Electrochemistry Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry

Transcription

1 2012 Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry

2 Electricity from Chemistry Many chemical reactions involve the transfer of electrons between atoms or ions electron transfer reactions all single displacement, and combustion reactions some synthesis and decomposition reactions The flow of electrons is associated with electricity Basic research into the nature of this relationship may, in the near future, lead to cheaper, more efficient ways of generating electricity fuel cell 2 Copyright 2011 Pearson Education, Inc.

3 Electrochemical Reactions In electrochemical reactions, electrons are transferred from one species to another Pearson Education, Inc.

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

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 Pearson Education, Inc.

6 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 Pearson Education, Inc.

7 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 Pearson Education, Inc.

8 Oxidation Reduction Reactions where electrons are transferred from one atom to another are called oxidation reduction reactions redox reactions for short Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced 2 Na(s) + Cl 2 (g) 2 Na + Cl (s) Na Na e oxidation Cl e 2 Cl reduction Oil Rig 8

9 Rules for Assigning Oxidation States Rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl 2 = 0 in 2 Na(s) + Cl 2 (g) 2. monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = 1 in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = 1 in NaCl, (+1) + ( 1) = 0 9

10 Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = 2 in NO 3, (+5) + 3( 2) = 1 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl 2 10

11 Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority Nonmetal Oxidation State Example F 1 CF 4 H +1 CH 4 O 2 CO 2 Group 7A 1 CCl 4 Group 6A 2 CS 2 Group 5A 3 NH 3 11

12 Example: Determine the oxidation states of all the atoms in a propanoate ion, C 3 H 5 O 2 There are no free elements or free ions in propanoate, so the first rule that applies is Rule 3b (C 3 ) + (H 5 ) + (O 2 ) = 1 Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order H = +1 O = 2 (C 3 ) + 5(+1) + 2( 2) = 1 (C 3 ) = 2 C = ⅔ Note: unlike charges, oxidation states can be fractions! 12

13 Practice Assign an oxidation state to each element in the following Br 2 K + LiF CO 2 SO 2 4 Na 2 O 2 Br = 0 (Rule 1) K = +1 (Rule 2) Li = +1 (Rule 4a) & F = 1 (Rule 5) O = 2 (Rule 5) & C = +4 (Rule 3a) O = 2 (Rule 5) & S = +6 (Rule 3b) Na = +1 (Rule 4a) & O = 1 (Rule 3a) 13

14 Example: Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reaction Reducing Agent Oxidizing Agent Fe + MnO H + Fe 3+ + MnO H 2 O Oxidation Reduction 14

15 Practice Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions Sn 4+ + Ca Sn 2+ + Ca Ca is oxidized, Sn 4+ is reduced Ca is the reducing agent, Sn 4+ is the oxidizing agent F 2 + S SF S is oxidized, F is reduced S is the reducing agent, F 2 is the oxidizing agent 15

16 Balancing Oxidation-Reduction Equations Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method Pearson Education, Inc.

17 Half-Reactions We generally split the redox reaction into two separate half-reactions a reaction just involving oxidation or reduction the oxidation half-reaction has electrons as products the reduction half-reaction has electrons as reactants 3 Cl 2 + I + 3H 2 O 6 Cl + IO H oxidation: I IO e reduction: Cl e 2 Cl 17

18 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 Pearson Education, Inc.

19 The Half-Reaction Method 3. Balance each halfreaction. 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 Pearson Education, Inc.

20 The Half-Reaction Method 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 Pearson Education, Inc.

21 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) 2012 Pearson Education, Inc.

22 The Half-Reaction Method First, we assign oxidation numbers: MnO 4 + C 2 O 4 2 Mn 2+ + CO 2 Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized Pearson Education, Inc.

23 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 2CO Pearson Education, Inc.

24 Oxidation Half-Reaction C 2 O 4 2 2CO 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 2CO 2 + 2e 2012 Pearson Education, Inc.

25 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 2012 Pearson Education, Inc.

26 Reduction Half-Reaction MnO 4 Mn H 2 O To balance the hydrogen, we add 8H + to the left side: 8H + + MnO 4 Mn H 2 O 2012 Pearson Education, Inc.

27 Reduction Half-Reaction 8H + + MnO 4 Mn H 2 O To balance the charge, we add 5e to the left side: 5e + 8H + + MnO 4 Mn H 2 O 2012 Pearson Education, Inc.

28 Combining the Half-Reactions Now we evaluate the two half-reactions together: C 2 O 4 2 2CO 2 + 2e 5e + 8H + + 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: 2012 Pearson Education, Inc.

29 Combining the Half-Reactions 5C 2 O CO e 10e + 16H + + 2MnO 4 2Mn H 2 O When we add these together, we get: 10e + 16H + + 2MnO 4 + 5C 2 O 2 4 2Mn H 2 O + 10CO 2 +10e 2012 Pearson Education, Inc.

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

31 Example: Balancing redox reactions in acidic solution 1. assign oxidation states and determine element oxidized and element reduced 2. separate into oxidation & reduction halfreactions Fe 2+ + MnO 4 Fe 3+ + Mn oxidation ox: Fe 2+ Fe 3+ reduction red: MnO 4 Mn 2+ 31

32 Example: Balancing redox reactions in acidic solution 3. balance halfreactions by mass a) first balance atoms other than O and H b) balance O by adding H 2 O to side that lacks O c) balance H by adding H + to side that lacks H Fe 2+ Fe 3+ MnO 4 Mn 2+ MnO 4 Mn H 2 O MnO 4 + 8H + Mn H 2 O 32

33 Example: Balancing redox reactions in acidic solution 4. balance each half-reaction with respect to charge by adjusting the numbers of electrons a) electrons on product side for oxidation b) electrons on reactant side for reduction Fe 2+ Fe e MnO 4 + 8H + Mn H 2 O MnO 4 + 8H e Mn H 2 O 33

34 Example: Balancing redox reactions 5. balance electrons between halfreactions 6. add halfreactions, canceling electrons and common species 7. Check that numbers of atoms and total charge are equal in acidic solution Fe 2+ Fe e } x 5 MnO 4 + 8H e Mn H 2 O 5 Fe 2+ 5 Fe e MnO 4 + 8H e Mn H 2 O 5 Fe 2+ + MnO 4 + 8H + Mn H 2 O + 5 Fe 3+ reactant side Element product side 5 Fe 5 1 Mn 1 4 O 4 8 H charge

35 Practice Balance the following equation in acidic solution I + Cr 2 O 7 2 Cr 3+ + I 2 35

36 Practice Balancing redox reactions 1. assign oxidation states and determine element oxidized and element reduced 2. separate into oxidation & reduction halfreactions I + Cr 2 O 7 2 I 2 + Cr oxidation ox: I I 2 reduction red: Cr 2 O 7 2 Cr 3+ 36

37 Practice Balancing redox reactions 3. balance halfreactions by mass a) first balance atoms other than O and H b) balance O by adding H 2 O to side that lacks O c) balance H by adding H + to side that lacks H ox: I I 2 ox: 2 I I 2 red: Cr 2 O 2 7 Cr 3+ red: Cr 2 O Cr 3+ red: Cr 2 O 2 7 2Cr 3+ +7H 2 O Cr 2 O H + 2Cr 3+ +7H 2 O 37

38 Practice Balancing redox reactions 4. balance each half-reaction with respect to charge by adjusting the numbers of electrons a) electrons on product side for oxidation b) electrons on reactant side for reduction 2 I I 2 + 2e Cr 2 O H + + 6e 2Cr 3+ +7H 2 O Cr 2 O H + 2Cr 3+ +7H 2 O 38

39 Practice Balancing redox reactions 5. balance electrons between halfreactions 6. add halfreactions, canceling electrons and common species 7. check 26 I I3 2 I+ 2e 2 + 6e } x 3 Cr 2 O H + + 6e 2Cr 3+ +7H 2 O Cr 2 O H I 2Cr 3+ +7H 2 O + 3 I 2 reactant side Element product side 2 Cr 2 6 I 6 7 O 7 14 H charge +6 39

40 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 Pearson Education, Inc.

41 Example: Balance the equation: I (aq) + MnO 4 (aq) I 2(aq) + MnO 2(s) in basic solution assign oxidation states separate into halfreactions I (aq) + MnO 4 (aq) I 2(aq) + MnO 2(s) Oxidation Reduction ox: I (aq) I 2(aq) red: MnO 4 (aq) MnO 2(s) 41

42 Example: Balance the equation: I (aq) + MnO 4 (aq) I 2(aq) + MnO 2(s) in basic solution balance half- ox: 2 ox: I I (aq) I2 (aq) I I 2(aq) I I 2(aq) I 2(aq) halfreactions by red: 4 red: H + + MnO (aq) + MnO 4 (aq) 44 (aq) MnO MnO 2(s) 2(s) H 2 O H (l) 2 O 2 (l) (l) reactions mass by first, then by mass elements O by in other adding then base, H than by H 2 O and adding O H + 4 H 2 O (aq) + MnO 4 (aq) MnO 2(s) + 2 H 2 O (l) + 4 OH (aq) neutralize the H + 2 H 2 O (l) + MnO 4 (aq) MnO 2(s) + 4 OH (aq) with OH 4 H + (aq) + 4 OH (aq) + MnO 4 (aq) MnO 2(s) + 2 H 2 O (l) + 4 OH (aq) 42

43 Example: Balance the equation: I (aq) + MnO 4 (aq) I 2(aq) + MnO 2(s) in basic solution balance Halfreactions by charge Balance electrons between halfreactions ox: 2 I (aq) I + 2 e 2(aq) red: MnO + 3 e MnO 2(s) + 4 OH 4 (aq) + 2 H 2 O (l) MnO 2(s) + 4 OH (aq) (aq) ox: 2 I (aq) I 2(aq) + 2 e } x3 red: MnO 4 (aq) + 2 H 2 O (l) + 3 e MnO 2(s) + 4 OH (aq) }x2 ox: 6 I (aq) 3 I 2(aq) + 6 e red: 2 MnO 4 (aq) + 4 H 2 O (l) + 6 e 2 MnO 2(s) + 8 OH (aq) 43

44 Example: Balance the equation: I (aq) + MnO 4 (aq) I 2(aq) + MnO 2(s) in basic solution add the halfrxns ox: 6 I (aq) 3 I 2(aq) + 6 e red: 2 MnO 4 (aq) + 4 H 2 O (l) + 6 e 2 MnO 2(s) + 8 OH (aq) tot: 6 I (aq)+ 2 MnO 4 (aq) + 4 H 2 O (l) 3 I 2(aq) + 2 MnO 2(s) + 8 OH (aq) check Reactant Count Product Count Element 6 I 6 2 Mn 2 12 O 12 8 H 8 8 charge 8 44

45 Practice Balance the following equation in basic solution I + CrO 4 2 Cr 3+ + I 2 45

46 Practice Balancing redox reactions 1. assign oxidation states and determine element oxidized and element reduced 2. separate into oxidation & reduction halfreactions I + CrO 4 2 I 2 + Cr oxidation ox: I I 2 reduction red: CrO 4 2 Cr 3+ 46

47 Practice Balancing redox reactions 3. balance halfreactions by mass a) first balance atoms other than O and H b) balance O by adding H 2 O to side that lacks O c) balance H by adding H + to side that lacks H d) if in basic solution, neutralize the H + with OH ox: I I 2 ox: 2 I I 2 red: CrO 4 2 Cr 3+ red: CrO 4 2 Cr 3+ +4H 2 O CrO H + Cr 3+ +4H 2 O CrO OH Cr 3+ +4H 2 O +8OH 4 +8H 2 O Cr 3+ +4H 2 O +8OH CrO H 2 O Cr 3+ +8OH 47

48 Practice Balancing redox reactions 4. balance each half-reaction with respect to charge by adjusting the numbers of electrons a) electrons on product side for oxidation b) electrons on reactant side for reduction 2 I I 2 + 2e CrO 2 Cr 3+ +8OH 4 +4H 2 O+ 3e Cr 3+ +8OH 48

49 Practice Balancing redox reactions 5. balance electrons between halfreactions 6. add halfreactions, canceling electrons and common species 7. check 26 I I3 2 I+ 2e 2 + 6e } x 3 2CrO H 2 O+ 6e 2Cr OH 4 +4H 2 O+ 3e Cr 3+ +8OH } x 2 2CrO H 2 O + 6 I 2Cr OH + 3 I 2 reactant side Element product side 2 Cr 2 6 I 6 16 O H charge 10 49

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

51 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 or a galvanic cell Pearson Education, Inc.

52 A typical cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode. Think Red Cat Voltaic Cells 2012 Pearson Education, Inc.

53 Voltaic Cell the salt bridge is required to complete the circuit and maintain charge balance 53

54 Electrochemical Cells Oxidation and reduction half-reactions are kept separate in half-cells Electron flow through a wire along with ion flow through a solution constitutes an electric circuit Requires a conductive solid electrode to allow the transfer of electrons through external circuit metal or graphite requires ion exchange between the two half-cells of the system electrolyte 54

55 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 Pearson Education, Inc.

56 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. Voltaic Cells 2012 Pearson Education, Inc.

57 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. Voltaic Cells 2012 Pearson Education, Inc.

58 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 Pearson Education, Inc.

59 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 Pearson Education, Inc.

60 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 Pearson Education, Inc.

61 Cell Potential Cell potential is measured in volts (V). 1 V = 1 J C 2012 Pearson Education, Inc.

62 Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated Pearson Education, Inc.

63 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) 2012 Pearson Education, Inc.

64 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 Pearson Education, Inc.

65 Cell Potentials For the oxidation in this cell, E red = 0.76 V For the reduction, E red = V 2012 Pearson Education, Inc.

66 Cell Potentials E cell = E red (cathode) E red (anode) = V ( 0.76 V) = V 2012 Pearson Education, Inc.

67 Which Way Will Electrons flow? Potential Energy Zn Zn e Cu e Cu Electron Flow Zn Zn e E = Cu Cu e E = 0.34 Under standard conditions, zinc has a stronger tendency to oxidize than copper Therefore the electrons flow from zinc; making zinc the anode 67

68 Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials Pearson Education, Inc.

69 Oxidizing and Reducing Agents The greater the difference between the two, the greater the voltage of the cell Pearson Education, Inc.

70 70

71 71

72 Example: Calculate E cell for the reaction at 25 C Al (s) + NO 3 (aq) + 4 H + (aq) Al 3+ (aq) + NO (g) + 2 H 2 O (l) separate the reaction into the oxidation and reduction half-reactions find the E for each halfreaction and sum to get E cell ox (anode): Al (s) Al 3+ (aq) + 3 e red: NO 3 (aq) + 4 H + (aq) + 3 e NO (g) + 2 H 2 O (l) E ox of Al = E 3+ red of Al = V E red of NO3 = V E cell = (+1.66 V) + (+0.96 V) = V 72

73 Practice Calculate E cell for the reaction at 25 C IO 3 (aq) + 6 H + (aq) + 5 I (aq) 3 I 2 (s) + 3 H 2 O(l) Reduction Half-Reaction E red, V F 2 (g) + 2e 2 F (aq) IO 3 (aq) + 6 H + + 5e ½I 2 (s) + 3H 2 O(l) Ag + (aq) + 1e Ag(s) I 2 (s) + 2e 2 I (aq) Cu 2+ (aq) + 2e Cu(s) Cr 3+ (aq) + 1e Cr 2+ (aq) 0.50 Mg 2+ (aq) + 2e Mg(s)

74 Practice Calculate E cell for the reaction at 25 C 2 IO H I 6 I H 2 O separate the reaction into the oxidation and reduction half-reactions find the E for each halfreaction and sum to get E cell ox (anode): 2 I (s) I 2(aq) + 2 e red: IO 3 (aq) + 6 H + (aq) + 5 e ½ I 2(s) + 3 H 2 O (l) E ox of I = E red of I = 0.54 V E red of IO3 = V 2 E cell = ( 0.54 V) + (+1.20 V) = V 74

75 Practice Sketch and label a voltaic cell in which one half-cell has Ag(s) immersed in 1 M AgNO 3, and the other half-cell has a Pt electrode immersed in 1 M Cr(NO 3 ) 2 and 1 M Cr(NO 3 ) 3 Write the half-reactions and overall reaction, and determine the cell potential under standard conditions. Reduction Half-Reaction E red, V F 2 (g) + 2e 2 F (aq) IO 3 (aq) + 6 H + + 5e ½I 2 (s) + 3H 2 O(l) Ag + (aq) + 1e Ag(s) I 2 (s) + 2e 2 I (aq) Cu 2+ (aq) + 2e Cu(s) Cr 3+ (aq) + 1e Cr 2+ (aq) 0.50 Mg 2+ (aq) + 2e Mg(s)

76 e e e salt bridge e anode = Pt cathode = Ag Cr 2+ Cr 3+ Ag + ox: Cr 2+ (aq) Cr 3+ (aq) + 1 e E = V red: Ag + (aq) + 1 e Ag(s) E = V tot: Cr 2+ (aq) + Ag + (aq) Cr 3+ (aq) + Ag(s) E = V 76

77 Cell Notation Shorthand description of a voltaic cell Electrode electrolyte electrolyte electrode Oxidation half-cell on the left, reduction halfcell on the right Single = phase barrier if multiple electrolytes in same phase, a comma is used rather than often use an inert electrode Double line = salt bridge 77

78 Voltaic Cell Anode = Zn(s) the anode is oxidized to Zn 2+ Cathode = Cu(s) Cu 2+ ions are reduced at the cathode Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) 78

79 Zn(s) Zn 2+ (1 M) H + (1 M) H 2 (g) Pt(s) 79

80 Predicting Spontaneity of Redox Reactions A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction lower on the table If paired the other way, the reverse reaction is spontaneous Cu 2+ (aq) + 2 e Cu(s) E red = V Zn 2+ (aq) + 2 e Zn(s) E red = 0.76 V Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Cu(s) + Zn 2+ (aq) Cu 2+ (aq) + Zn(s) spontaneous nonspontaneous 80

81 separate the reaction into the oxidation and reduction half-reactions look up the relative positions of the reduction half-reactions Example: Predict if the following reaction is spontaneous under standard conditions Fe (s) + Mg 2+ (aq) Fe 2+ (aq) + Mg (s) ox: red: red: red: Fe (s) Fe 2+ (aq) + 2 e Mg 2+ (aq) + 2 e Mg (s) Fe 2+ (aq) + 2 e Fe (s) Mg 2+ (aq) + 2 e Mg (s) because Mg 2+ reduction is below Fe 2+ reduction, the reaction is NOT spontaneous as written 81

82 the reaction is spontaneous in the reverse direction sketch the cell and label the parts oxidation occurs at the anode; electrons flow from anode to cathode Mg (s) + Fe 2+ (aq) Mg 2+ (aq) + Fe (s) ox: red: Mg (s) Mg 2+ (aq) + 2 e Fe 2+ (aq) + 2 e Fe (s) 82

83 Practice Decide whether each of the following will be spontaneous as written or in the reverse direction F 2(g) + 2 I (aq) I 2(s) + 2 F (aq) spontaneous as written Mg (s) + 2 Ag + (aq) Mg 2+ (aq) + 2 Ag (s) spontaneous as written Cu 2+ (aq) + 2 I (aq) I 2(s) + Cu (s) spontaneous in reverse Cu 2+ (aq) + 2 Cr 2+ (aq) Cu (s) + 2 Cr 3+ (aq) spontaneous as written Reduction Half-Reaction F 2 (g) + 2e 2 F (aq) IO 3 (aq) + 6 H + + 5e I 2 (s) + 3H 2 O(l) Ag + (aq) + 1e Ag(s) I 2 (s) + 2e 2 I (aq) Cu 2+ (aq) + 2e Cu(s) Cr 3+ (aq) + 1e Cr 2+ (aq) Mg 2+ (aq) + 2e Mg(s) 83

84 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 2012 Pearson Education, Inc.

85 Example: Calculate G for the reaction I 2(s) + 2 Br (aq) Br 2(l) + 2 I (aq) Given: Find: Conceptual Plan: I 2(s) + 2 Br (aq) Br 2(l) + 2 I (aq) DG, (J) E ox, E red E cell G Relationships: Solve: Answer: ox: 2 Br (aq) Br 2(l) + 2 e E = 1.09 V red: I 2(l) + 2 e 2 I (aq) E = V tot: I 2(l) + 2Br (aq) 2I (aq) + Br 2(l) E = 0.55 V because G is +, the reaction is not spontaneous in the forward direction under standard conditions 85

86 Practice Calculate G for the reaction at 25 C 2IO 3 (aq) + 12H + (aq) + 10 I (aq) 6I 2 (s) + 6H 2 O(l) Reduction Half-Reaction E red, V F 2 (g) + 2e 2 F (aq) IO 3 (aq) + 6 H + + 5e ½I 2 (s) + 3H 2 O(l) Ag + (aq) + 1e Ag(s) I 2 (s) + 2e 2 I (aq) Cu 2+ (aq) + 2e Cu(s) Cr 3+ (aq) + 1e Cr 2+ (aq) 0.50 Mg 2+ (aq) + 2e Mg(s)

87 Practice Calculate G for the reaction at 25 C 2IO 3 (aq) + 12H + (aq) + 10 I (aq) 6 I 2 (s) + 6 H 2 O(l) Given: Find: Conceptual Plan: 2IO 3 (aq) +12H + (aq)+10i (aq) 6 I 2(s) + 6 H 2 O(l) G, (J) E ox, E red E cell G Relationships: Solve: ox : 2 I (s) I 2 (aq) + 2 e Eº = 0.54 V red: IO 3 (aq) + 6 H + (aq) + 5 e ½ I 2 (s) + 3 H 2 O(l) Eº = 1.20 V tot: 2IO 3 (aq) + 12H + (aq) + 10I (aq) 6I 2(s) + 6H 2 O (l) Eº = 0.66 V Answer: because G is, the reaction is spontaneous in the forward direction under standard conditions 87 Tro: Chemistry: A Molecular Approach, 2/e

88 E cell, G and K For a spontaneous reaction one that proceeds in the forward direction with the chemicals in their standard states - G < 1 (negative) E > 1 (positive) K > 1 G = RTlnK = nfe cell n is the number of electrons F = Faraday s Constant = 96,485 C/mol e 88

89 Example: Calculate K at 25 C for the reaction Cu (s) + 2 H + (aq) H 2(g) + Cu 2+ (aq) Given: Find: Conceptual Plan: Relationships: Cu (s) + 2 H + (aq) H 2(g) + Cu 2+ (aq) K E ox, E red E cell K Solve: ox: Cu (s) Cu 2+ (aq) + 2 e red: 2 H + (aq) + 2 e H 2(aq) E = V tot: Cu (s) + 2H + (aq) Cu 2+ (aq) + H 2(g) E = 0.34 V Answer: E = 0.34 V because K < 1, the position of equilibrium lies far to the left under standard conditions 89 Tro: Chemistry: A Molecular Approach, 2/e

90 Practice Calculate K for the reaction at 25 C 2IO 3 (aq) + 12H + (aq) + 10 I (aq) 6I 2 (s) + 6H 2 O(l) Reduction Half-Reaction E red, V F 2 (g) + 2e 2 F (aq) IO 3 (aq) + 6 H + + 5e ½I 2 (s) + 3H 2 O(l) Ag + (aq) + 1e Ag(s) I 2 (s) + 2e 2 I (aq) Cu 2+ (aq) + 2e Cu(s) Cr 3+ (aq) + 1e Cr 2+ (aq) 0.50 Mg 2+ (aq) + 2e Mg(s)

91 Practice Calculate K for the reaction at 25 C 2IO 3 (aq) + 12H + (aq) + 10 I (aq) 6I 2 (s) + 6H 2 O(l) Given: Find: Conceptual Plan: Relationships: 2 IO 3 (aq) +12 H + (aq)+10 I (aq) 6 I 2(s) + 6 H 2 O (l) K E ox, E red E cell K Solve: ox : 2 I (s) I 2 (aq) + 2e Eº = 0.54 V red: IO 3 (aq) + 6 H + (aq) + 5e ½ I 2 (s) + 3H 2 O(l) Eº = 1.20 V tot: 2IO 3 (aq) + 12H + (aq) + 10I (aq) 6I 2(s) + 6H 2 O (l) Eº = 0.66 V Answer: because K >> 1, the position of equilibrium lies far to the right under standard conditions 91

92 Cell Potential when Ion Concentrations Are Not 1 M We know there is a relationship between the reaction quotient, Q, the equilibrium constant, K, and the free energy change, Gº Changing the concentrations of the reactants and products so they are not 1 M will affect the standard free energy change, Gº Because Gº determines the cell potential, E cell, the voltage for the cell will be different when the ion concentrations are not 1 M 92

93 E cell When Ion Concentrations are Not 1 M 93

94 E cell When Ion Concentrations are Not 1 M The standard cell potential, Eo, corresponds to the standard conditions of Q = 1. As the system approaches equilibrium, the magnitude (i.e., absolute value) of the cell potential decreases, reaching zero at equilibrium (when Q K). Deviations from standard conditions that take the cell further from equilibrium than Q = 1 will increase the magnitude of the cell potential relative to E. Deviations from standard conditions that take the cell closer to equilibrium than Q = 1 will decrease the magnitude of the cell potential relative to E. In concentration cells, the direction of spontaneous electron flow can be determined by considering the direction needed to reach equilibrium.

95 Deriving the Nernst Equation 95

96 Electrochemical Cells In all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathode In voltaic cells anode is the source of electrons and has a ( ) charge cathode draws electrons and has a (+) charge In electrolytic cells electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery electrons are forced toward the anode, so it must have a source of electrons, the terminal of the battery 96

97 97

98 Electrolysis Electrolysis is the process of using electrical energy to break a compound apart Electrolysis is done in an electrolytic cell Electrolytic cells can be used to separate elements from their compounds 98

99 Electrolysis In electrolysis we use electrical energy to overcome the energy barrier of a non-spontaneous reaction, allowing it to occur The reaction that takes place is the opposite of the spontaneous process 2 H 2 (g) + O 2 (g) 2 H 2 O(l) spontaneous 2 H 2 O(l) 2 H 2 (g) + O 2 (g) electrolysis Some applications are (1) metal extraction from minerals and purification, (2) production of H 2 for fuel cells, (3) metal plating 99

100 Electrolytic Cells The electrical energy is supplied by a Direct Current power supply AC alternates the flow of electrons so the reaction won t be able to proceed Some electrolysis reactions require more voltage than E cell predicts. This is called the overvoltage 100

101 Electrolytic Cells The source of energy is a battery or DC power supply The + terminal of the souce is attached to the anode The terminal of the source is attached to the cathode Electrolyte can be either an aqueous salt solution or a molten ionic salt Cations in the electrolyte are attracted to the cathode and anions are attracted to the anode Cations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidized 101

102 Electrolysis of Pure Compounds The compound must be in molten (liquid) state Electrodes normally graphite Cations are reduced at the cathode to metal element Anions oxidized at anode to nonmetal element 102

103 Electrolysis of Water 103

104 Electroplating In electroplating, the work piece is the cathode Cations are reduced at cathode and plate to the surface of the work piece The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution. 104

105 Current Current is the number of electrons that flow through the system per second unit = Ampere 1 A of current = 1 Coulomb of charge flowing by each second 1 A = x electrons per second Electrode surface area dictates the number of electrons that can flow larger batteries produce larger currents 105

106 Stoichiometry of Electrolysis In an electrolytic cell, the amount of product made is related to the number of electrons transferred essentially, the electrons are a reactant The number of moles of electrons that flow through the electrolytic cell depends on the current and length of time 1 Amp = 1 Coulomb of charge/second 1 mole of e = 96,485 Coulombs of charge Faraday s constant 106

107 Faraday s Laws The first law of electrolysis We can link this product to the mass of metal produced at the cathode as being directly proportional to the quantity of electricity passed through the cell. This relationship is known as Faraday s first law of electrolysis. It may be written as: m

108 Example: Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction Au 3+ (aq) + 3 e Au(s) Given: Find: 3 mol e : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, g Conceptual Plan: t(s), amp charge (C) mol e mol Au g Au Relationships: Solve: Check: units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e 108

109 Practice Calculate the amperage required to plate 2.5 g of gold in 1 hour (3600 sec) Au 3+ (aq) + 3 e Au(s) 109

110 Practice Calculate the amperage required to plate 2.5 g of Au in 1 hour (3600 sec) Au 3+ (aq) + 3 e Au(s) Given: Find: 3 mol e : 1 mol Au, mass = 2.5 g, time = 3600 s current, A Conceptual Plan: g Au mol Au mol e charge amps Relationships: Solve: Check: units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e 110

111 Applications of Oxidation-Reduction Reactions 2012 Pearson Education, Inc.

112 Batteries 2012 Pearson Education, Inc.

113 Alkaline Batteries 2012 Pearson Education, Inc.

114 Hydrogen Fuel Cells 2012 Pearson Education, Inc.

115 Corrosion and 2012 Pearson Education, Inc.

116 Corrosion Prevention 2012 Pearson Education, Inc.