are more readily oxidised (e.g. iron) than others, and the ions of others more readily reduced (e.g. copper). Placing the metals in order of ease of oxidation is known as the activity series of metals, as shown in Figure 5.1. Increasing ease of reduction of M + Gold Platinum Silver Copper Lead Tin Nickel Iron Zinc Increasing ease of oxidation of M FIGURE 5.1 The activity series of metals The higher up the activity series, the less likely a metal is to be oxidised. Said another way, if metal X is placed in contact with a solution of the ions of metal Y, a reaction will only occur if metal X is lower on the table. Hydrogen is often added to this list, even though it is not a metal. It fits in this list between copper and lead. We will see shortly a less obvious reason for its inclusion, but for the moment, suffice to say, that we can tell whether a metal will react with HCl or H 2 SO 4 (be oxidised by H + ) by whether it is above or below hydrogen: those above will not, those below will. Chemistry 2 5.5

6 EXAMPLES 1. What will happen when an iron nail is dipped into a copper (II) solution? Since iron is lower than copper in this activity series, iron will be oxidised to iron (II) and the electrons will reduce copper ions to copper metal. A copper coating will appear on the nail. 2. Can HCl dissolve copper? Copper is higher on the table than hydrogen. Therefore, copper will not be dissolved in HCl. PRACTICAL WORK You are required to test the reactions of silver, copper, zinc, iron and lead in solutions of the ions of these metals (use iron (II), not iron (III)). This need only be done in test tubes. Ensure that the metals are clean by polishing with emery paper. In some cases, changes will be immediately apparent while still in the test tube. Where there is no visible change, the piece of metal will have removed, and wiped with paper towel to see whether any metallic residue is removed, which will be a sign of a reaction. Q1. Draw up a table for your results, with the metals in columns and the ions in rows. Indicate positive or negative reactions clearly (e.g. + and ). Q2. Can see find a sequence of reactivity for the metal ions, i.e. did one metal ion react with all the other metals, did one metal ion react with no metals? PRACTICE QUESTIONS 8. Compare the sequence of reduction halfreactions for metal ions in Table 5.1 with those in the activity series. (a) Where would gold and platinum fit in Table 5.1? (b) Where would the dividing line for metals affected or not affected by oxygen be drawn? (c) Suggest why platinum and gold are called noble metals. 9. Predict what would happen if an iron nail was exposed to a copper salt in the presence of moisture. 5.5 VOLTAGE AND ELECTRON TRANSFER When a species is oxidised, it transfers electrons to the oxidant. This, at a molecular level, is like a flow of electrical current. After all, what is electrical current but a flow of electrons from one place to another. In a redox reaction in a beaker, the electrons go directly from molecule to another. If, however, we could separate the two processes reduction and oxidation so that electrons had to flow between one beaker and another via a wire, then we have the capacity to generate a measurable (and useful) electrical current. In a modified form, all our batteries (in your calculator, car etc) fit this description: a chemical reaction producing electrical current. In the laboratory, how would such a setup work? Figure 5.2 shows a typical apparatus, using copper, zinc and solutions of their ions. Table 5.2 defines some key terminology. Chemistry 2 5.6

7 connecting wire Zn Pt Zn 2+ connection to complete circuit H + FIGURE 5.2 A typical electrochemical cell Thus, electrons flow from the zinc electrode through the external wire to the platinum where they are used to reduce H + to H 2. Over time, the following observations could be made: the zinc will lose mass, the concentration of zinc ions in solution will increase, hydrogen gas will bubble off, and the concentration of hydrogen ions in solution will decrease. TABLE 5.2 Important terminology Electrochemical cell Half cell Electrode A redox reaction where the halfreactions are separated and the electrons are forced to travel through external connections in the form of current One of the halves of an electrochemical cell a conducting material through electrons flow in electrochemical reactions there must always be two electrodes PRACTICE QUESTION 10. Identify the following components of the apparatus in Figure 5.2: electrochemical cell each halfcell each electrode Voltage Measurements Electrons flow in an electrochemical cell because of a voltage (or potential) difference between one electrode and another. Electrons flow from the electrode with the greater ability to lose them. It is like water in a bucket half way up a hill. If you tip the bucket over, the water runs downhill, not uphill, because of the laws of gravity. Some species have a greater need to lose electrons than others. The potential (or ability) of a halfcell to lose or gain electrons can be measured, but not by itself. Another halfcell must be present to complete the circuit. This halfcell has its own potential, so any measurements of potential (voltage) can only be relative between the two halfcells. If this causes you a problem, imagine yourself standing on a wall between two bodies of water (the two halfcells). You have a measuring stick (a voltmeter), which can reach the surface of each body of water. You want to measure the depth of each body of water, but your measuring stick isn t long enough. However, you can measure the difference in their depths (potential difference). Figure 5.3 illustrates this idea. Chemistry 2 5.7

8 height difference FIGURE 5.3 The depth of water analogy for potential difference Electrode potentials are dependent on concentration, which means their measurement can be used to tell something about the composition of a solution. Voltage measurements are among the most important methods used in analytical instrumentation to determine the composition of solutions. 5.6 APPLICATIONS OF ELECTRON TRANSFER REACTIONS Chemical analysis makes considerable use of redox reactions and their associated electrochemical measurements. Potassium permanganate and sodium thiosulfate are important redox titrants. Permanganate, an oxidant, is used to analyse iron levels in samples, once the iron has been prereduced to iron (II). Thiosulfate, a reductant with some peculiar properties, only reacts usefully with elemental iodine. However, a number of species can oxidise iodide ion to elemental iodine, allowing the titration with thiosulfate. You have met a ph meter and electrode in this and other modules. Simply put, it and the solution being tested form an electrochemical cell. It produces a reading by an electrochemical response at the electrode surface. Other socalled ionselective electrodes have been built to perform the same function, but which are selective for different ions, such as fluoride and sodium. Industrially, electrolysis (the forced reversal of nonspontaneous electrochemical processes) is an important method of producing a number of otherwise difficult chemicals. For example, aluminium metal is obtained from the reduction in molten conditions of an aluminium fluoride compound. Sodium metal and chlorine gas can be produced by the electrolysis of molten sodium chloride. Water cannot be present in either of these processes because it will be preferentially used up before the desired materials. However, the most important commercial application of electrochemical processes is the battery. Whether it is a battery that runs your watch, your calculator or your car, it produces its power from chemical reactions between two electrodes. The only differences between a commercial battery and the zinc/hydrogen cell used to illustrate the key points in Section 5.5 are the chemicals involved and the design (to minimise the amount of solution slopping around). You may have heard of the terms aerobic and anaerobic. They describe conditions (especially in water) where oxygen is plentiful or absent, respectively. Aerobic conditions are oxidising because of the oxygen, while anaerobic conditions are reducing. This will affect the fate of chemicals that find their way into the different environments. For example, carboncontaining materials (eg dead plants) are oxidised to carbon dioxide in aerobic conditions, but reduced to methane and other hydrocarbons in anaerobic environments this is how the oil deposits formed. Not all redox processes are desirable. Corrosion is the oxidation of metals by atmospheric oxygen gas or by other oxidising agents: rusting of iron is the most common example. Large amounts of money are invested in attempting to minimise the effects of this process. Space does not allow further discussion of this process here. The redox potential of a solution or soil (known as the ORP) is an important measure of its potential corrosiveness, and is simply measured with a voltmeter. Chemistry 2 5.8

9 WHAT YOU NEED TO BE ABLE TO DO define terminology associated with redox reactions distinguish the reactive species in redox reactions write balanced half and overall equations for redox processes draw a complete electrochemical cell and explain the function of its components explain how a voltage difference applies between halfcells outline important applications of redox reactions Chemistry 2 5.9