Chapter 20 Electrochemistry 1

Electrochemical Reactions In electrochemical reactions, electrons are transferred from one species to another. 2

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

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. 4

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. 5

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. 6

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. 7

Assigning Oxidation Numbers 3. 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 when bonded to a metal, +1 when bonded to a nonmetal. 8

Assigning Oxidation Numbers ◦ Fluorine always has an oxidation number of − 1. ◦ The other halogens have an oxidation number of − 1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. 9

Assigning Oxidation Numbers 4. 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. 10

SAMPLE EXERCISE 20.1 What Chemical Reactions Occur in a Battery? Identify the substances that are oxidized and reduced, and indicate which are oxidizing agents and which are reducing agents. The nickel-cadmium (nicad) battery, a rechargeable “dry cell” used in battery-operated devices, uses the following redox reaction to generate electricity: 11

PRACTICE EXERCISE Identify the oxidizing and reducing agents in the oxidation-reduction reaction 12

Balancing Oxidation-Reduction Equations Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method. 13

Balancing Oxidation-Reduction Equations This involves treating (on paper only) 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. 14

Half-Reaction Method 1. Assign oxidation numbers to determine what is oxidized and what is reduced. 2. Write the oxidation and reduction half- reactions. 15

Half-Reaction Method 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. 16

Half-Reaction Method 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. 17

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) 18

Half-Reaction Method First, we assign oxidation numbers. MnO 4 − + C 2 O 4 2-  Mn 2+ + CO Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized. 19

Oxidation Half-Reaction C 2 O 4 2 −  CO 2 To balance the carbon, we add a coefficient of 2: C 2 O 4 2 −  2 CO 2 20

Oxidation Half-Reaction C 2 O 4 2 −  2 CO 2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side. C 2 O 4 2 −  2 CO e − 21

Reduction Half-Reaction MnO 4 −  Mn 2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. MnO 4 −  Mn H 2 O 22

Reduction Half-Reaction MnO 4 −  Mn H 2 O To balance the hydrogen, we add 8 H + to the left side. 8 H + + MnO 4 −  Mn H 2 O 23

Reduction Half-Reaction 8 H + + MnO 4 −  Mn H 2 O To balance the charge, we add 5 e − to the left side. 5 e − + 8 H + + MnO 4 −  Mn H 2 O 24

Combining the Half-Reactions Now we evaluate the two half-reactions together: C 2 O 4 2 −  2 CO e − 5 e − + 8 H + + MnO 4 −  Mn H 2 O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2. 25

Combining the Half-Reactions 5 C 2 O 4 2 −  10 CO e − 10 e − + 16 H MnO 4 −  2 Mn H 2 O When we add these together, we get: 10 e − + 16 H MnO 4 − + 5 C 2 O 4 2 −  2 Mn H 2 O + 10 CO e − 26

Combining the Half-Reactions 10 e − + 16 H MnO 4 − + 5 C 2 O 4 2 −  2 Mn H 2 O + 10 CO e − The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16 H MnO 4 − + 5 C 2 O 4 2 −  2 Mn H 2 O + 10 CO 2 27

SAMPLE EXERCISE 20.2 Balancing Redox Equations in Acidic Solution Complete and balance this equation by the method of half-reactions : 28

29

PRACTICE EXERCISE Complete and balance the following equations using the method of half-reactions. Both reactions occur in acidic solution. 30

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. 31

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: 32

33

PRACTICE EXERCISE Complete and balance the following equations for oxidation-reduction reactions that occur in basic solution: 34

Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released. 35

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. 36

Voltaic Cells A typical cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode. 37

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. 38

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. 39

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. 40

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. 41

SAMPLE EXERCISE 20.4 Reactions in a Voltaic Cell is spontaneous. A solution containing K 2 Cr 2 O 7 and H 2 SO 4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes. The oxidation-reduction reaction 42

(a) Indicate which reaction occurs at the anode and which at the cathode. (b) Which electrode is consumed in the cell reaction? (c) Which electrode is positive? PRACTICE EXERCISE The two half-reactions in a voltaic cell are 43

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. 44

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. 45

Cell Potential Cell potential is measured in volts (V). 1 V = 1 JCJC 46

Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated. 47

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 −  H 2 (g, 1 atm) 48

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. 49

Cell Potentials For the oxidation in this cell, For the reduction, E red = −0.76 V  E red = V  50

Cell Potentials E cell  = E red  (cathode) − E red  (anode) = V − (−0.76 V) = V 51

SAMPLE EXERCISE 20.5 Calculating E red from E cell ºº 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: 52

The standard emf for this cell is 1.46 V. Using the data in Table 20.1, calculate for the reduction of In 3+ to In +. PRACTICE EXERCISE A voltaic cell is based on the half-reactions 53

SAMPLE EXERCISE 20.6 Calculating E cell from E red ºº Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction 54

PRACTICE EXERCISE Using data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction : 55

SAMPLE EXERCISE 20.7 From Half-Reactions to Cell EMF By using the data in Appendix E, determine (a) the half-reactions that occur at the cathode and the anode, and (b) the standard cell potential. A voltaic cell is based on the following two standard half-reactions : 56

PRACTICE EXERCISE A voltaic cell is based on a Co 2+ /Co half-cell and an AgCl/Ag half-cell. (a) What reaction occurs at the anode? (b) What is the standard cell potential? 57

Oxidizing and Reducing Agents The greater the difference between the two, the greater the voltage of the cell. 58

Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials. 59

SAMPLE EXERCISE 20.8 Determining the Relative Strengths of Oxidizing Agents Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: NO 3 – (aq), Ag + (aq), Cr 2 O 7 2– (aq). PRACTICE EXERCISE Using Table 20.1, rank the following species from the strongest to the weakest reducing agent: I – (aq), Fe(s), Al(s). 60

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 61

Free Energy Under standard conditions,  G  = − nFE  62

SAMPLE EXERCISE 20.9 Spontaneous or Not? Using standard reduction potentials (Table 20.1), determine whether the following reactions are spontaneous under standard conditions. 63

PRACTICE EXERCISE Using the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions: 64

Nernst Equation Remember that  G =  G  + RT ln Q This means − nFE = − nFE  + RT ln Q 65

Nernst Equation Dividing both sides by − nF, we get the Nernst equation: E = E  − RT nF ln Q or, using base-10 logarithms, E = E  − RT nF log Q 66

Nernst Equation At room temperature (298 K), Thus the equation becomes E = E  − n log Q RT F = V 67

Concentration Cells Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. The more dilute is the anode so its [Ni 2+ ] can increase. The more concentrated is the cathode so its [Ni 2+] can decrease. For such a cell,would be 0, but Q would not. E cell  Therefore, as long as the concentrations are different, E will not be 0. (Until they become equal) 68

SAMPLE EXERCISE Determining  G ° and K What are the values of E°,  G°, and K when the reaction is written in this way? (a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free- energy change,  G°, and the equilibrium constant, K, at room temperature (T = 298 K) for the reaction (b) Suppose the reaction in part (a) was written 69

(a) What is the value of n? (b) Use the data in Appendix E to calculate  G°. (c) Calculate K at T = 298 K. PRACTICE EXERCISE For the reaction 70

SAMPLE EXERCISE Voltaic Cell EMF Under Nonstandard Conditions Calculate the emf at 298 K generated by the cell described in Sample Exercise 20.4 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. 71

PRACTICE EXERCISE Calculate the emf generated by the cell described in the practice exercise accompanying Sample Exercise 20.6 when [Al 3+ ] = 4.0  10 –3 M and [I – ] = M. 2 Al + 3 I 2  2 Al I 1- E o = 2.20 V 72

SAMPLE EXERCISE Concentrations in a Voltaic Cell If the voltage of a Zn–H + cell (like that in Figure 20.11) is 0.45 V at 25°C when [Zn 2+ ] = 1.0 M and atm, what is the concentration of H + ? 73

PRACTICE EXERCISE What is the pH of the solution in the cathode compartment of the cell pictured in Figure when atm, [Zn 2+ ] in the anode compartment is 0.10 M, and cell emf is V? 74

[Zn 2+ ] = 0.10 M V P(H 2 )= 1.0 atm 75

SAMPLE EXERCISE pH of a Concentration Cell A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has atm and an unknown concentration of H + (aq). Electrode 2 is a standard hydrogen electrode ([H + ] = 1.00 M, atm). At 298 K the measured cell voltage is V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate[H + ] for the solution at electrode 1. What is its pH? 76

PRACTICE EXERCISE A concentration cell is constructed with two Zn(s)-Zn 2+ (aq) half-cells. The first half-cell has [Zn 2+ ] = 1.35 M, and the second half-cell has [Zn 2+ ] = 3.75  10 –4 M. (a) Which half-cell is the anode of the cell? (b) What is the emf of the cell? 77

Electric Current and Electrolysis Coulombs = Joules/Volts Coulombs = Amperes X Seconds (q = I * t) Faraday’s Constant = 96,500 Coulombs/mole of e- 78

SAMPLE EXERCISE Aluminum 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. 79

PRACTICE EXERCISE (a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl 2 is Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00  10 3 s. (b) How many seconds would be required to produce 50.0 g of Mg from MgCl 2 if the current is A? 80

SAMPLE EXERCISE Calculating Energy in Kilowatt-hours Calculate the number of kilowatt-hours of electricity required to produce 1.0  10 3 kg of aluminum by electrolysis of Al 3+ if the applied voltage is 4.50 V. 81

PRACTICE EXERCISE Calculate the number of kilowatt-hours of electricity required to produce 1.00 kg of Mg from electrolysis of molten MgCl 2 if the applied emf is 5.00 V. Assume that the process is 100% efficient. 82

Applications of Oxidation- Reduction Reactions 83

Batteries 84

Alkaline Batteries 85

Hydrogen Fuel Cells 86

Corrosion and… 87

…Corrosion Prevention 88

SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together The K sp at 298 K for iron(II) fluoride is 2.4  10 –6. (a) Write a half-reaction that gives the likely products of the two-electron reduction of FeF 2 (s) in water. (b) Use the K sp value and the standard reduction potential of Fe 2+ (aq) to calculate the standard reduction potential for the half-reaction in part (a). (c) Rationalize the difference in the reduction potential for the half-reaction in part (a) with that for Fe 2+ (aq). 89