Oxidation-Reduction Reactions and Introduction to Electrochemistry

ADVANCED PLACEMENT CHEMISTRY Oxidation-Reduction Reactions and Introduction to Electrochemistry Students will be able to: identify oxidation and reduction of chemical species; identify oxidants and reductants describe the nature of a redox reaction use the activity series of the metals to predict the outcome of a proposed redox reaction balance redox reactions by the half-reaction ion method characterize species as oxidizers or reducers, and predict the redox nature of species, including predicting the products of potential oxidations and reductions relate oxidation and reduction ability to the periodic table describe the fundamentals of an electrochemical cell, and identify the essential characteristics of a cell determine the cell potential for a given cell determine reduction potentials for half-reactions used in cells propose and construct spontaneous electrochemical cells use shorthand notation for describing electrochemical cells identify the nature of batteries and how their structure relates to their function This booklet covers the material in Assignment Sets F and G. This booklet includes material from Chapters 4 and 20. 41

Part 1: OXIDATION-REDUCTION REACTIONS Understanding chemistry requires a complete understanding of oxidation-reduction reactions, which are reactions that occur with a transfer of electrons from one species to another. For example, look at the reaction here, which occurs in acid solution: 2 H + (aq) + Zn(s) Zn 2+ (aq) + H2(g) During a redox reaction, one species increases its oxidation number, while another species decreases its oxidation number. The species that decreases its oxidation number is said to have been reduced, while the species that increases its oxidation number is said to have been oxidized. All redox reactions have both an oxidized and reduced species. A reduction is the result of gaining electrons the increase in electrons provides a less positive value for the oxidation number. That is, the oxidation number is reduced. This is the reduction half-reaction. An oxidation is the result of losing electrons the decrease in electrons provides a more positive value for the oxidation number. That is, the oxidation number is increased. This is the oxidation half-reaction. +1 0 2 H + (aq) + Zn(s) Zn 2+ (aq) + H2(g) 0 +2 In this example equation, the oxidation number of zinc has increased from zero to positive two; zinc has been oxidized, which is due to the loss of electrons. The oxidation number of hydrogen has gone from positive one to zero; hydrogen has been reduced, which is due to the gain of electrons. In this case, zinc is the reducing agent (reductant) a substance that loses electrons, and hydrogen is the oxidizing agent (oxidant) a substance that gains electrons. The vocabulary can be quite difficult to sort here s a succinct statement that uses all of these new redox terms: A reducing agent loses electrons to reduce the oxidation number on another species it is itself oxidized. An oxidizing agent gains electrons to increase the oxidation number on another species it is itself reduced. The terms oxidized and reduced refer to the oxidation numbers reduction involves a decrease, while oxidation involves an increase. There are two particular reactions that are of interest currently: the oxidation of metals by acids and the oxidation of metals by salts. The processes that occur are as described above: one chemical species is oxidized (a pure metal), and another species is reduced (either hydrogen ion or a metal ion). 42

Oxidation of metals by acids The oxidation of a metal by an acid results in the formation of a positive metal ion and hydrogen gas. Many of the reactions that you will perform in lab are of this type. One of the most common is the oxidation of magnesium metal by hydrochloric acid: Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) (molecular equation) Mg(s) + 2 H + (aq) Mg 2+ (aq) + H2(g) (net ionic equation) So, what s happening? You now know that in a redox reaction one species is reduced while another is oxidized. In the case above, valence electrons of magnesium are being transferred to hydrogen ion, resulting in the reduction of hydrogen from +1 to 0 as H2. Table 2. Activity Series of Selected Elements. Elements reduction equations are shown in this table. Also shown are the elements reduction potentials. Relative to the reduction of hydrogen ion to hydrogen, a reduction potential provides a relative measure of the ease with which a particular reduction will occur. Magnesium metal is being oxidized from 0 to +2 as it loses electrons. Oxidation of metals by salts Reduction Element reaction Ered Element Reduction reaction Ered Li Li + + e Li 3.05 Fe Fe 2+ + 2e Fe 0.440 L K + + e K 2.925 Ni Ni 2+ + 2e Ni 0.28 Ba Ba 2+ + 2e Ba 2.90 Sn Sn 2+ + 2e Sn 0.136 Ca Ca 2+ + 2e Ca 2.87 Pb Pb 2+ + 2e Pb 0.126 Na Na + + e Na 2.71 H 2H + + 2e H2 0.00 Mg Mg 2+ + 2e Mg 2.37 Cu Cu 2+ + 2e Cu +0.337 Al Al 3+ + 3e Al 1.66 Ag Ag + + e Ag +0.799 Mn Mn 2+ + 2e Mn 1.18 Hg Hg 2+ + 2e Hg +0.854 Zn Zn 2+ + 2e Zn 0.763 Pt Pt 2+ + 2e Pt +1.188 Cr Cr 3+ + 3e Cr 0.74 Au Au 3+ + 3e Au +1.52 The oxidation of a metal by a salt occurs when a more active metal is placed in a solution containing a less active metal. The less active metal is reduced; the more active metal is oxidized (Fig. 14). The discussion of active in the sense of metals and redox reactions fundamentally involves a discussion of the ease with which an elemental metal sample will lose electrons: a more active metal will readily lose electrons to a less active metal. That is, a more active metal will be oxidized by a less active metal, and a less active metal will be reduced by a more active metal. We shall now turn our attention to discussing how we can quantitatively measure the activity of metals. 43

Figure 14. Zinc is more active than copper. When we place a piece of Zn in a solution of CuSO4, the zinc atoms on the metal strip will lose their electrons to the copper atoms in solution, and solid copper will deposit on the strip; the less active metal undergoes reduction, while the more active metal undergoes oxidation. Reduction, Cu 2+ (aq) + 2 e- Cu(s), is occurring in the top of the figure, and oxidation, Zn(s) Zn 2+ (aq) + 2 e-,, is occurring at the bottom of the figure. REDUCTION POTENTIALS, Ered We will simplify our discussion by simply stating that there is a potential energy difference between elemental metal and its ions in solution. We can compare this potential energy difference for any metal and its ions to that of the potential energy difference between hydrogen ion and elemental hydrogen to obtain a relative potential energy difference, which is measured in volts, V. You can see in Table 2 that the potential energy difference or simply, the potential for 2H + + 2e - H2 is zero, which is a defined potential. Moreover, we can further simplify the discussion of potentials by always considering the potential for a reduction reaction, even if a reaction occurs as an oxidation. This provides us the following: A standard reduction potential measures the electromotive force of a particular reduction relative to the reduction of hydrogen ion. A more negative reduction potential indicates that the force behind a particular reduction is less than that of another, and a more positive reduction potential indicates that the force behind a particular reduction is greater than that of another. Thus, the more positive a reduction potential, the more likely a reaction is to occur as a reduction. Thus, if one considers a pair of reduction reactions, the reduction that has the more positive reduction potential value will occur as a reduction, while the reaction with the more negative reduction potential will occur as an oxidation; i.e., in a battle between two reductions, the reduction with a more positive reduction potential wins. This provides the basis for the activity series of the metals, which allows us to predict whether or not a reaction will occur notice that the most active metals have large negative reduction potential values, which corresponds to the likelihood of undergoing oxidation. 44

Practice 3.1 Consider a solid piece of lead placed into a solution containing the copper(ii) ion. We are interested in determining whether or not the lead will reduce copper(ii) ion into elemental copper and itself be oxidized to lead(ii). That is, will the following reaction occur? Pb(s) + Cu 2+ (aq) Pb 2+ (aq) + Cu(s) First, determine the values of each reduction s potential, and then compare the values of Ered. It is wise to write both reactions as reductions to ensure you look up the correct potential. Consider a solid piece of Ag placed into a solution containing the chromium(iii) ion. Will a chemical reaction occur? In this case, first write the proposed chemical reaction before beginning. Practice 3.2 A solution contains the iron(ii) ion. Will the addition of either zinc or nickel cause the removal of iron(ii) ion from solution? Be sure to write all proposed reactions before beginning. Clearly show that the use of reduction potentials supports your answer. 45

Practice 3.3 Use reduction potentials to determine whether each of the reactions below will occur. Write the proposed reaction for each first, and write the net ionic equation for any reaction that occurs. Justify your response in each case with reduction potentials. 1) iron is dropped in an aqueous solution of nickel nitrate 2) to a solution of lead(ii) nitrate, a piece of silver metal is added 3) hydrogen gas is bubbled through a solution of copper(ii) chloride 4) barium is added to cold water 5) platinum is placed in hydrochloric acid Consider a solution containing silver ion, copper(i) ion and tin(iv) ion. Can you propose a series of solid metal additions that will selectively precipitate each ion from the solution? That is, how could you use reduction potentials to separate this mixture? (Assume that any new ions that form are removed by another method after their formation.) 46

E = Ered(reduction) Ered(oxidation) We conclude this statement from the observations: If the value of E (the potential difference) between the reduction and the oxidation is positive, then the reaction will occur spontaneously; otherwise, the reaction will not occur spontaneously. Notice, importantly, that we use the reduction potential when applying this equation whether the process occurs as a reduction or an oxidation. Part 2: OXIDIZING AND REDUCING AGENTS The relative values of Ered can tell us whether or not species will serve as oxidizing agents or as reducing agents; we have already said that the more positive the Ered value for a half-reaction, the greater the tendency for the reactant of the half-reaction to be reduced and, therefore, to oxidize another species. For example, recall the oxidation of zinc by hydrogen: 2 H + (aq) + Zn(s) Zn 2+ (aq) + H2(g) In this reaction, zinc is oxidized (Zn Zn 2+ ), while hydrogen is reduced (H + H2). The value of Ered for the reduction of hydrogen ion to hydrogen atom is 0.00 V; the value of Ered for the reduction of zinc ion to zinc atom (the reverse of the oxidation of zinc atom) is -0.76 V. Because the value 0.00 V is more positive than the value -0.76 V, we can see that hydrogen is more likely to be reduced, which is to be an oxidizing agent. Oxidizing Agents strong oxidizers have more positive values for Ered. Oxidizing agents are species that can cause the oxidation of another species; strong oxidizing agents are easily reduced. A strong oxidizer must be able to take electrons from another species thus, the halogens, oxygen and many oxyanions (like permanganate ion, nitrate ion, dichromate ion) are strong oxidizers. They are easily reduced to their reduced forms: fluorine atom fluorine ion, chlorine atom chlorine ion, and dichromate ion chromium(iii) [from chromium(vi)]. Other common oxidizing agents include peroxides and some metal ions. Strong oxidizers are weak reducers. Reducing Agents strong reducers have less positive values for Ered. Reducing agents are species that can easily lose electrons to other species. Among common strong reducers are the metal atoms, which can easily lose electrons to reduce another species; the metal atoms are themselves oxidized. Strong reducing agents are weak oxidizers. Redox and the Periodic Table It is possible to use the periodic table to qualitatively discuss the oxidation or reduction character of elements: species that can accept electrons from another species are oxidizers, as they can be reduced. Species that can provide electrons to another species are reducers, as they can be oxidized. Besides just free element, consider whether or not one of the species in a compound or ion can accept or donate electrons, too. For example, perhaps a metal ion in a compound can be reduced or oxidized to a different ion. 47

COMMON OXIDIZING AND REDUCING AGENTS Common oxidizing agents these species will cause a loss of electrons in another species, and they will themselves gain electrons. Oxidizing agents undergo reduction. MnO4 - in acid solution Mn 2+ MnO2 in acid solution Mn 2+ MnO4 in neutral or base solution MnO2(s) Cr2O7 2- in acidic solution Cr 3+ HNO3 (conc) NO2 HNO3 (dilute) NO H2SO4 (hot, conc) SO2 metallic ions metallous ions (lower oxidation state) free halogens halide ions Na2O2 NaOH HClO4 Cl H2O2 H2O S2O8 2- SO4 2- CrO4 2- Cr 3+ Common reducing agents these species will cause a gain of electrons in another species, and they will themselves lose electrons. Additional reducing agents will be added. Reducing agents undergo oxidation. halide ions free halogens free metals metal ions sulfite ions or SO2 sulfate ions nitrite ions nitrate ions free halogens, dilute basic solution hypohalite ions free halogens, conc basic solution halate ions metallous ions metallic ions (higher oxidation state) C2O4 2- CO2 MnO2 in base solution MnO4 You must learn to recognize the relationships between ionization energy, electron affinity, valence structure, reduction potentials and atomic character in order to evaluate problems surrounding electrochemistry and reactions this is not a time for memorization; rather, it is your last best chance to take the time to learn this chemistry before we continue into more challenging material. Your assignment sets will include items helping you with this. 48

Practice 3.4 Identify the following as good oxidizers or good reducers. For your selection, write the product. (A) oxygen (B) chlorine (C) sodium (D) Cr2O7 2- (E) iodine Rank the following from the strongest reducer to the weakest reducer without using the table of potentials. Then, look up the values to check your answer. Al, Fe, Br Predict whether the following species are likely better reducers or better oxidizers. Be prepared to explain your selection. For each, write the reduction or oxidation reaction. Br Br2 NO3 MnO4 Use reduction potentials to determine the value of the potential difference when the following reactions occur. Silver nitrate solution is dripped over a solid piece of copper A piece of solid magnesium is placed into a solution of hydrochloric acid. The following reaction occurs as written. Write the products of the reaction. Identify the oxidizing agent and the reducing agent. MnO4 (aq) + Fe 2+ (aq)+ H + (aq)? 49

Part 3: BALANCING REDOX EQUATIONS You are undoubtedly familiar with balancing equations to observe the conservation of mass. However, a new consideration for redox reactions is the balancing of charge, too; more complex reactions will not balance as to charge and mass quite so easily. We can balance redox reactions by the half-reaction method. The half-reaction methods involves separating the oxidation half of the reaction from the reduction half of the reaction, balancing each separately for atoms and charge, and then putting the equations back together it is much easier than it sounds when reading the preceding sentence! The half-reaction method works because the number of electrons gained by the reduction half-reaction must equal the number of electrons lost by the oxidation half-reaction. When balancing redox reactions, we are typically discussing reactions that are occurring in acidic or basic solutions. The character of solution is important to balance redox equations in this manner! 50

Balancing redox in acid solution Identify the oxidation half reaction and the reduction half reaction Cu + NO3 NO2 + Cu 2+ Write two half-reactions: one for the oxidation and one for the reduction Balance all atoms except O and H in both half-reactions Balance oxygen by adding one water molecule for each oxygen atom needed Balance hydrogen by adding H + as needed to either reaction. Balance charges by adding electrons to the reactions. This will result in adding e- to the left side of one reaction and the right side of the other. Multiply through the equation(s) as needed to cancel the electrons. The electrons MUST cancel. Add the reactions cancel any common ions or molecules; simplify if possible. Write complete equation. 51

Balancing redox in base solution Identify the oxidation half reaction and the reduction half reaction Pb(OH)4 2- + ClO PbO2 + Cl Write two half-reactions: one for the oxidation and one for the reduction Balance all atoms except O and H in both half-reactions Balance oxygen by adding one water molecule for each oxygen atom needed Balance hydrogen by adding H + as needed to either reaction. Balance charges by adding electrons to the reactions. This will result in adding e- to the left side of one reaction and the right side of the other. Multiply through the equation(s) as needed to cancel the electrons. The electrons MUST cancel. Add the reactions cancel any common ions or molecules; simplify if possible. Add an OH - ion for each H + ion shown. Combine them to form water. Add the same number of OH - to the other side, too. Cancel common water molecules. This step is performed in basic solution only to eliminate the hydrogen ions present in the equation. Why? There will not be an excess of H+ in basic solution! Write complete equation. 52

Practice 3.5 Balance the following redox reactions by the half-reaction method. CN (aq) + MnO4 (aq) CNO (aq) + MnO2(s) in acid solution Cu(s) + NO3 (aq) Cu 2+ (aq) + NO2(g) in acid solution Cr(OH)3(s) + ClO (aq) CrO4 2- (aq) + Cl2(g) in base solution 53

Part 4: ELECTROCHEMISTRY VOLTAIC CELLS Recall that in a spontaneous redox reaction there is a potential energy difference. When the reaction occurs, energy is released, and this energy can be harnessed to perform electrical work (although, because of the Second Law, we cannot harness the maximum work possible, which we shall see when studying thermodynamics). A voltaic cell (or galvanic cell) is a device that allows the transfer of electrons between the reactants of a redox reaction without contact between the reactants. The energy released as the electrons travel is harnessed to perform work. The Structure of a Voltaic Cell Figure 15. A spontaneous voltaic cell. In a typical voltaic cell, two metals are placed in solutions of their ions 1 ; e.g., zinc metal is placed in a solution of zinc ions, and copper metal is placed in solution of copper ions. Because the two metals are not in contact with one another, the redox reaction cannot occur until a method of allowing for the flow of electrons is provided. This can be accomplished by connecting the two metal pieces with wire. The wire provides a conduit for the transfer of electrons from one electrode to the other. An electrode is a sample of metal that is connected to another by an external circuit here, the wire is the external circuit. In one half-cell of a voltaic cell, a reduction reaction occurs. Here, the reduction reaction is Cu 2+ (aq) Cu(s). In the other half-cell of a voltaic cell, an oxidation reaction occurs. Here, the oxidation reaction is Zn(s) Zn 2+ (aq). The compartments of a voltaic cell are given names: the anode compartment and the cathode compartment. 1 For metals that take on several oxidation states, one could use a platinum or graphite electrode at the cathode to cause the reduction of the metallic ion to an ion with a lower state. These are not common to see in AP Chemistry, but we will work a few in free-response practice in order to present a complete discussion. 54

Reduction occurs at the cathode, and oxidation occurs at the anode. Thus, electrons are traveling from the anode to the cathode in order that cations in the cathode are reduced to elemental metal. This means that at the anode elemental metal is being oxidized. By convention, the cathode is labeled positive, while the anode is labeled negative; however, the electrodee compartments are neutral. In fact, if the electrode compartments take on electrical charge, then the voltaic cell ceases to operate. There are five essential pieces of a voltaic cell: : A reduction half-cell; this is the cathode An oxidation half-cell; this is the anode A conduit for the transfer of electrons; this is the circuit A method of opening and closing the circuit; this might be a switch A source of positive and negative ions that move into the electrode compartments to maintain neutral solution; this is called a salt bridge, and it is filled with a salt that can provide ions to the compartments (a porous disk may be used; this allows for the transfer of ions through the disk) The Operation of a Voltaic Cell A) Zinc metal at the anode is losing electrons. This removes the zinc atoms from the electrode and increases the zinc ions in solution. The mass of the anode decreases as the atoms are lost to solution. The positive charge in the compartment solution is increasing as zinc cations are released. D) A constant transfer of ions from the salt bridge into the electrode compartments occurs in order that the compartments maintain a neutral character. B) The electrons travel through the circuit to the cathode. C) Copper ions in the cathode compartment s solution are gaining electrons. This removes copper ions from solution, which are deposited on the copper electrode. The mass of the cathode increases as atoms of metal are deposited. The negative charge in the compartment solution is increasing as copper cations are removed. 55

A voltaic cell will operate as long as there is a potential difference between the anode and cathode. For a spontaneous voltaic cell to operate in the desired direction, the value of the reduction potential at the cathode is more positive than the reduction potential of the anode, because the value of the cell potential must be positive. This is not unlike the single-replacement discussion on previous pages: E = Ered(reduction) Ered(oxidation) Ecell = Ered(reduction) Ered(oxidation) Ecell = Ered(cathode) Ered(anode) Notice that we always use the values of the reduction potentials even though one reaction occurs as an oxidation. You can determine the cathode and anode of a cell by looking at the reduction potentials for the two half-reactions the more positive reduction potential occurs as a reduction, which means it is the cathode. You can also use the value of Ecell to determine the values of reduction potentials for reactions by rearranging the equation for Ecell. Practice 3.6 The following reaction occurs in a voltaic cell: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s). What is the value of the Ecell? A voltaic cell operates spontaneously; it is constructed of cells containing Ni(NO3)2(aq) and Ni(s) and CuNO3(aq) and Cu(s). A) Determine the half-reactions occurring in each cell, write the balanced reaction and identify the species undergoing oxidation and the species undergoing reduction. Identify the number of electrons being transferred. B) Which reaction is occurring at the cathode? Justify your choice. C) In which compartment is the mass of the electrode increasing? Justify your choice. D) What is the value of the cell potential? 56

A voltaic cell operates spontaneously with Ecell = +1.46 V. The reactions in the cell are shown here: In + (aq) In 3+ (aq) + 2e Br2(g) + 2e 2Br (aq) A) Write the balanced overall reaction, and indicate the number of mol electrons transferred. B) Which reaction is occurring at the cathode? Which reaction is occurring at the anode? Justify each choice. C) Using a table of reduction potentials and the value of cell emf provided in the question, determine the value of Ered for the reduction of indium(iii) to indium(i). Practice 3.7 Consider the following reaction: Cr2O7 2- (aq) + I (aq) Cr 3+ (aq) + I2(s) Balance the reaction in acid solution, determine the anode and cathode half-reactions if this is used in a voltaic cell, and determine the value of Ecell. (In the case of substances such as these i.e., not metals the electrodes are generally made of graphite or an inert metal such as Pt. Thus, instead of deposition of metal, solution species will change.) 57

Two reduction reactions are shown here. Using the data for Ered for the reactions, determine the reactions that occur if these are used in a voltaic cell. Identify the reaction that occurs at the cathode and the reaction that occurs at the anode. Justify your choice. Then, use the blank set-up below to completely structure the operating voltaic cell. Theree are several things that should always be labeled when constructing a cell. Cd 2+ (aq) + 2e- Cd(s) Sn 2+ (aq) + 2e- Sn(s) Later, we will see that the concentration of the reactants and the temperature of the cell. For now, we are working with standard cells, which are cells at 298 K using 1 M solutions. Shorthand Notation for Voltaic Cells We can easily represent the details of a voltaic cell by using shorthand notation to describe the cell. We write the oxidation half-cell on the left, separating the reactant from the product with a single bar. We show the separation of cells by using a double bar, which represents the salt bridge, and then show the reduction half- cell in the same manner as we represented the anode reaction. Practice 3.8 Show the cadmium-tin cell on Page 58 in shorthand notation. Part 5: APPLICATIONS OF ELECTROCHEMISTRY: BATTERIES, CORROSION AND ELECTROLYSIS Redox titration, batteries, corrosion and electrolysis will be explored in a laboratory setting, and more on electrochemistry will be added when we study thermodynamics later. 58