Oxidation Reduction and Electrochemistry. David A. Katz Department of Chemistry Pima Community College

Transcription

1 Oxidation Reduction and Electrochemistry David A. Katz Department of Chemistry Pima Community College

2 Oxidation Reduction Reactions In an oxidation reduction (Redox) reaction, electrons are transferred from one species to another. For example, in a single replacement reaction Cu (s) + 2 AgNO 3 (aq) 2 Ag (s) + Cu(NO 3 ) 2 (aq) The Cu atoms lose electrons to form Cu 2+ in the Cu(NO 3 ) 2 (aq) and the Ag + gains electrons to form metallic Ag

3 Oxidation Reduction Reactions This can be more easily observed by writing the net ionic equation for the reaction: Cu (s) + 2 Ag + (aq) 2 Ag (s) + Cu 2+ (aq) The metallic Cu atoms are uncombined, so they are considered to have an oxidation number of zero. The initial combined Ag + ions are in a +1 oxidation state. Each Cu atom will lose 2 electrons to 2 Ag + ions The resulting Ag atoms are considered to have an oxidation number of zero

4 Oxidation Reduction Reactions Cu (s) + 2 Ag + (aq) 2 Ag (s) + Cu 2+ (aq)

5 Oxidation Numbers In order to keep track of what loses electrons and what gains them, we list the oxidation numbers of each element.

6 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.

7 Oxidation and Reduction A species is reduced when it gains electrons. Here, each of the H + gains an electron and they combine to form H 2.

8 Oxidation and Reduction The species that contains the element that is reduced is the oxidizing agent. H + oxidizes Zn by taking electrons from it. The species that contains the element that is oxidized is the reducing agent. Zn reduces H + by giving it electrons.

9 Assigning Oxidation Numbers 1. Elements in their elemental form have an oxidation number of The oxidation number of a monatomic ion is the same as its charge.

10 Assigning Oxidation Numbers 3. The oxidation number of metals depends on their position in the periodic table Group IA elements are +1 Group IIA elements are +2 Group IIIA elements are +3 Group IVA metals are usually +2 or +4 Group VA metals are usually +3 or +5

11 Assigning Oxidation Numbers 4. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. Oxygen always has an oxidation number of 2, except in the peroxide ion in which it has an oxidation number of 1. Hydrogen is always 1 when bonded to a metal Hydrogen is +1 when bonded to a nonmetal.

12 Assigning Oxidation Numbers 4. Nonmetals (continued). Fluorine always has an oxidation number of 1. The halogens (Cl, Br, and I)have an oxidation number of 1 when they are negative The halogens (Cl, Br, and I) will have positive oxidation numbers in oxyanions (ClO, ClO 2, ClO 3, etc.)

13 Assigning Oxidation Numbers 5. The sum of the oxidation numbers in a neutral compound is The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

14 Balancing Oxidation Reduction Equations Oxidation reduction equations may be difficult to balance. Generally, the easiest way to balance the equation of an oxidation reduction reaction is via the halfreaction method. This involves treating the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.

15 Balancing Redox Equations by the Half Reaction Method 1. Assign oxidation numbers to the elements in the equation 2. Determine what is oxidized and what is reduced. 3. Write the oxidation and reduction halfreactions.

16 Balancing Redox Equations by the Half Reaction Method 4. 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. 5. Multiply the half reactions by integers so that the electrons gained and lost are the same.

17 Balancing Redox Equations by the Half Reaction Method 6. Add the half reactions, subtracting things that appear on both sides. 7. Make sure the equation is balanced according to mass. 8. Make sure the equation is balanced according to charge.

18 Balancing Redox Equations by 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) (acidic solution)

19 Balancing Redox Equations by the Half Reaction Method First, assign oxidation numbers (remember oxygen is 2) MnO 4 + C 2 O 4 2 Mn 2+ + CO 2 The oxidation number of manganese goes from +7 to +2, it is reduced. The oxidation number of carbon goes from +3 to +4, it is oxidized.

20 Balancing Redox Equations by the Half Reaction Method Determine the oxidizing and reducing agents MnO 4 + C 2 O 4 2 Mn 2+ + CO 2 The MnO 4 is the oxidizing agent The C 2 O 4 2 is the reducing agent

21 Balancing Redox Equations by the Half Reaction Method The oxidation half reaction is C 2 O 4 2 CO 2 Balance the half reaction C 2 O CO 2 This balances both the carbon atoms and the oxygen atoms

22 Balancing Redox Equations by the Half Reaction Method Balance the charge by adding 2 electrons to the right side. C 2 O CO e

23 Balancing Redox Equations by the Half Reaction Method The reduction half reaction is MnO 4 Mn 2+ The manganese is balanced In order to balance the 4 oxygens, we add 4 water molecules to the right side. MnO 4 Mn H 2 O

24 Balancing Redox Equations by the Half Reaction Method MnO 4 Mn H 2 O The addition of water on the right side of the equation included hydrogen atoms. To balance the 8 hydrogens, add 8 H + to the left side. 8 H + + MnO 4 Mn H 2 O

25 Balancing Redox Equations by the Half Reaction Method 8 H + + MnO 4 Mn H 2 O Balance the charges, add 5 e to the left side. 5 e + 8 H + + MnO 4 Mn H 2 O

26 Balancing Redox Equations by the Half Reaction Method Before the two half reactions can be added together, the number of electrons lost must be equal to the electrons gained: C 2 O CO e (2 e lost) 5 e + 8 H + + MnO 4 Mn H 2 O (5 e gained) To attain the same number of electrons on each side, multiply the first reaction by 5 and the second by 2. 5 C 2 O CO e 10 e + 16 H MnO 4 2 Mn H 2 O

27 Balancing Redox Equations by the Half Reaction Method 5 C 2 O CO e 10 e + 16 H MnO 4 2 Mn H 2 O Add these half reactions 10 e + 16 H MnO C 2 O Mn H 2 O + 10 CO e Cancel out the electrons from both sides to get 16 H MnO C 2 O Mn H 2 O + 10 CO 2

28 Balancing Redox Equations in Basic Solution If a reaction occurs in basic solution, use OH and H 2 O instead of the H + and H 2 O used in acid solution The easiest method is to balance the equation 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 results in water on both sides of the equation, subtract water from each side as a final step.

29 Balancing Redox Equations in Basic Solution Complete and balance the following redox equation that takes place in basic solution CN (aq) + MnO 4 (aq) CNO (aq) + MnO 2 (s) solution) (basic First, assign oxidation numbers (remember oxygen is 2) CN (aq) + MnO 4 (aq) CNO (aq) + MnO 2 (s) Note: In an anion such as CN, C, which comes first, would be positive and N would be negative. If negative, N would be 3.

30 Balancing Redox Equations in Basic Solution CN (aq) + MnO 4 (aq) CNO (aq) + MnO 2 (s) The oxidation number of carbon goes from +2 to +4, it is oxidized. (C lost 2 e ) The oxidation number of manganese goes from +7 to +4, it is reduced. (Mn gained 3 e ) MnO 4 is the oxidizing agent CN is the reducing agent

31 Balancing Redox Equations in Basic Solution The reduction half reaction is MnO 4 MnO 2 Balance the half reaction MnO 4 MnO 2 Balance the oxygen atoms, add H + and H 2 O 4 H + + MnO 4 MnO H 2 O

32 Balancing Redox Equations in Basic Solution Balance the charges, add 3 e to the left side 3 e + 4 H + + MnO 4 MnO H 2 O

33 Balancing Redox Equations in Basic Solution The oxidation half reaction is CN CNO Since the C atoms are balanced, balance the oxygen. Add H 2 O to the left side and H + to the right side. H 2 O + CN CNO + 2 H +

34 Balancing Redox Equations in Basic Solution H 2 O + CN CNO + 2 H + Balance the charge, add 2 e to the right side H 2 O + CN CNO + 2 H e

35 Balancing Redox Equations in Basic Solution Before the two half reactions can be added together, the number of electrons lost must be equal to the electrons gained: H 2 O + CN CNO + 2 H e (2 e lost) 3 e + 4 H + + MnO 4 MnO H 2 O (3 e gained) To attain the same number of electrons on each side, multiply the first equation by 3 and the second by 2. 3 H 2 O + 3 CN 3 CNO + 6 H e 6 e + 8 H MnO 4 2 MnO H 2 O

36 Balancing Redox Equations in Basic Solution Add the half reactions 3 H 2 O + 3 CN + 8 H MnO 4 3 CNO + 6 H MnO H 2 O Cancel out the H + and H 2 O to get 2H CN + 2 MnO 4 3CNO + 2MnO 2 + H 2 O

37 Balancing Redox Equations in Basic Solution This reaction, however, takes place in basic solution 2H CN + 2 MnO 4 3CNO + 2MnO 2 + H 2 O Add 2 OH to both sides to cancel out the H + 2 OH +2 H CN + 2 MnO 4 3CNO + 2MnO 2 + H 2 O + 2 OH (Remember that H + + OH H 2 O) 2 H 2 O + 3 CN + 2 MnO 4 3CNO + 2MnO 2 + H 2 O + 2 OH The balanced equation is 3 CN + 2 MnO 4 + H 2 O 3CNO + 2MnO OH

38 A History of Electricity/Electrochemistry Thales of Miletus ( B.C.) is credited with the discovery that amber when rubbed with cloth or fur acquired the property of attracting light objects. The word electricity comes from "elektron" the Greek word for amber. Otto von Guericke ( ) invented the first electrostatic generator in It was made of a sulphur ball which rotated in a wooden cradle. The ball itself was rubbed by hand and the charged sulphur ball had to be transported to the place where the electric experiment was carried out. Thales of Miletus Otto von Guericke

39 Eventually, a glass globe replaced the sulfur sphere used by Guericke Later, large disks were used

40 Ewald Jürgen von Kleist ( ), invented the Leyden Jar in 1745 to store electric energy. The Leyden Jar contained water or mercury and was placed onto a metal surface with ground connection. In 1746, the Leyden jar was independently invented by physicist Pieter van Musschenbroek ( ) and/or his lawyer friend Andreas Cunnaeus in Leyden/the Netherlands Leyden jars could be joined together to store large electrical charges

41 In 1752, Benjamin Franklin ( ) demonstrated that lightning was electricity in his famous kite experiment In 1780, Italian physician and physicist Luigi Aloisio Galvani ( ) discovered that muscle and nerve cells produce electricity. Whilst dissecting a frog on a table where he had been conducting experiments with static electricity, Galvani touched the exposed sciatic nerve with his scalpel, which had picked up an electric charge. He noticed that the frog s leg jumped.

42 Count Alessandro Giuseppe Antonio Anastasio Volta ( ) developed the first electric cell, called a Voltaic Pile, in A voltaic pile consist of alternating layers of two dissimilar metals, separated by pieces of cardboard soaked in a sodium chloride solution or sulfuric acid. Volta determined that the best combination of metals was zinc and silver Volta s electric pile (right) A Voltaic pile at the Smithsonian Institution, (far right)

43 In 1800, English chemist William Nicholson ( ) and surgeon Anthony Carlisle ( ) separated water into hydrogen and oxygen by electrolysis. Johann Wilhelm Ritter ( ) repeated Nicholson s separation of water into hydrogen and oxygen by electrolysis. Soon thereafter, Ritter discovered the process of electroplating. He also observed that the amount of metal deposited and the amount of oxygen produced during an electrolytic process depended on the distance between the electrodes Humphrey Davy ( ) utilized the voltaic pile, in 1807, to isolate elemental potassium by electrolysis which was soon followed by sodium, barium, calcium, strontium, magnesium. William Nicholson Johann Wilhelm Ritter Humphrey Davy

44 Michael Faraday ( ) began his career in 1813 as Davy's Laboratory Assistant. In 1834, Faraday developed the two laws of electrochemistry: The First Law of Electrochemistry The amount of a substance deposited on each electrode of an electrolytic cell is directly proportional to the amount of electricity passing through the cell. The Second Law of Electrochemistry The quantities of different elements deposited by a given amount of electricity are in the ratio of their chemical equivalent weights.

45 Faraday also defined a number of terms: The anode is therefore that surface at which the electric current, according to our present expression, enters: it is the negative extremity of the decomposing body; is where oxygen, chlorine, acids, etc., are evolved; and is against or opposite the positive electrode. The cathode is that surface at which the current leaves the decomposing body, and is its positive extremity; the combustible bodies, metals, alkalies, and bases are evolved there, and it is in contact with the negative electrode. Many bodies are decomposed directly by the electric current, their elements being set free; these I propose to call electrolytes... Finally, I require a term to express those bodies which can pass to the electrodes, or, as they are usually called, the poles. Substances are frequently spoken of as being electro negative or electro positive, according as they go under the supposed influence of a direct attraction to the positive or negative pole...i propose to distinguish such bodies by calling those anions which go to the anode of the decomposing body; and those passing to the cathode, cations; and when I have occasion to speak of these together, I shall call them ions. the chloride of lead is an electrolyte, and when electrolyzed evolves the two ions, chlorine and lead, the former being an anion, and the latter a cation.

46 John Frederic Daniell ( ), professor of chemistry at King's College, London. Daniell's research into development of constant current cells took place at the same time (late 1830s) that commercial telegraph systems began to appear. Daniell's copper battery (1836) became the standard for British and American telegraph systems. In 1839, Daniell experimented on the fusion of metals with a 70 cell battery. He produced an electric arc so rich in ultraviolet rays that it resulted in an instant, artificial sunburn. These experiments caused serious injury to Daniell's eyes as well as the eyes of spectators. Ultimately, Daniell showed that the ion of the metal, rather than its oxide, carries an electric charge when a metal salt solution is electrolyzed. Left: An early Daniell Cell Right:Daniell cells used by Sir William Robert Grove, 1839.

47 Voltaic Cells In spontaneous oxidationreduction (redox) reactions, electrons are transferred and energy is released.

48 If the reaction is separated into two parts, we can use that energy to do work by making the electrons flow through an external device. This type of setup is called a voltaic cell. Voltaic Cells

49 Voltaic Cells This is a typical voltaic cell A strip of zinc metal is immersed in a solution of Zn(NO 3 ) 2 A strip of copper metal is immersed in a solution of Cu(NO 3 ) 2 The two solutions are connected by a salt bridge containing NaNO 3 The oxidation occurs at the anode (Zn) The reduction occurs at the cathode (Cu)

50 Voltaic Cells The salt bridge is used to prevent electrons flowing directly from the zinc to the copper The salt bridge consists of a U shaped tube that contains a salt solution, sealed with porous plugs, or an agar gel solution of the salt The salt bridge keeps the charges balanced and forces the electron to move through the wire Cations move toward the cathode. Anions move toward the anode.

51 In the cell, 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. Eventually, if the cell is used for a long time, the anode (zinc) will dissolve Voltaic Cells

52 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. Eventually, if the cell is used for a long time, all the copper ions will plate onto the copper cathode Voltaic Cells

53 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.

54 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.

55 Cell Potential Cell potential is measured in volts (V). 1 V = 1 Where J = Joules C = Coulombs Recall that 1 electron has a charge of 1.6 x C J C

56 Standard Reduction Potentials The cell potential is the difference between two electrode potentials. By convention, electrode potentials are written as reductions Reduction potentials for most common electrodes are tabulated as standard reduction potentials.

57 Standard Hydrogen Electrode Electrode potentials are referenced to a standard hydrogen electrode (SHE). By definition, the reduction potential for hydrogen is 0 V: 2 H + (aq, 1M) + 2 e H 2 (g, 1 atm)

58 Standard Cell Potentials The cell potential at standard conditions is calculated E cell = E red (cathode) E red (anode) Substance reduced Substance oxidized Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

59 Cell Potentials Oxidation: E red = V Reduction: E red = V

60 Cell Potentials E cell = E red (cathode) E red (anode) = V ( 0.76 V) = V Generally, most of the common cells used, on the average, generate approximately 1.5 V

61 Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials.

62 Oxidizing and Reducing Agents The greater the difference between the two, the greater the voltage of the cell.

63 Free Energy G for a redox reaction can be found by using the equation G = nfe where: n is the number of moles of electrons transferred F is a constant, the Faraday. 1 F = 96,485 C/mol = 96,485 J/V mol E = The standard cell potential in V Under standard conditions, G = nfe

64 Remember that This means Nernst Equation G = G + RT ln Q nfe = nfe + RT ln Q Dividing both sides by nf, we get the Nernst equation: RT E = E ln Q nf or, using base 10 logarithms, RT E = E log Q nf

65 Nernst Equation At room temperature (298 K), and R = J/mol K F = 96,485 J/V mol RT F = V The final form of the Nernst Equation becomes E = E n log Q

66 Walther Hermann Nernst ( ) Nernst's early studies in electrochemistry were inspired by Arrhenius' dissociation theory of ions in solution. In 1889 he elucidated the theory of galvanic cells by assuming an "electrolytic pressure of dissolution" which forces ions from electrodes into solution and which was opposed to the osmotic pressure of the dissolved ions. Also, in 1889, he showed how the characteristics of the current produced could be used to calculate the free energy change in the chemical reaction producing the current. This equation, known as the Nernst Equation, relates the voltage of a cell to its properties. Independently of Thomson, he explained why compounds ionize easily in water. The explanation, called the Nernst Thomson rule, holds that it is difficult for charged ions to attract each other through insulating water molecules, so they dissociate.

67 Concentration Cells 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.

68 Applications of Oxidation Reduction Reactions

69 Why Study Electrochemistry? Batteries Corrosion Industrial production of chemicals such as Cl 2, NaOH, F 2 and Al Biological redox reactions The heme group

70 BATTERIES Primary, Secondary, and Fuel Cells

71 Batteries Most commercial batteries produce 1.5 V. To get a higher voltage, batteries are joined together.

72 Dry Cell Battery Primary battery uses redox reactions that cannot be restored by recharge. Anode (-) Zn Zn e - Cathode (+) 2 NH e - 2 NH 3 + H 2

73 Alkaline Battery Nearly same reactions as in common dry cell, but under basic conditions. Anode ( ): Zn + 2 OH ZnO + H 2 O + 2e Cathode (+): 2 MnO 2 + H 2 O + 2e Mn 2 O OH

74 Alkaline Batteries

75 Lead Storage Battery Secondary battery Uses redox reactions that can be reversed. Can be restored by recharging

76 Lead Storage Battery Anode (-) E o = V Pb + HSO 4- PbSO 4 + H + + 2e- Cathode (+) E o = V PbO 2 + HSO H + + 2e- PbSO H 2 O

77 Ni-Cad Battery Anode (-) Cd + 2 OH - Cd(OH) 2 + 2e- Cathode (+) NiO(OH) + H 2 O + e- Ni(OH) 2 + OH -

78 Fuel Cells: H 2 as a Fuel Fuel cell reactants are supplied continuously from an external source. Cars can use electricity generated by H 2 /O 2 fuel cells. H 2 carried in tanks or generated from hydrocarbons.

79 Hydrogen Air Fuel Cell

80 Hydrogen Fuel Cells

81 H 2 as a Fuel Comparison of the volumes of substances required to store 4 kg of hydrogen relative to car size.

82 Storing H 2 as a Fuel One way to store H 2 is to adsorb the gas onto a metal or metal alloy.

83 Electrolysis Using electrical energy to produce chemical change. Sn 2+ (aq) + 2 Cl (aq) Sn (s) + Cl 2(g)

84 Electrolysis of Aqueous NaOH Electric Energy f Chemical Change Anode (+) 4 OH - O 2 (g) + 2 H 2 O + 4e- Cathode (-) 4 H 2 O + 4e- 2 H OH - E o for cell = V Anode Cathode

85 Electrolysis Electric Energy f Chemical Change Electrolysis of molten NaCl. Here a battery pumps electrons from Cl to Na +. NOTE: Polarity of electrodes is reversed from batteries. + Anode electrons BATTERY Cl - Na + Cathode

86 Electrolysis of Molten NaCl See Figure 20.18

87 Electrolysis of Molten NaCl electrons Anode (+) BATTERY 2 Cl - Cl 2 (g) + 2e- + Anode Cathode Cathode (-) Cl - Na + Na + + e- Na E o for cell (in water) = E c E a = 2.71 V (+1.36 V) = 4.07 V (in water) External energy needed because E o is ( ).

88 Electrolysis of Aqueous NaCl Anode (+) electrons 2 Cl - Cl 2 (g) + 2e- Cathode (-) 2 H 2 O + 2e- H OH - E o for cell = V + Anode BATTERY Cathode Note that H 2 O is more easily reduced than Na +. Cl - Na + H 2 O Also, Cl is oxidized in preference to H 2 O because of kinetics.

89 Electrolysis of Aqueous NaCl Cells like these are the source of NaOH and Cl 2. In 1995: 25.1 x 10 9 lb Cl 2 and 26.1 x 10 9 lb NaOH Also the source of NaOCl for use in bleach.

90 Electrolysis of Aqueous NaI Anode (+): 2 I - I 2 (g) + 2e- Cathode (-): 2 H 2 O + 2e- H OH - E o for cell = V

91 Electrolysis of Aqueous CuCl 2 Anode (+) 2 Cl - Cl 2 (g) + 2e- electrons Cathode (-) Cu e- Cu + Anode BATTERY Cathode E o for cell = V Note that Cu is more easily reduced than Cl - Cu 2+ H 2 O either H 2 O or Na +.

92 Electrolytic Refining of Copper Impure copper is oxidized to Cu 2+ at the anode. The aqueous Cu 2+ ions are reduced to Cu metal at the cathode. The copper at the cathode is over 99% pure

93 Producing Aluminum 2 Al 2 O C f 4 Al + 3 CO 2 Charles Hall ( ) developed electrolysis process. Founded Alcoa.

94 Corrosion and

95 Corrosion Prevention