Chapter Nineteen. Electrochemistry

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1 Chapter Nineteen Electrochemistry 1

2 Electrochemistry The study of chemical reactions through electrical circuits. Monitor redox reactions by controlling electron transfer REDOX: Shorthand for REDuction-OXidation Use redox reactions to experimentally measure: Reaction progress (kinetics) Composition (equilibrium constants) Energy changes (thermodynamics) 2

3 Oxidation-Reduction Reactions Oxidation-reduction reactions (REDOX reaction) occur when electrons are transferred from one reactant to another during a chemical reaction. There is a change in oxidation number for both substances Oxidation Number: Theoretical charge on an ion Oxidation is the process where the oxidation number increases. Electrons are lost from the substance Reduction is the process where the oxidation number decreases. Electrons are gained by the substance Oxidation and reduction always accompany each other; Neither can occur alone 3

Redox Reaction: Magnesium Burning Oxidation: Mg (s) Mg 2+ + 2e-

4 Redox Reaction: Magnesium Burning Oxidation: Mg (s) Mg e- Reduction: 1/2O 2 (g) + 2e- O 2- Reaction: Mg (s) + 1/2O 2 (g) MgO(s) 4

5 LEO the lion says GER LEO Lose Electrons Oxidation GER Gain Electrons Reduction 5

6 Oxidation Number Rules: See Chapter 4 The rule earlier in the list always takes precedence. 1) ON = 0 for a compound or ionic charge for an ion 2)ON = +1 for IA elements and ON = +2 2A elements 3) ON= -2 for oxygen 4)ON= -1 for 7A elements If both elements in 7A, then the one higher in the list is -1 5)ON = -2 for 6A elements other than oxygen 6) ON = -3 for 5A elements (very shaky!!!) 6

7 Common Oxidation Numbers Must be able to determine charges on each element 7

8 Balancing Redox Reactions: Balance Elements 1. Write equation for the ions in acid solution: Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr Divide equation into 2 half reactions Fe 2+ Fe 3+ Cr 2 O 7 2-2Cr Balance all elements except H and O 4. Balance O with H 2 O Fe 2+ Fe 3+ Cr 2 O 7 2-2Cr 3+ Fe 2+ Fe Balance H with H + Fe 2+ Fe 3+ Cr 2 O 7 2-2Cr H 2 O Cr 2 O H + 2Cr H 2 O 8

9 Balancing Redox Reactions: Charge 6. Balance charge Fe 2+ Fe e- 6e- + Cr 2 O H + 2Cr H 2 O Oxidation: lose electron Reduction: gain electrons 7. Multiply by an integer to equalize # electrons 6Fe 2+ 6Fe e- 6e- + Cr 2 O H + 2Cr H 2 O 8. Add reactions together and cancel like species 6Fe e- + Cr 2 O H + 2Cr H 2 O+ 6Fe e- 9. Check balance of final # of atoms and charge 6Fe 2+ + Cr 2 O H + 2Cr H 2 O+ 6Fe 3+ (12+) + (2-) + (14+) = +24 (6+) + (0) + (18+) = +24 9

10 Balancing Redox Reactions: Basic Reactions Need OH - instead of H + in final equation 1. Balance as if the reaction is in acidic solution. 6Fe 2+ + Cr 2 O H + 2Cr H 2 O+ 6Fe Add OH- to both sides to match H + 14OH - + 6Fe 2+ + Cr 2 O H + 2Cr H 2 O+ 6Fe OH - 3. Combine OH- and H+ to make water 14H 2 O+ 6Fe 2+ + Cr 2 O 7 2-2Cr H 2 O+ 6Fe OH - 4. Cancel water from both sides 7H 2 O+ 6Fe 2+ + Cr 2 O 7 2-2Cr Fe OH - 5. Check to see if balanced 10

11 Galvanic Cells 11

Parts of an Electrochemical Cell Ionic Solutions Provide ions to transfer charge Solution + Electrode= Half-cell Electrodes Anode: oxidation occurs Cathode: reduction occurs Salt bridge.

12 Parts of an Electrochemical Cell Ionic Solutions Provide ions to transfer charge Solution + Electrode= Half-cell Electrodes Anode: oxidation occurs Cathode: reduction occurs Salt bridge. Keeps 2 half-cells connected Ions flow, but solution doesn t Metal wires Connect the electrodes to the terminals of the voltmeter. Provide means of transporting electrons between electrodes Voltmeter Measures the electron flow in the system 12

13 Galvanic Cell: Daniell Cell 13

14 Types of Cells Voltaic or Galvanic Cell Net oxidation/reduction reaction is spontaneous Convert energy to useful work Batteries Electrolytic Cell Net redox reaction is non-spontaneous Work is done on the cell, energy must be supplied: Recharge a battery, electroplating 14

15 Standard Reduction Potentials 15

16 Cell Diagram Reaction: Zn (s) + Cu 2+ Zn 2+ + Cu Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) anode solution salt bridge solution cathode Single bar ( ) Divides reduced from oxidized Show phase differences Double bars ( ) Represent salt bridge Divides redox half reactions Goes left to right Anode written first Remainder in order of actual cell 16

17 Cell Diagram Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) anode solution salt bridge solution cathode Anode (oxidation) reaction: Zn(s) Zn 2+ (aq) + 2e Solid zinc (Zn(s)) oxidized to Zn 2+ (aq) Solid zinc is the physical electrode Cathode (reduction) reaction: Cu 2+ (aq) + 2e Cu(s) Cu 2+ (aq) is reduced to solid copper (Cu(s)) Solid copper is the physical electrode Net Reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) 17

Standard Electrode Potentials Standard Cell Potentials E o cell = E o reduction -E o oxidation Standard Hydrogen Electrode (SHE) 2H + (aq) + 2e H 2 (g) Standard Electrode Potential,

18 Standard Electrode Potentials Standard Cell Potentials E o cell = E o reduction -E o oxidation Standard Hydrogen Electrode (SHE) 2H + (aq) + 2e H 2 (g) Standard Electrode Potential, E o red [H + ] = 1.00 M P H2 = 1.00atm T = 298K E o red= 0.000V Calculate potential of oxidation half reactions(e o oxid) E o oxid= E o red -E o cell = 0.000V - E o cell = -E o cell 18

19 Spontaneous: Galvanic Cell Pt H 2 (g) H + (aq) Cu 2+ (aq) Cu(s) H 2 (g) + Cu 2+ (aq) 2H + (aq) + Cu(s) Cell Potential: E o cell = E o red -E o oxid E o cell = +.340V Anode Reaction: H 2 (g) 2H + (aq) + 2e E o oxid = V Cathode Reaction: Cu 2+ (aq) + 2e Cu(s) E o red = +.340V + E o cell; Reaction is spontaneous as written (Galvanic) 19

20 Nonspontaneous: Electrolytic Cell Pt H 2 (g) H + (aq) Zn 2+ (aq) Zn(s) H 2 (g)+ Zn 2+ (aq) 2H + (aq) + Zn(s) Cell Potential: E o cell = E o red -E o oxid Anode Reaction: H 2 (g) 2H + (aq) + 2e Cathode Reaction: Zn 2+ (aq) + 2e Zn(s) E o cell = V E o oxid = 0.00 V E o red = -.763V -E o cell: Reaction nonspontaneous (electrolytic) 20

21 Calculating Cell Potentials Daniell cell: Cu 2+ (aq) + Zn (s) Cu (s) + Zn 2+ (aq) Cell Reactions Reduction: Cu 2+ (aq) + 2e Cu(s) Oxidation: Zn(s) Zn 2+ (aq) + 2e From Reduction Tables Cu 2+ E o red = V Zn 2+ E o ox = 0.76V Cell potential, E o cell E o cell= E o red -E o oxid = +0.34V - (-0.76 V) E o cell= +1.10V 21

22 25 o C For a Spontaneous Reaction Reduction reaction is the more positive, higher in table Oxidation reaction is lower in table and needs to be reversed 22

23 Redox Rules 1. E 0 is for the reaction as written 2. The more positive E 0 the greater the tendency for the substance to be reduced 3. Half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed 4. Changing stoichiometric coefficients of a half-cell reaction does not change E 0 Intensive Property: Amount doesn t matter 23

24 Will Br 2 (l) spontaneously oxidize Fe 2+ (aq)? If so, what are the net cell reaction and the E o cell? Figure out what is being asked: Oxidize: Cause to lose e-: Fe 2+ (aq) Fe 3+ (aq) Write and balance chemical equation: 2Fe 2+ (aq) + Br 2 (l) 2Fe 3+ (aq) + 2 Br (aq) Determine half-reactions and get E o from table: Br 2 (l) + 2e 2Br (aq) reduction E o = V Fe 2+ (aq) Fe 3+ (aq) + e oxidation Fe 3+ (aq) + e Fe 2+ (aq E o = V from table Cell Potential E o cell= E o reduction-e o oxidation= +1.07V V = +0.30V E o cell>0 Spontaneous reaction 24

25 Will I (aq) reduce Cr 3+ (aq) to the free metal. If so, what are the net reaction and the E o cell? Figure out what is being asked: Reduce: Cause something to gain e-: Cr 3+ (aq) + 3e- Cr(s) Write and balance chemical equation: 2Cr 3+ (aq) + 6I (aq) 2Cr(s) + 3I 2 (s) Cell Reactions Oxidation: 2I (aq) I 2 (s) + 2e Table: I 2 (s) + 2e 2I (aq) E o = V Reduction: Cr 3+ (aq) + 3e Cr(s) E o = V Cell Potential E o cell= E o reduction -E o oxidation = (-0.74) - (+.53V) = -1.27V E o <0 cell Non-spontaneous Reaction NR 25

26 Thermodynamics of Redox Reactions 26

27 Gibb s Energy and Cell Potentials Electrical work, w w (joules)= Total cell charge (coulombs) x cell potential (V) Total cell charge = # mol electrons(n) x (coulombs/mol e-s) Faraday s constant, F= Coulombs/1 mol e-s Gibb's Energy:G G =w max = nfe cell At standard conditions: G o = nfe o cell = -RTlnK (Voltaic cells are spontaneous, E cell must be positive) Cell potential, E cell nfe o cell = -RTlnK so E o cell=( RT/nF)lnK 27

28 Summarizing the Important Relationships ΔG K E cell Spontaneity Direction (-) >1 (+) Spontaneous More products At equilibrium Products= reactants (+) <1 (-) Nonspontaneous More reactants 28

29 Find K eq at 25 o C for the reaction; Sn(s) + 2Cu 2+ (aq) Sn 2+ (aq) + 2Cu + (aq) Balance net reaction: Sn(s) + 2Cu 2+ (aq) Sn 2+ (aq) + 2Cu + (aq) Cell Reactions Oxidation: Sn(s) Sn 2+ (aq) + 2e E o ox= V Reduction : 2Cu 2+ (aq) + 2e - 2Cu + (aq) E o red= V Link E o cell and K and fill in constants E o cell = (RT/nF) lnk eq lnk eq = E o cell(nf/rt) R = J/mol K n = 2 E o cell= V T = 25 C = 298K F = coul/mol Solve for K eq ln K eq = 22.6 K eq = e 22.6 = 7x

30 Concentration and Cell Potential 30

31 The Nernst Equation: Cell Potentials at Nonstandard Conditions From Thermodynamics: G = G o + RT ln Q Linking Thermodynamics & Electrochemistry G = nfe and G o = nfe o Substitute into first equation to remove G : ( nf)e = ( nf)e o + RT ln Q Divide by nf to get the Nernst Equation: E= E o cell - (RT/nF) ln Q Can now change temperature and concentrations 31

32 Concentration Cells: Daniell Cell at Nonstandard Conditins Electrochemical cell that does not have 1M solutions Find the EMF at the following conditions: 1. [Cu 2+ ] = 1.00 M, [Zn 2+ ] = M 2. [Cu 2+ ] = 0.10 M, [Zn 2+ ] = 0.90 M Daniell Cell Zn(s) + Cu 2+ (aq) Cu(s) + Zn 2+ (aq) E o cell = V Nernst Equation E= E o cell - (RT / nf) ln Q Q = [Cu(s)] [Zn 2+ (aq)] / [Zn(s)] [Cu 2+ (aq)] = [Zn 2+ /Cu 2+ ] Plugging in from equation E= E o cell - (RT / nf) ln [Zn 2+ /Cu 2+ ] 32

33 Concentration Cells: Daniell Cell at Nonstandard Conditions Solve for E E 8.314J / molk *298K [ Zn 1.10V ln 2 2mol *96485coul / mol Cu Conditions of First Cell: [Cu 2+ ] = 1.00 M, [Zn 2+ ] = M E = 1.37 V Conditions of Second Cell [Cu 2+ ] = 0.10 M [Zn 2+ ] = 0.90 M E = 1.07 V 2 ] ] 33

34 Changing Zinc Concentration After the initial drop, concentration has a minimal effect on voltage 34

35 Galvanic Cells: Batteries & Corrosion 35

The Dry Cell (Disposable Batteries) Portable electronic devices Zn(s) ZnCl 2(aq), NH 4 Cl (aq) Mn 2 O 3(aq) MnO 2 (s) C(s) Oxidation: Zn Zn 2+ (aq)+ 2e- Reduction: 2MnO 2 + 2NH 4 + + 2e- Mn 2 O 3

36 The Dry Cell (Disposable Batteries) Portable electronic devices Zn(s) ZnCl 2(aq), NH 4 Cl (aq) Mn 2 O 3(aq) MnO 2 (s) C(s) Oxidation: Zn Zn 2+ (aq)+ 2e- Reduction: 2MnO 2 + 2NH e- Mn 2 O 3 (s) + 2NH 3 + H 2 O Cell Reaction: Zn+2MnO 2 + 2NH 4 + Zn 2+ +Mn 2 O 3 +2NH 3 + H 2 O Cell Voltage Calculated:+ 1.36V Produced: + 1.5V Irreversible Reaction Cell is dead when reactants are used up 36

Button Batteries: Mercury Battery Pacemakers, hearing aids, watches Zn ZnO, OH- HgO, OH- Hg(s) Fe(s) Oxidation: Zn + 2OH- ZnO + H 2 O +2e- Reduction: HgO + H 2 O + 2e- Hg + 2OH- Cell

37 Button Batteries: Mercury Battery Pacemakers, hearing aids, watches Zn ZnO, OH- HgO, OH- Hg(s) Fe(s) Oxidation: Zn + 2OH- ZnO + H 2 O +2e- Reduction: HgO + H 2 O + 2e- Hg + 2OH- Cell Reaction: Zn+ HgO + H 2 O ZnO + Hg Cell Voltage Constant OH- composition More constant voltage Reported voltage: + 1.5V Irreversible Reaction Cell is dead when reactants are used up 37

Lead Storage Battery Car and boat batteries Pb H 2 SO 4 (38%) H 2 SO 4 (38%) PbO 2 Oxidation: Pb + SO 2-4 PbSO 4 +2e- Reduction: PbO 2 + 4H + + SO 2-4 + 2e- PbSO 4 + 2H 2 O Cell Reaction: Pb +

38 Lead Storage Battery Car and boat batteries Pb H 2 SO 4 (38%) H 2 SO 4 (38%) PbO 2 Oxidation: Pb + SO 2-4 PbSO 4 +2e- Reduction: PbO 2 + 4H + + SO e- PbSO 4 + 2H 2 O Cell Reaction: Pb + PbO 2 + 4H + + SO 2-4 2PbSO 4 + 2H 2 O Cell Voltage Voltage: + 2.0V Usually 6 cells=12v Reversible Reaction Reaction reverses when engine is running Battery recharges Electrolytic cell reaction 38

39 Fuel Cells Galvanic cells: Continually renew reactants C-Ni (catalyst) H 2 OH- (aq) OH- (aq) O 2 C-Ni (catalyst) Oxidation: 2H 2 + 4OH - 4H 2 O +4e- Reduction: O 2 + 2H 2 O + 4e- 4OH - Cell Reaction: 2H 2 + O 2 2H 2 O Cell Voltage Voltage: V Nonreversible Reaction Need fresh reactants Removal of products Need electrocatalysts 39

67V 4Fe 2+ + O 2 + (4+2x)H 2 O 2Fe 2 O 3 xh 2 O + 8H + E >0 Cell Reaction Spontaneous reactions

40 Corrosion: Deterioration of metal through an electrochemical process Rust: Fe 2 O 3 xh 2 O Oxidation of metals: Rust, tarnish, patina 2Fe + O 2 + 4H + 2Fe H 2 O E =1.67V 4Fe 2+ + O 2 + (4+2x)H 2 O 2Fe 2 O 3 xh 2 O + 8H + E >0 Cell Reaction Spontaneous reactions Salts increase rate Cathodic protection More reactive metal protects (Zn) Reacts in place of protected metal (Fe) 40

41 Electrolysis and Electrometallurgy 41

42 Electrolytic Cells Non-spontaneous processes driven by the application of an external power supply Side reactions from solvent and dissolved ions Determine reactions that are most spontaneous Least Negative E cell 42

43 A 1 M solution of potassium iodide is electrolyzed under acidic conditions.what are the products? Dissociation Reaction: Resulting Products: KI(aq) K + (aq) + I (aq) K +, I -, H +, H 2 O Cation Reduction: K + (aq) + e K(s) Acid Reduction to H 2 : 2H + (aq) + 2e H 2 (g) Anion Oxidation: 2I (aq) I 2 (s) + 2e H 2 O Oxidation to O 2 : 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e Half-Cell Potentials from table K + (aq) + e K(s) E o = 2.92 V Reduction 2I (aq) I 2 (s) + 2e E o = V Oxidation 2H + (aq) + 2e H 2 (g) E o = 0.00 V Reduction 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e E o = V Oxidation 43

44 A 1 M solution of potassium iodide is electrolyzed under acidic conditions. What are the products? Less Negative Reactions Reduction: 2H + (aq) + 2e H 2 (g) E o = 0.00 V Oxidation: 2I (aq) I 2 (s) + 2e E o = V Net Reaction 2H + (aq) + 2I (aq) I 2 (s) + H 2 (g) Standard Potential E o = (+0.54) = 0.54V need at least.54v to start reaction Products Solid Iodine:I 2 (s) and Hydrogen gas: H 2 (g) 44

copper Impure copper anode More reactive impurities oxidized (ions) Less reactive

45 Producing Products by Electrolysis Electroplating Coating one metal onto another Silver or gold over iron or steel Cheaper Product often more durable Purification of copper Impure copper anode More reactive impurities oxidized (ions) Less reactive impurities drop to bottom Gold, silver, etc. now separated Copper built up on cathode 99.5% pure 45

46 Quantitative Electrolysis Use Electrolytic Cells to find stoichiometry Measure the charge passed through the cell The current (amps, A) is the rate of charge flow 1amp = 1 coulomb per second. nf = At n = number of moles of electrons F = Faraday's constant = C/mol A = current in amps = 1 coulomb/sec t = time in seconds 46

Quantitative Electrolysis of Water: 2H 2 O O 2 + 2H 2 Water is electrolyzed in a cell at 25 ma for 15 minutes. How many ml of oxygen gas are produced at 1 atm & 25 C? A 1A 25mAx 1000mA 2.

47 Quantitative Electrolysis of Water: 2H 2 O O 2 + 2H 2 Water is electrolyzed in a cell at 25 ma for 15 minutes. How many ml of oxygen gas are produced at 1 atm & 25 C? A 1A 25mAx 1000mA 2.5x10 2 A t 60sec 2 15min x 9.0x10 1min sec 2 2 nf At (2.5x10 C / s)(9.0x10 sec) 22. 5C n C F 22.5C 2.3x C / mol 4 mol electrons mol O 1mole 4mol O x2.3x10 molelectrons 5.8x electrons mol 2 O 2 nrt 5 Latm V ( 5.8x10 molo 2)( )(298K) L 1. 4mL P molk 47

Electrochemistry. Chapter 18. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Electrochemistry. Chapter 18. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Electrochemistry Chapter 18 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Electrochemical processes are oxidation-reduction reactions in which: the energy