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Chapter 20 Electrochemistry

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1 Chapter 20 Electrochemistry
Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 20 Electrochemistry John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

2 Electrochemical Reactions
In electrochemical reactions, electrons are transferred from one species to another. Metals tend to lose electrons and are oxidized, non metals tend to gain electrons and are reduced.

3 LEO GER Losing Electrons is Oxidation. Gaining Reduction

4 OIL RIG Oxidation Is Loss. Reduction Gain.

5 REDOX REACTIONS Are reduction – oxidation reactions.
Electrons are transferred. When an atom is losing electrons its O.N. increases. It is being oxidized. When an atom gains electrons its O.N. decreases. It is being reduced

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

7 Assigning Oxidation Numbers
Elements in their elemental form have an ON= 0. The oxidation number of a monatomic ion is its charge. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1. Hydrogen is +1 except in metal hydrides when is −1 . Fluorine always has an oxidation number of −1.

8 Assigning Oxidation Numbers
The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. The sum of the oxidation numbers in a neutral compound is 0.

9 Oxidation and Reduction
A species is oxidized when it loses electrons. Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

10 Oxidation and Reduction
A species is reduced when it gains electrons. Here, each of the H+ gains an electron and they combine to form H2.

11 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.

12 Oxidation-Reduction Reactions
Zn added to HCl yields the spontaneous reaction Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g). The oxidation number of Zn has increased from 0 to 2+. The oxidation number of H has reduced from 1+ to 0. Zn is oxidized to Zn2+ while H+ is reduced to H2. H+ causes Zn to be oxidized and is the oxidizing agent. Zn causes H+ to be reduced and is the reducing agent. Note that the reducing agent is oxidized and the oxidizing agent is reduced.

13 Balancing Oxidation-Reduction Reactions
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. Half Reactions Half-reactions are a convenient way of separating oxidation and reduction reactions.

14 Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq)
Half Reactions The half-reactions for Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq) are Sn2+(aq)  Sn4+(aq) +2e- 2Fe3+(aq) + 2e-  2Fe2+(aq) Oxidation: electrons are products. Reduction: electrons are reactants. Loss of Gain of Electrons is Electrons is Oxidation Reduction

15 Balancing Equations by the Method of Half Reactions
Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple). MnO4- is reduced to Mn2+ (pale pink) while the C2O42- is oxidized to CO2. The equivalence point is given by the presence of a pale pink color. If more KMnO4 is added, the solution turns purple due to the excess KMnO4.

16 What is the balanced chemical equation?
1. Write down the two half reactions. 2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H+. d. If it is in basic solution, remove H+ by adding OH- e. Finish by balancing charge by adding electrons. Multiply each half reaction to make the number of electrons equal. Add the reactions and simplify. Check!

17 MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)
Half-Reaction Method Consider the reaction between MnO4− and C2O42− : MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)

18 Balancing Equations by the Method of Half Reactions
Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple). MnO4- is reduced to Mn2+ (pale pink) while the C2O42- is oxidized to CO2. The equivalence point is given by the presence of a pale pink color. If more KMnO4 is added, the solution turns purple due to the excess KMnO4.

19 Half-Reaction Method MnO4− + C2O42-  Mn2+ + CO2
First, we assign oxidation numbers. MnO4− + C2O42-  Mn2+ + CO2 +7 +3 +4 +2 Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized.

20 The two incomplete half reactions are MnO4-(aq)  Mn2+(aq)
For KMnO4 + Na2C2O4: The two incomplete half reactions are MnO4-(aq)  Mn2+(aq) C2O42-(aq)  2CO2(g) 2. Adding water and H+ yields 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O There is a charge of 7+ on the left and 2+ on the right. Therefore, 5 electrons need to be added to the left: 5e- + 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O

21 10e- + 16H+ + 2MnO4-(aq)  2Mn2+(aq) + 8H2O
In the oxalate reaction, there is a 2- charge on the left and a 0 charge on the right, so we need to add two electrons: C2O42-(aq)  2CO2(g) + 2e- 3. To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying gives: 10e- + 16H+ + 2MnO4-(aq)  2Mn2+(aq) + 8H2O 5C2O42-(aq)  10CO2(g) + 10e-

22 16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g)
4. Adding gives: 16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g) 5. Which is balanced!

23 Balancing in Basic Solution
If a reaction occurs in 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.

24 Examples – Balance the following oxidation-reduction reactions:
Cr (s) + NO3- (aq)  Cr3+ (aq) + NO (g) (acidic) Al (s) + MnO4- (aq)  Al3+ (aq) + Mn2+ (aq) (acidic) PO33- (aq) + Mn4- (aq)  PO43- (aq) + MnO2 (s) (basic) H2CO (aq) + Ag(NH3)2+ (aq)  HCO3- (aq) + Ag (s) + NH3 (aq) (basic)

25 Section 20-3 20-4 Voltaic Cells – Spontaneous reactions

26 Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

27 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.

28 Voltaic Cells A typical cell looks like this.
The oxidation occurs at the anode. The reduction occurs at the cathode.

29 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.

30 Voltaic Cells 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.

31 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.

32 Voltaic Cells 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.

33 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.

34 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 Ecell.

35 Cell Potential Cell potential is measured in volts (V).
One volt is the potential difference required to impart 1J of energy to a charge of 1 coulomb. (1 electron has a charge of 1.6 x Coulombs). The potential difference between the 2 electrode provides the driving force that pushes the electron through the external circuit.

36 Cell Potential or Electromotive Force (emf)
The “pull” or driving force on the electrons. ELECTROMOTIVE FORCE EMF CAUSES THE ELECTRON MOTION! 1 V = 1 J C

37 Standard Reduction Potentials
Reduction potentials for many electrodes have been measured and tabulated.

38 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) + 2 e−  H2 (g, 1 atm)

39 Standard Cell Potentials
The cell potential at standard conditions can be found through this equation: Ecell = Ered (cathode) − Ered (anode) Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

40 For the reaction to be SPONTANEOUS
E has to be positive

41 Cell Potentials Ered = −0.76 V  Ered = +0.34 V 
For the oxidation in this cell, For the reduction, Ered = −0.76 V Ered = V

42 Cell Potentials Ecell  = Ered (cathode) − (anode)
= V − (−0.76 V) = V

43 2Zn2+(aq) + 4e-  2Zn(s), Ered = -0.76 V.
Since Ered = V we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous. The oxidation of Zn with the SHE is spontaneous. Changing the stoichiometric coefficient does not affect Ered. Therefore, 2Zn2+(aq) + 4e-  2Zn(s), Ered = V. Reactions with Ered > 0 are spontaneous reductions relative to the SHE.

44 Reactions with Ered < 0 are spontaneous oxidations relative to the SHE.
The larger the difference between Ered values, the larger Ecell. In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode). Recall


46 Spontaneity of Redox Reactions
In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode) since Or More generally, for any electrochemical process A positive E indicates a spontaneous process (galvanic cell). A negative E indicates a nonspontaneous process.

47 Cell Potential Calculations
To Calculate cell potential using Standard Reduction Potentials: 1. One reaction and its cell potential’s sign must be reversed--it must be chosen such that the overall cell potential is positive. 2. The half-reactions must often be multiplied by an integer to balance electrons--this is not done for the cell potentials.

48 Cell Potential Calculations Continued
Fe3+(aq) + Cu(s) ----> Cu2+(aq) + Fe2+(aq) Fe3+(aq) + e > Fe2+(aq) Eo = 0.77 V Cu2+(aq) + 2 e > Cu(s) Eo = 0.34 V Reaction # 2 must be reversed.

49 Cell Potential Calculations Continued
2 (Fe3+(aq) + e > Fe2+(aq)) Eo = V Cu(s) ----> Cu2+(aq) + 2 e Eo = V 2Fe3+(aq) + Cu(s) ----> Cu2+(aq) + 2Fe2+(aq) Eo = 0.43 V

50 Oxidizing and Reducing Agents
The greater the difference between the two, the greater the voltage of the cell. REMEMBER TO REVERSE THE SIGN OF THE SPECIE THAT GETS OXIDIZED (THE ONE BELOW!!!!!)

51 Section 4.5 – 4.6 Free energy and redox rx Cell EMF under nonstandard conditions Nerst Equation- Concentration cells Redox Applications - Batteries and fuel cells – Corrosion prevention HW 47, 51 a, 59, 63

52 Oxidizing and Reducing Agents
The strongest oxidizers have the most positive reduction potentials. O.A. PULL electrons The strongest reducers have the most negative reduction potentials. R.A. PUSH their electrons

53 Examples – Sketch the cells containing the following reactions
Examples – Sketch the cells containing the following reactions. Include E°cell, the direction of electron flow, direction of ion migration through salt bridge, and identify the anode and cathode: Cr3+ + Cl2 (g)  Cr2O72- + Cl- Cu2+ + Mg (s)  Mg2+ + Cu (s) IO3- + Fe2+  Fe3+ + I2 Zn (s) + Ag+  Zn2+ + Ag

54 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, and F is a constant, the Faraday. 1 F = 96,485 C/mol = 96,485 J/V-mol

55 Free Energy and Cell Potential
Under standard conditions G = nFE n = number of moles of electrons F = Faraday = 96,485 coulombs per mole of electrons = 96,485 J/Vmol E = V

56 Find the free energy for the reaction of
Zn + Cu2+ ---> Zn2+ + Cu

57 Effect of Concentration on Cell EMF
The Nernst Equation A voltaic cell is functional until E = 0 at which point equilibrium has been reached. The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. The Nernst equation relates emf to concentration using and noting that

58 The Nernst Equation This rearranges to give the Nernst equation: The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K: (Note that change from natural logarithm to base-10 log.) Remember that n is number of moles of electrons.

59 Cell EMF and Chemical Equilibrium
A system is at equilibrium when G = 0. From the Nernst equation, at equilibrium and 298 K (E = 0 V and Q = Keq):

60 Examples Calculate E°cell for the following reaction: Zn (s) + Cu2+  Zn2+ + Cu (s) Calculate Ecell if [Zn2+] = 1.88 M and [Cu2+] = M IO3- (aq) + Fe2+ (aq)  Fe3+ (aq) + I2 (aq) Calculate Ecell if [IO3-] = 0.20 M, [Fe2+] = 0.65 M, [Fe3+] = 1.0 M, and [I2] = 0.75 M

61 Concentration Cells Ecell 
Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell, would be 0, but Q would not. Ecell Therefore, as long as the concentrations are different, E will not be 0.

62 Concentration Cells We can use the Nernst equation to generate a cell that has an emf based solely on difference in concentration. One compartment will consist of a concentrated solution, while the other has a dilute solution. Example: 1.00 M Ni2+(aq) and 1.00 10-3 M Ni2+(aq). The cell tends to equalize the concentrations of Ni2+(aq) in each compartment. The concentrated solution has to reduce the amount of Ni2+(aq) (to Ni(s)), so must be the cathode.

63 Concentration Cells

64 PROBLEM Use the Nernst equation to determine the (emf) at 25oC of the cell. Zn(s)│Zn2+( M ║Cu2+(10.0M)│Cu(s). The standard emf of this cell is 1.10 V.

65 Applications of Oxidation-Reduction Reactions

66 Batteries A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series.

67 Batteries

68 A lead storage battery consists of a lead anode, lead
dioxide cathode, and an electrolyte of 38% sulfuric acid.

69 Pb(s) +H2SO4(aq) ---> PbSO4(aq) + 2H+(aq) + 2e-
Lead Storage Battery Anode reaction: Pb(s) +H2SO4(aq) ---> PbSO4(aq) + 2H+(aq) + 2e- Cathode reaction: PbO2(s)+H2SO4(aq)+ 2e-+2H+(aq)->PbSO4(aq)+2H2O(l) Overall reaction: Pb(s)+ PbO2(s) + 2H2SO4(aq)-->PbSO4(aq)+ 2H2O (l)

70 Common dry cell and its components.

71 2NH4+(aq) + 2MnO2(s) + 2e-  Mn2O3(s) + 2NH3(aq) + 2H2O(l)
Alkaline Battery Anode: Zn cap: Zn(s)  Zn2+(aq) + 2e- Cathode: MnO2, NH4Cl and C paste: 2NH4+(aq) + 2MnO2(s) + 2e-  Mn2O3(s) + 2NH3(aq) + 2H2O(l) The graphite rod in the center is an inert cathode. For an alkaline battery, NH4Cl is replaced with KOH.

72 Anode: Zn powder mixed in a gel:
Zn(s)  Zn2+(aq) + 2e- Cathode: reduction of MnO2.


74 Direct production of electricity from fuels occurs in a fuel cell.
Fuel Cells Direct production of electricity from fuels occurs in a fuel cell. On Apollo moon flights, the H2-O2 fuel cell was the primary source of electricity. Cathode: reduction of oxygen: 2H2O(l) + O2(g) + 4e-  4OH-(aq) Anode: 2H2(g) + 4OH-(aq)  4H2O(l) + 4e-

75 Mercury battery used in calculators.

76 Fuel Cells . . . galvanic cells for which the reactants are continuously supplied. 2H2(g) + O2(g)  2H2O(l) anode: 2H2 + 4OH  4H2O + 4e cathode: 4e + O2 + 2H2O  4OH

77 Hydrogen Fuel Cells

78 Ion selective electrodes are glass electrodes that measures
a change in potential when [H+] varies. Used to measure pH.

79 Corrosion Corrosion of Iron
Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen. Cathode: O2(g) + 4H+(aq) + 4e-  2H2O(l). Anode: Fe(s)  Fe2+(aq) + 2e-. Dissolved oxygen in water usually causes the oxidation of iron. Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3.xH2O(s).

80 Preventing Corrosion of Iron
Oxidation occurs at the site with the greatest concentration of O2. Preventing Corrosion of Iron Corrosion can be prevented by coating the iron with paint or another metal. Galvanized iron is coated with a thin layer of zinc.


82 Zinc protects the iron since Zn is the anode and Fe the cathode:
Zn2+(aq) +2e-  Zn(s), Ered = V Fe2+(aq) + 2e-  Fe(s), Ered = V With the above standard reduction potentials, Zn is easier to oxidize than Fe.


84 Preventing Corrosion of Iron
To protect underground pipelines, a sacrificial anode is added. The water pipe is turned into the cathode and an active metal is used as the anode. Often, Mg is used as the sacrificial anode: Mg2+(aq) +2e-  Mg(s), Ered = V Fe2+(aq) + 2e-  Fe(s), Ered = V


86 Corrosion Some metals, such as copper, gold, silver and platinum, are relatively difficult to oxidize. These are often called noble metals. About 1/5 of all iron and steel produced each year is used to replace rusted metal.

87 Self-protecting Metals
Some metals such as aluminum, copper, and silver form a protective coating that keeps them from corroding further. The protective coating for iron and steel flakes away opening new layers of metal to corrosion.

88 Prevention of Corrosion
Coating--painting or applying oil to keep out oxygen and moisture. Galvanizing--dipping a metal in a more active metal -- galvanized steel bucket. Alloying -- mixing metals with iron to prevent corrosion -- stainless steel. Cathodic protection -- attaching a more active metal. Serves as sacrificial metal--used to protect ships, gas lines, and gas tanks.

89 Redox Titrations Same as any other titration.
the permanganate ion is used often because it is its own indicator. MnO4- is purple, Mn+2 is colorless. When reaction solution remains clear, MnO4- is gone. Chromate ion is also useful, but color change, orangish yellow to green, is harder to detect.

90 Example The iron content of iron ore can be determined by titration with standard KMnO4 solution. The iron ore is dissolved in excess HCl, and the iron reduced to Fe+2 ions. This solution is then titrated with KMnO4 solution, producing Fe+3 and Mn+2 ions in acidic solution. If it requires mL of M KMnO4 to titrate a solution made with g of iron ore, what percent of the ore was iron?

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