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1 Chapter 20 Principles of Reactivity: Electron Transfer Reactions.

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1 1 Chapter 20 Principles of Reactivity: Electron Transfer Reactions

2 2 I.Oxidation-Reduction Reactions A.Balancing oxidation-reduction reactions Balancing Redox Equations by the Oxidation Number Change Method (Other Stuff page) Balancing Redox Equations by the Oxidation Number Change Method II.Voltaic (Galvanic) Cells A.Cell construction B.Cell potential C.Effect of concentration on cell potential D.Commercial voltaic cells III.Electrolytic Cells A.Electrolysis B.Quantitative aspects of electrolysis

3 3 I. Oxidation-Reduction Reactions Review: 2 Na + Cl 2  2 NaCl (0) (+1)(–1) oxidation = increase in oxidation number (loss of electrons) reduction = decrease in oxidation number (gain of electrons) e.g., Assign oxidation numbers to the species being oxidized and reduced in the following equation, and label the oxidizing agent and reducing agent. NaI + 3 HOCl  NaIO HCl

4 4 I. Oxidation-Reduction Reactions A. Balancing oxidation-reduction reactions 1. Determine the oxidation numbers of the species being oxidized and reduced (and make sure there are the same number of each on each side). 2. Balance the changes in oxidation numbers by multiplying each species by the appropriate coefficient (i.e., balance the electrons gained and lost). 3. Balance charges with:H + in acidic solution OH – in basic solution 4. Balance H (and O!) with H 2 O. hard way: ion-electron (half-reaction) method (text) easy way: oxidation number change method (Web - Other Stuff)

5 5 I. Oxidation-Reduction Reactions A. Balancing oxidation-reduction reactions e.g., PH 3 + I 2  H 3 PO 2 + I – (acidic solution) e.g., MnO 4 – + H 2 SO 3  Mn 2+ + SO 4 2– (acidic)

6 6 I. Oxidation-Reduction Reactions A. Balancing oxidation-reduction reactions e.g., Cl 2  Cl – + ClO 3 – (basic) e.g., CrO 2 – + S 2 O 8 2–  CrO 4 2– + SO 4 2– (basic)

7 7 II. Voltaic (Galvanic) Cells Produce electricity: chemical energy  electrical energy 2Na + + 2Cl –  2Na + Cl 2  G >> 0 requires input of electrical energy (electrolysis) 2Ag + + Ni  2Ag + Ni 2+  G < 0 produces energy but with Ag + and Ni in contact, we can’t generate electricity; electrons just flow from Ni to Ag + have to use a voltaic (galvanic) cell…

8 8 II. Voltaic (Galvanic) Cells A. Cell construction Ni Ni(NO 3 ) 2 (aq) Ag AgNO 3 (aq) Ni  Ni e – (oxidation) Ni 2+ Ag + + e –  Ag (reduction) Ag + –––––––––––– e–e– e–e– salt bridge KCl in gelatin allows electrolytic conduction without mixing K+K+ Cl – net: Ni + 2Ag +  Ni Ag cathode (+) anode (–) electrical device

9 9 II. Voltaic (Galvanic) Cells A. Cell construction NiAg Ni 2+ Ag + salt bridge or porous partition Shorthand notation: Ni | Ni 2+ || Ag + | Ag Potentiometer (voltmeter) measures cell potential (voltage or electromotive force) depends on: species in redox equation concentrations temperature cell potential, E = actual cell voltage standard cell potential, Eº = voltage at standard state 25ºC 1 atm pressure 1 M concentration

10 10 II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº a. Eº = tendency for a species to be reduced Ag + + e –  AgEº(Ag + ) Ni e –  NiEº(Ni 2+ ) can’t measure directly Can only measure difference in a voltaic cell: 2Ag + + 2e –  2Ag Eº(Ag + ) Ni  Ni e – –Eº(Ni 2+ ) (not 2 x) net: Ni + 2Ag +  Ni Ag Eº cell = Eº(Ag + ) – Eº(Ni 2+ ) = 1.05 V General: Eº cell = Eº(species reduced) – Eº(species oxidized)

11 11 II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº b. standard: hydrogen electrode at standard state 1 atm H 2 1 M H + Pt electrode by definition: at standard state, the reduction 2H + + 2e –  H 2 has Eº = V (exactly)

12 12 II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº c. standard reduction potentials - measure others against the standard hydrogen electrode: DVM Ag 1 M Ag + 1 M H + 1 atm H 2 Find: cathode:2Ag + + 2e –  2Ag anode:H 2  2H + + 2e – Eº cell = 0.80 V Since Eº cell = Eº(Ag + ) – Eº(H + ) 0.80 V = Eº(Ag + ) – 0.00 V  Eº(Ag + ) = V (i.e., more easily reduced than H + )

13 13 II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº c. standard reduction potentials Find: cathode: 2H + + 2e –  H 2 anode: Ni  Ni e – Eº cell = 0.25 V Since Eº cell = Eº(H + ) – Eº(Ni 2+ ) 0.25 V = 0.00 V – Eº(Ni 2+ )  Eº(Ni 2+ ) = –0.25 V (i.e., less easily reduced than H+ or Ag + ) Ni | Ni 2+ (1 M) || H + (1 M) | H 2

14 14 II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº d. determining cell potentials Ag + + e –  AgEº = V Ni e –  NiEº = –0.25 V Ni | Ni 2+ (1 M) || Ag + (1 M) | Ag more easily reduced Eº cell = Eº(Ag + ) – Eº(Ni 2+ ) = V – (–0.25 V) = V  2Ag + + 2e –  2Ag Ni  Ni e – Ni + 2Ag +  Ni Ag

15 15 II. Voltaic (Galvanic) Cells B. Cell potential 2. spontaneity of redox reactions F 2 + 2e – Ag + + e – 2H + + 2e – Ni e – Li + + e – 2F – Ag H 2 Ni Li Eº V V 0.00 V –0.25 V –3.05 V easiest to reduce (strongest oxidant) hardest to reduce (weakest oxidant) hardest to oxidize (weakest reductant) easiest to oxidize (strongest reductant) A species on the left will react spontaneously with a species on the right that is below it in the table. Or: The species with the more positive Eº will be reduced, and the species with the more negative Eº will be oxidized (Eº cell always > 0).

16 16 II. Voltaic (Galvanic) Cells B. Cell potential 2. spontaneity of redox reactions Given the two half reactions below, what is the net cell reaction? What is Eº? Draw a galvanic cell using these half cells and label the anode and cathode, their charges, and the direction electrons flow in the circuit. Fe e –  FeEº = –0.04 V Zn e –  ZnEº = –0.76 V

17 17 II. Voltaic (Galvanic) Cells B. Cell potential 3. cell potential and free energy  G sys = –w surr w = # e – s  energy e – = coulombs  joules coulomb (= joules) = coulombs  volts coulombs = nF(n = # of moles of e – s in redox reaction) (F = 96,500 coulombs/mol)  w = nFE  G = –nFE  Gº = –nFEº Eº cell > 0,  Gº < 0, spontaneous (voltaic) Eº cell 0, nonspontaneous (electrolytic)

18 18 II. Voltaic (Galvanic) Cells B. Cell potential 3. cell potential and free energy e.g., Zn | Zn 2+ (1 M) || Fe 3+ (1 M) | Fe 2Fe Zn  2Fe + 3Zn 2+ Eº cell = 0.72 V What is  Gº for the cell?

19 19 II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 1. Nernst equation  G =  Gº + RTlnQ –nFE = –nFEº + RTlnQ E = Eº – RTlnQ nF at 25ºC: E = Eº – n logQ Nernst equation

20 20 II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 1. Nernst equation e.g., Ni | Ni 2+ (0.05 M) || Ag + (0.01 M) | Ag Ni + 2Ag +  Ni Ag Eº = 1.05 V

21 21 II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 2. applications a. measuring K sp e.g., AgCl(s)Ag + + Cl – K sp = [Ag + ][Cl – ] Find: E = 0.53 V;What is K sp for AgCl? Ni + 2Ag +  Ni Ag Eº = 1.05 V DVM AgNi AgCl(s) 0.10 M Cl – [Ag + ] = ? 1.0 M Ni 2+

22 22 II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 2. applications b. measuring pH Ag 1.0 M Ag + 1 atm H 2 lemon juice [H + ] = ? DVM Find E = 0.94 V; What is pH? H 2 + 2Ag +  2H + + 2Ag Eº = 0.80 V

23 23 II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 2. applications A cell was constructed using the standard hydrogen electrode ([H + ] = 1.0 M) in one compartment and a lead electrode in a 0.10 M K 2 CrO 4 solution in contact with undissolved PbCrO 4 in the other. The potential of the cell was measured to be 0.51 V with the Pb electrode as the anode. Determine the K sp of PbCrO 4 from this data. (Pb e –  Pb Eº = –0.13 V)

24 24 II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 2. applications A galvanic cell was constructed with a Cu electrode in a solution of 1.0 M Cu 2+ in one compartment and a hydrogen electrode immersed in a sample of a soft drink. The cell potential was measured to be V. What was the pH of the soft drink? (Cu e –  Cu Eº = V)

25 25 II. Voltaic (Galvanic) Cells D. Commercial voltaic cells anode: Pb + SO 4 2–  PbSO 4 + 2e – cathode: PbO 2 + 4H + + SO 4 2– + 2e –  PbSO 4 + H 2 O discharge: H 2 SO 4 consumed, H 2 O produced dilutes electrolyte solution can measure with densitometer discharge charge cell: Pb + PbO 2 + 4H + + 2SO 4 2– 2PbSO 4 + 2H 2 OE ~ 2 V (6 cells in series)

26 26 II. Voltaic (Galvanic) Cells D. Commercial voltaic cells anode: Zn + 2OH –  ZnO + H 2 O + 2e – cathode: 2MnO 2 + H 2 O + 2e –  Mn 2 O 3 + 2OH – cell: Zn + 2MnO 2  ZnO + Mn 2 O 3 E ~ 1.5 V Zinc cup anode Graphite cathode Moist paste of MnO 2, KOH and H 2 O Porous partition

27 27 II. Voltaic (Galvanic) Cells D. Commercial voltaic cells Anode cap Partition Cathode: Ag 2 O paste Anode: Zn and KOH Gasket Cell can anode: Zn + 2OH –  ZnO + H 2 O + 2e – cathode: Ag 2 O + H 2 O + 2e –  2Ag + 2OH – cell: Zn + Ag 2 O  ZnO + 2Ag E ~ 1.5 V

28 28 III. Electrolytic cells A. Electrolysis electrical energy  chemical energy e.g., NaCl(l)  Na (l) + Cl 2 (g)  G >> 0 cell: 2Na + + 2Cl –  2Na + Cl 2 Eº = Eº(Na + ) - Eº(Cl 2 ) = (-2.71) - (1.36) = V  Gº = +786 kJ/mol

29 29 III. Electrolytic cells A. Electrolysis e.g., NaCl(aq)  ? Eº cathode:2H 2 O + 2e –  H 2 + OH – V (reduction)Na + + e –  Na-2.71 V H 2 O more easily reduced than Na + Eº anode:Cl 2 + 2e –  2Cl – V (oxidation)O 2 + 4H + + 4e –  2H 2 O+1.23 V H 2 O more easily oxidized than Cl – net:(2H 2 O + 2e –  H 2 + OH – ) x 2 2H 2 O  O 2 + 4H + + 4e – 2H 2 O  2H 2 + O 2 E

30 30 III. Electrolytic cells A. Electrolysis e.g., CuCl 2 (aq)  ? Eº cathode:Cu e –  Cu+0.34 V (reduction) 2H 2 O + 2e –  H 2 + OH – V Cu 2+ more easily reduced than H 2 O net:(Cu e –  Cu) x 2 2H 2 O  O 2 + 4H + + 4e – 2Cu H 2 O  2Cu + O 2 + 4H + E Eº anode:Cl 2 + 2e –  2Cl – V (oxidation)O 2 + 4H + + 4e –  2H 2 O+1.23 V H 2 O more easily oxidized than Cl –

31 31 IV. Electrolytic cells B. Quantitative aspects of electrolysis Units of charge:1 faraday (F) = 1 mol e – s 1 coulomb = 1 amp ·1 sec (A·s) experimentally:1 F = 96,500 C e.g., How many moles of Na and Cl 2 are produced in the electrolysis of NaCl(l) when a current of 25 A is applied for 8.0 hours?

32 32 IV. Electrolytic cells B. Quantitative aspects of electrolysis e.g., How long would it take to deposit 21.4 g of Ag from a solution of AgNO 3 using a current of 10.0 A?


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