Oxidation- Reduction Chemistry

Transcription

1 Oxidation- Reduction Chemistry Chem 36 Spring 2002 Definitions Redox reactions involve electron transfer: Lose e - = Oxidation Cu (s) + 2 Ag + (aq) fi Cu 2+ (aq) + 2 Ag (s) Gain e - = Reduction 2 1

2 Half-Reactions Consider each process indivually: Oxidation Cu (s) Cu 2+ (aq) + 2 e - Reduction [Ag + + e - Ag (s) ] x 2 Overall: Cu (s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag (s) Oxidized (reducing agent) Reduced (oxidizing agent) 3 Balancing Redox Reactions The Half-Reaction Method Three Steps: 1. Determine net ionic equations for both half-reactions 2. Balance half-reactions with respect to mass and charge 3. Combine so as that electrons cancel 4 2

3 Example SO 3 + H + + MnO 4- SO 4 + Mn 2+ + H 2 O 1. Write Skeleton Half-Reactions Oxidation SO 3 SO 4 Reduction MnO 4- Mn Mass Balance SO 3 + H 2 O SO 4 + 2H + MnO H + Mn H 2 O Add H 2 O to side needing oxygen Add H + to balance hydrogen 5 Example: Continued 3. Charge Balance (use electrons) SO 3 + H 2 O SO 4 + 2H + + 2e - MnO H + + 5e - Mn H 2 O 4. Combine! [SO 3 + H 2 O SO 4 + 2H + + 2e - ] x 5 [MnO H + + 5e - Mn H 2 O] x 2 5SO 3 + 5H 2 O + 2MnO H e - 5SO H e - + 2Mn H 2 O 6 3

4 Example: Simplify and Verify Collecting and cancelling gives: 5SO 3 + 2MnO H + fi 5SO 4 + 2Mn H 2 O Verify! Mass Balance Charge Balance Done! 7 In Basic Solution? First: Balance as though in acid Next: Add OH - to both sides of reaction in an amount equal to the amount of H + Change: H + + OH - to H 2 O Verify! 8 4

5 Example (Basic Solution) Cr(OH) 3 + OCl - + OH - CrO 4 + Cl - + H 2 O In Acid Solution 2Cr(OH) 3 + 3OCl - 2CrO 4 + 3Cl - + H 2 O + 4H + + 4OH - Add OH- + 4OH - Make Water 2Cr(OH) 3 + 3OCl - + 4OH - 2CrO 4 + 3Cl - + 5H 2 O 9 Example: Simplify and Verify Collecting and cancelling gives: 5SO 3 + 2MnO H + fi 5SO 4 + 2Mn H 2 O Verify! Mass Balance Charge Balance Done! 10 5

6 In Basic Solution? First: Balance as though in acid Next: Add OH - to both sides of reaction in an amount equal to the amount of H + Change: H + + OH - to H 2 O Verify! 11 Example (Basic Solution) Cr(OH) 3 + OCl - + OH - CrO 4 + Cl - + H 2 O In Acid Solution 2Cr(OH) 3 + 3OCl - 2CrO 4 + 3Cl - + H 2 O + 4H + + 4OH - Add OH- + 4OH - Make Water 2Cr(OH) 3 + 3OCl - + 4OH - 2CrO 4 + 3Cl - + 5H 2 O 12 6

7 Back to Cu/Ag Reaction How can we predict reaction spontaneity? Cu (s) + 2 Ag + (aq) fi Cu 2+ (aq) + 2 Ag (s) versus Cu 2+ (aq) + 2 Ag (aq) fi Cu (s) + 2 Ag + (s) X G? K? Direction/Extent of Reaction? 13 The Galvanic Cell Set up, literally, two half-cells for the reaction: Anode (ox) Cu V e - e - NO 3 - Na + Ag Cathode (red) Ag NO Cu 2+ NO 3 - Cell Potential: volts 1.0 M Cu(NO 3 ) M AgNO 3 Cu (s) Cu 2+ (aq) + 2e - Ag + + e - Ag (s) 14 7

8 Electromotive Force What is the Cell Potential? Recall: 1 Joule = 1 volt x 1 coulomb (work) (potential) (charge) Driving Force of reaction The Electromotive Force (EMF) E o cell (all species in standard states) Galvanic Cell shorthand: Cu (s) Cu 2+ (aq) Ag + (aq) Ag (s) E o cell = volts Anode (oxidation) Cathode (reduction) 15 Another system Now, consider this cell: Cu (s) Cu 2+ (aq) Zn 2+ (aq) Zn (s) E o cell = volts What does a negative potential mean? Electrons want to flow from Cathode (Zn) to Anode (Cu) So, reaction is spontaneous in the reverse direction: Cu 2+ (aq) + Zn (s) fi Cu (s) + Zn 2+ (aq) 16 8

9 Pulling Rank Rank Ag, Cu and Zn based on their ability to cause reduction: Cu reduces Ag + Zn reduces Cu 2+ So, as reducing agents: Zn > Cu > Ag How can we quantify this? 17 Quantifying Reduction By definition Measure E o cell for reduction ½-cells with a reference ½-cell: 2H + (aq) + 2e - fi H 2 (g) E o H = volts Standard Hydrogen Electrode (SHE) Tabulate as Standard Reduction Potentials (oxidizing power): Ag > Cu > Zn E o v v v 18 9

10 Calculating E o cell Cu (s) Cu 2+ (aq) Ag + (aq) Ag (s) Anode (oxidation) Cu (s) Cu 2+ (aq) + 2e - E o ox = -Eo Cu = v Cathode (reduction) Ag + + e - Ag (s) E o red = Eo Ag = v 2 x [Ag + + e - Ag (s)] v Cu (s) Cu 2+ (aq) + 2e v Cu (s) + 2Ag + (aq) Cu 2+ (aq) + 2 Ag (s) v = E o cell Or: E o cell = E o Cathode - Eo Anode (right) (left) reduction potentials 19 The Thermodynamics Connection! Recall that: Which gives: work = charge x potential w electr = n FE cell mol e - /mol rxn Faraday s Constant = x 10 4 C/mol e - For a reversible system at constant Temp and Pressure: w max = -DG So: DG o = -n FE o cell 20 10

11 What about K? Simple! G o = -RTLnK = -n FE o cell Rearranging gives: E o cell = RTLnK n F At K: E o cell = (0.0592/n)LogK 21 Example 2Al (s) + 3Cu 2+ (aq) 3Cu (s) + 2Al 3+ (aq) E o cell? DGo? K? E o cell : 2 x [Al(s) Al 3+ (aq) + 3e - ] -(-1.66 v) 3 x [Cu 2+ (aq) + 2e - Cu (s)] v 2Al (s) + 3Cu 2+ (aq) 3Cu (s) + 2Al 3+ (aq) v reaction is spontaneous 22 11

12 Now for G o DG o = -n FE o cell G o = -(6 mol e-/mol)( x 10 4 C/mol e - )(2.00 v) G o = x 10 6 C-v/mol G o = x 10 3 kj/mol Joules 23 Equilibrium at last Two paths: From G o (= -RTLnK) From E o cell = (0.0592/n)LogK LogK = (E o cell )n Log K = (2.00 v)(6 mol e - ) = K = = x =

13 Non-Standard States? Recall: Since: Substituting: DG = DG o + RTLnQ G = -n FE cell E cell = E o cell - RT LnQ n F The Nernst Equation At K: E cell = E o cell - (0.0592/n)LogQ 25 Nernst Example Pt (s) Fe 2+ (0.10 M), Fe 3+ (0.20 M) Ag + (1.0 M) Ag (s) Calculate E cell First, determine E o cell : Ox Fe 2+ Fe 3+ + e - -(0.771 v) Red Ag + + e - Ag (s) v Ag + + Fe 2+ Ag (s) + Fe v E o cell 26 13

14 Example Continues Next, find value of Q: Q = [Fe 3+ ] = (0.20) = 2.0 [Ag + ][Fe 2+ ] (1.0)(0.10) Lastly, into the Nernst Equation: E cell = E o cell - (0.0592/n)LogQ E cell = (0.0592/1)Log(2.0) E cell = = volts E cell = volts 27 Nernst (huh!)... What is it good for? Concentration Determinations Pt, H 2 (1 atm) H + (x M) H + (1 M) H 2 (1 atm), Pt Unknown [H + ] (Anode) Std Hydrogen Electrode (Cathode) Ox H 2 (g) 2H + (aq) + 2e - Red 2H + (aq) + 2e - H 2 (g) 2H + (aq) 2H + (aq) (1 M) (x M) concentration cell 28 14

15 Using Nernst E cell = E o cell - (0.0592/n)LogQ Q = [H + ] 2 anode = [H+ ] 2 anode [H + ] 2 cathode 1 M E cell = (0.0592/2)Log[H + ] 2 anode E cell = -2(0.0592) Log[H + ] anode 2 E cell = Log [H + ] anode E cell = ph 29 More Uses for Nernst Equilibrium Constant Determinations Pb (s) Pb 2+ (sat d PbCl 2 (aq)) Pb 2+ (0.100 M) Pb (s) Another concentration cell Ox Pb (s) Pb 2+ (aq) + 2e - Red Pb 2+ (aq) + 2e - Pb (s) Pb 2+ (0.100 M) Pb 2+ (sat d) Measure E cell : volts What is K sp for PbCl 2? 30 15

16 The Nernst Solution Nernst will provide [Pb 2+ ]: E cell = E o cell - (0.0592/n)LogQ Q = [Pb 2+ ] anode = [Pb 2+ ] sat d [Pb 2+ ] cathode M = (0.0592/2)Log([Pb 2+ ]/0.100) [Pb 2+ ] = 1.6 x 10-2 M 31 Finally, Equilibrium Solubility Equilbrium: PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) S 2S Substitute into K sp expression: K sp = [Pb 2+ ][Cl - ] 2 = S(2S) 2 = 4S 3 K sp = 4(1.6 x 10-2 ) 3 = 1.6 x

17 Batteries What happens if we allow current to flow in a Galvanic Cell? Oxidation and Reduction reactions occur Concentrations change Q K (equilibrium) E cell 0 (dead battery!) Is the reaction reversible? Battery can be recharged Apply a potential sufficient to drive the reverse reaction 33 Example: Pb-Acid Battery Anode: Pb-Sb alloy grid filled with spongy Pb Cathode: PbO 2 coating Pb + PbO 2 + 4H + + 2SO 4 2PbSO 4 + 2H 2 O E cell 2.0 volts As it discharges: PbSO 4 coats electrodes Electrolyte gets diluted 34 17

18 Charge it up! Apply a potential > E cell (opposite polarity) Forces the reverse reaction: PbSO 4 Pb + PbO 2 + H + + SO 4 When battery discharges completely: PbSO 4 completely covers electrodes Can t recharge! Dead Battery When battery is overcharged: No PbSO 4 left to react Electrolysis of water H 2 and O 2 form Ruins Pb and PbO 2 coatings Explosive! 35 Fuel Cells Just an Open system battery Products and reactants are continuously replenished 36 18

19 Electrolysis Force a nonspontaneous reaction to occur by applying a potential: Greater than E cell Opposite polarity Why Do It? Useful chemistry can happen Can quantify the extent of the reaction 37 Quantifying Electrolysis: Faraday s Laws 1. The amount of chemical change is proportional to the quantity of electrical charge that passes through an electrolytic cell Measure current: 1 Ampere = 1 Coulomb/sec Get charge: # Coulombs = current (Amps) x time (sec) 2. A given quantity of electricity produces the same number of equivalents of any substance in an electrolysis process: 1 equivalent = 1 mol e - in a half-reaction Relate charge (Coulombs) to equivalents using Faraday s Constant (96,487 C/mol e - ) 38 19

20 Illustrative Example What mass of Cu is deposited if a current of 1.50 A flows for 1.00 hour in the electrolysis of a CuSO 4 solution? The reaction: Cu 2+ (aq) + 2e - Cu (s) time charge mol e - mol Cu g Cu 1.00 hr x 60 min x 60 sec x 1.5 C x 1 mol e - x 1 mol Cu x g Cu = 1 hr 1 min sec 96,487 C 2 mol e - 1 mol Cu = 1.78 g Cu 39 But what about... Aluminum? One of the most abundant elements (8% of the Earth s crust) Found almost exclusively as Al 3+ in nature Can t do electrolysis reduction in aqueous solution: Al 3+ (aq) + 3e - Al (s) 2H 2 O + 2e - H 2 + 2OH - E o = V E o = V This is what happens! 40 20

21 The Hall-Hêroult Process 1885: Charles Martin Hall 22 year-old student at Oberlin College Electrolysis in a molten salt (Na 3 AlF 6 at 1000 o C): Al 2 O 3 (s) + 3C (s) 2Al (l) + 3CO (g) Result: inexpensive Al metal LOTS of energy is stored in Al metal ~3 billion pounds of Al thrown away each year G (Hall Process) = kj/mol Al G (Al Recycling) = 25 kj/mol Al 41 Corrosion O 2 is a strong oxidizing agent and can oxidize many metals Rust! Fe + O 2 Solution? Coat Fe with a more easily oxidizable metal: Zn (galvanization) E o Zn = V versus Eo Fe = V Why doesn t Al oxidize ( rust )? E o Al = V It does! All Al metal has a thin coating of Al 2 O 3 Al 2 O 3 adheres to the surface and protects it 42 21