20.1 Oxidation States and Oxidation- Reduction Reactions

Lecture Presentation Chapter 20 Yonsei University 1 20.1 Oxidation States and Oxidation- Reduction Reactions is the branch of chemistry that deals with relationships between electricity and chemical reactions Chemical reactions in which the oxidation state of one or more substances change are called oxidation-reduction reactions (redox reactions). Recall: Oxidation involves loss of electrons (OIL). Reduction involves gain of electrons (RIG). Also: Oxidation involves an increase of an oxidation number. Reduction involves a decrease of an oxidation number. 2

Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers. 3 Electronegativity 4

5 Oxidation and Reduction A species is oxidized when it loses electrons. Here, zinc loses two electrons to go from neutral zinc metal to the Zn 2+ ion. A species is reduced when it gains electrons. Here, each of the H + gains an electron, and they combine to form H 2. 6

Oxidation and Reduction What is reduced is the oxidizing agent. H + oxidizes Zn by taking electrons from it. What is oxidized is the reducing agent. Zn reduces H + by giving it electrons. 7 Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents The nickel-cadmium (nicad) battery uses the following redox reaction to generate electricity: Cd(s) + NiO 2 (s) + 2 H 2 O(l) Cd(OH) 2 (s) + Ni(OH) 2 (s) Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and which is the reducing agent. Solution The oxidation state of Cd increases from 0 to +2, and that of Ni decreases from +4 to +2. Thus, the Cd atom is oxidized (loses electrons) and is the reducing agent. The oxidation state of Ni decreases as NiO 2 is converted into Ni(OH) 2. Thus, NiO 2 is reduced (gains electrons) and is the oxidizing agent. Practice Exercise Identify the oxidizing and reducing agents in the reaction 2 H 2 O(l) + Al(s) + MnO 4 (aq) Al(OH) 4 (aq) + MnO 2 (s) Answer: Al(s) is the reducing agent; MnO 4 is the oxidizing agent. 8

20.2 Balancing Redox Equations Recall the law of conservation of mass: The amount of each element present at the beginning of the reaction must be present at the end. Conservation of charge: Electrons are not lost in a chemical reaction. Some redox equations may be easily balanced by inspection. However, for many redox reactions we need to look carefully at the transfer of electrons. 9 Half-Reactions Sn 2+ (aq) + 2Fe 3+ (aq) Sn 4+ (aq) + 2Fe 2+ (aq) The oxidation half-reaction is: Sn 2+ (aq) Sn 4+ (aq) +2e The reduction half-reaction is: 2Fe 3+ (aq) + 2e 2Fe 2+ (aq) 10

The Half-Reaction Method 1. Assign oxidation numbers to determine what is oxidized and what is reduced. 2. Write the oxidation and reduction half-reactions. 3. Balance each half-reaction. a. Balance elements other than H and O. b. Balance O by adding H 2 O. c. Balance H by adding H +. d. Balance charge by adding electrons. 4. Multiply the half-reactions by integers so that the electrons gained and lost are the same. 5. Add the half-reactions, subtracting things that appear on both sides. 6. Make sure the equation is balanced according to mass. 7. Make sure the equation is balanced according to charge. 11 The Half-Reaction Method Consider the reaction between MnO 4 and C 2 O 4 2 : MnO 4 (aq) + C 2 O 4 2 (aq) Mn 2+ (aq) + CO 2 (aq) 12

MnO 4 (aq) + C 2 O 4 2 (aq) Mn 2+ (aq) + CO 2 (aq) 16H + (aq) + 2MnO 4 (aq) + 5C 2 O 4 2 (aq) 2Mn 2+ (aq) + 8H 2 O(l) + 10CO 2 (g) 13 Sample Exercise 20.2 Balancing Redox Equations in Acidic Solution Complete and balance this equation by the method of half-reactions: Cr 2 O 2 7 (aq) + Cl (aq) Cr 3+ (aq) + Cl 2 (g) (acidic solution) Solution 14 H + (aq) + Cr 2 O 7 2 (aq) + 6 Cl (aq) 2 Cr 3+ (aq) + 7 H 2 O(l) + 3 Cl 2 (g) Practice Exercise Complete and balance the following equations using the method of half-reactions. Both reactions occur in acidic solution. (a) Cu(s) + NO 3 (aq) Cu 2+ (aq) + NO 2 (g) (b) Mn 2+ (aq) + NaBiO 3 (s) Bi 3+ (aq) + MnO 4 (aq) Answers: (a) Cu(s) + 4 H + (aq) + 2 NO 3 (aq) Cu 2+ (aq) + 2 NO 2 (g) + 2 H 2 O(l) (b) 2 Mn 2+ (aq) + 5 NaBiO 3 (s) + 14 H + (aq) 2 MnO 4 (aq) + 5 Bi 3+ (aq) + 5 Na + (aq) + 7 H 2 O(l) 14

Balancing in Basic Solution If a reaction occurs in a basic solution, one can balance it as if it occurred in acid. Once the equation is balanced, add OH to each side to neutralize the H + in the equation and create water in its place. If this produces water on both sides, you might have to subtract water from each side. 15 Sample Exercise 20.3 Balancing Redox Equations in Basic Solution Complete and balance this equation for a redox reaction that takes place in basic solution: CN (aq) + MnO 4 (aq) CNO (aq) + MnO 2 (s) (basic solution) Solution 3 CN (aq) + H 2 O(l) + 2 MnO 4 (aq) 3 CNO (aq) + 2 MnO 2 (s) + 2 OH (aq) Practice Exercise Complete and balance the following equations for oxidation-reduction reactions that occur in basic solution: (a) NO 2 (aq) + Al(s) NH 3 (aq) + Al(OH) 4 (aq) (b) Cr(OH) 3 (s) + ClO (aq) CrO 2 4 (aq) + Cl 2 (g) Answers: (a) NO 2 (aq) + 2 Al(s) + 5 H 2 O(l) + OH (aq) NH 3 (aq) + 2 Al(OH) 4 (aq) (b) 2 Cr(OH) 3 (s) + 6 ClO (aq) 2 CrO 2 4 (aq) + 3 Cl 2 (g) + 2 OH (aq) + 2 H 2 O(l) 16

20.3 Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released. 17 Voltaic Cells We can use that energy to do work if we make the electrons flow through an external device. We call such a setup a voltaic cell. 18

Voltaic Cells The oxidation occurs at the anode. The reduction occurs at the cathode. 19 Electrochemical Cells Zn Zn 2+ +2e - Cu 2+ +2e - Cu 20

Voltaic Cells Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop. Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. Cations move toward the cathode. Anions move toward the anode. 21 Electrochemical Cells (Anode) 22

Electrochemical Cells (Cathode) 23 Voltaic Cells In the cell, then, electrons leave the anode and flow through the wire to the cathode. As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment. As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. The electrons are taken by the cation, and the neutral metal is deposited on the cathode. 24

Cell Notation The Daniel Cell reaction is: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Using cell notation to represent the cell: Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) anode salt bridge cathode 25 20.4 Cell Potentials Under Standard Conditions 26 Electromotive Force (emf) Water only spontaneously flows one way in a waterfall. Likewise, electrons only spontaneously flow one way in a redox reaction from higher to lower potential energy.

Electromotive Force (emf) The potential difference between the anode and cathode in a cell is called the electromotive force (emf). It is also called the cell potential and is designated E cell. Cell potential is measured in volts (V). 1 V = 1 J C Useful Units : 1C = 1A s 1J = 1V C 27 Standard Reduction Potentials Standard reduction potentials, Eº red for many electrodes have been measured and tabulated. 28

Standard Hydrogen Electrode Their values are referenced to a standard hydrogen electrode (SHE). By definition, the reduction potential for hydrogen is 0 V: 2 H + (aq, 1M) + 2e H 2 (g, 1 atm) 29 Standard Cell Potentials The cell potential at standard conditions can be found through this equation: E cell = E red (cathode) E red (anode) Because cell potential is based on the potential energy per unit of charge, it is an intensive property. 30

Cell Potentials For the oxidation in this cell, E red = 0.76 V - + For the reduction, E red = +0.34 V E cell = E red (cathode) E red (anode) 31 = +0.34 V ( 0.76 V) = +1.10 V Cell Potentials The greater the difference between the two, the greater the voltage of the cell. 32

Sample Exercise 20.5 Calculating from For the Zn-Cu 2+ voltaic cell shown in Figure 20.5, we have Given that the standard reduction potential of Zn 2+ to Zn(s) is 0.76 V, calculate the for the reduction of Cu 2+ to Cu: Cu 2+ (aq, 1 M) + 2 e Cu(s) Solution 33 Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials. 34

20.5 Free Energy and Redox Reactions G for a redox reaction can be found by using the equation G = nfe where n is the number of moles of electrons transferred, and F is a constant, the Faraday: C 96,485 mole J 96,485 (V)(mole 1F ) 35 Free Energy Under standard conditions, G = nfe Since G is related to the equilibrium constant, K, we can relate E to K: E G RT K RT nf nf nf 0 0 ln ln K 36

Sample Exercise 20.9 Determining Spontaneity Use Table 20.1 to determine whether the following reactions are spontaneous under standard conditions. (a) Cu(s) + 2 H + (aq) Cu 2+ (aq) + H 2 (g) (b) Cl 2 (g) + 2 I (aq) 2 Cl (aq) + I 2 (s) Solution (a) Cu 2+ (aq) + H 2 (g) Cu(s) + 2 H + (aq) E = +0.34 V, spontaneous (b) E = (1.36 V) (0.54 V) = +0.82 V spontaneous 37 20.6 Cell Potentials Under Nonstandard Conditions A voltaic cell is functional until E = 0 at which point equilibrium has been reached. The cell is then dead. The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. 38

Nernst Equation G = G + RT ln Q; nfe = nfe + RT ln Q Dividing both sides by nf, we get the Nernst equation: E = E RT nf ln Q or, using base-10 logarithms, E = E 2.303RT nf log Q 39 Nernst Equation At room temperature (298 K), 2.303RT F = 0.0592 V Thus, the equation becomes E = E 0.0592 n log Q 40

Concentration Cells 41 Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell, E cell would be 0, but Q would not. Therefore, as long as the concentrations are different, E will not be 0. Sample Exercise 20.11 Cell Potential under Nonstandard Conditions Calculate the emf at 298 K generated by a voltaic cell in which the reaction is Cr 2 O 7 2 (aq) + 14 H + (aq) + 6 I (aq) 2 Cr 3+ (aq) + 3 I 2 (s) + 7 H 2 O(l) when [Cr 2 O 7 2 ] = 2.0 M, [H + ] = 1.0 M, [I ] = 1.0 M, and [Cr 3+ ] = 1.0 10 5 M. Solution 42

Sample Exercise 20.12 Calculating Concentrations in a Voltaic Cell If the potential of a Zn-H + cell (like that in Figure 20.9) is 0.45 V at 25 C when [Zn 2+ ] = 1.0 M and P H2 = 1.0 atm, what is the H + concentration? Solution Zn(s) + 2 H + (aq) Zn 2+ (aq) + H2(g) n = 2 43 20.7 Batteries and Fuel Cells A battery is a portable, self-contained electrochemical power source consisting of one or more voltaic cells. Primary cells: cannot be recharged. Secondary cells: can be recharged from an external power source after its voltage has dropped. 44

Lead-Acid Battery Anode: Pb(s) + SO 4 2 (aq) PbSO 4 (s) + 2e E=0.127V Cathode: PbO 2 (s) + 4H + (aq) + SO 4 2 (aq) + 2e PbSO 4 (s) + 2H 2 O E=1.687V Pb(s)+ PbO 2 (s)+ 4H + (aq)+ 2SO 4 2 (aq) 2PbSO 4 (s)+ 2H 2 O 45 Leclanche cell 산화전극 : Zn(s) Zn 2+ (aq) + 2e 환원전극 : MnO 2 (s)+ 2NH 4+ (aq) + 2e Mn 2 O 3 (s)+ 2NH 3 + H 2 O Zn(s)+ MnO 2 (s)+ 2NH 4+ (aq) Zn 2+ (aq) + Mn 2 O 3 (s)+ 2NH 3 (g) + H 2 O 46

Alkaline Batteries Anode : Zn(s) + 2OH (aq) Zn(OH) 2 (aq) + 2e 47 Cathode : 2MnO 2 (s) + 2H 2 O(l) + 2e 2MnO(OH)(s) + 2OH (aq) V= 1.55 V Nickel-Cadmium Batteries Cathode : 2NiO(OH)(s) + 2H 2 O(l) + 2e 2Ni(OH) 2 (s) +2OH - (aq) Anode : Cd(s) + 2OH (aq) Cd(OH) 2 (s) + 2e Working Cd(s)+ 2NiO (s)+ 2H 2 O Cd(OH) 2 (s) + Ni(OH) 2 (s) E= 1.30 V Recharge Cd(OH) 2 (s) + Ni(OH) 2 (s) Cd(s)+ 2NiO 2 (s)+ 2H 2 O E=-1.3 V Cadmium is a toxic heavy metal. Other rechargeable batteries have been developed. NiMH batteries (nickel-metal-hydride). Li-ion batteries (lithium-ion batteries). 48

Hydrogen Fuel Cells 2H 2 (g) + 4OH (aq) 4H 2 O(l) + 4e 2H 2 O(l) + O 2 (g) + 4e 4OH (aq) 49 Hydrogen Economy 50

20.8 Corrosion 51 Galvanized iron Corrosion Prevention Zn 2+ (aq) +2e Zn(s) E red = -0.76 V Fe 2+ (aq) + 2e Fe(s) E red = -0.44 V cathodic protection :the sacrificial anode is destroyed Mg 2+ (aq) +2e Mg(s) E red = -2.37 V Fe 2+ (aq) + 2e Fe(s) E red = -0.44 V 52

Electrical Work Free energy is a measure of the maximum amount of useful work that can be obtained from a system. We know: G = w max and G = -nfe thus: w max = -nfe If E cell is positive, w max will be negative. Work is done by the system on the surroundings. 53 Electrical Work The emf can be thought of as being a measure of the driving force for a redox process. In an electrolytic cell an external source of energy is required to force the reaction to proceed. : w = nfe external To drive the nonspontaneous reaction, the external emf must be greater than E cell. From physics we know that work is measured in units of watts: 1 W = 1 J/s Electric utilities use units of kilowatt-hours: kwh = 3.6 10 6 J. 54

20.9 Electrolysis Electrolysis reactions: nonspontaneous reactions - Need an external current to force the reaction to proceed : electrolytic cells. In both cells (voltaic and electrolytic) - reduction: cathode, oxidation: anode. - In electrolytic cells, electrons forced to flow from anode to cathode. thus, anode (positive terminal) cathode (negative). -In voltaic cells, anode (negative ), cathode (positive). 55 Decomposition of molten NaCl Cathode: 2Na + (l) + 2e 2Na(l) E red = -2.714 V 2H 2 O + 2e H 2 (g) + 2 OH (aq) E red = - 0.828 V Anode: 2Cl (l) Cl 2 (g) + 2e : E ox = -1.360 V 56

Electrolytic Reactions Electrowinning cryolite 2Al 2 O 3 (l) + 3C(s) 4Al(l)+3CO 2 (g) Overvoltage E -2.1V 57 Electroplating 58 Active electrodes nickel strip steel strip

Quantitative Aspects of Electrolysis Faraday observed that the amount of current applied to a cell is directly proportional to the amount of metal deposited. n e IA ( ) ts ( ) I t M molofmetal : g F F mol of e - - A = ampere s = seconds M = molar mass F = 96,485 C/mol of e - For the case of M 2 2e M mol M 1 mol 2 e 1 n 2 e Useful Units : 1C = 1A s 1J = 1V C 59 Sample Exercise 20.14 Relating Electrical Charge and Quantity of Electrolysis Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl 3 if the electrical current is 10.0 A. Solution Al 3+ + 3 e Al 60

Problems 8, 16, 28, 36, 40, 48, 56, 68, 94 61