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Chapter 14

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Chapter 14

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  1. Chapter 14 Electrode Potentials

  2. 14-1 Redox Chemistry & Electricity Oxidation: a loss of electrons to an oxidizing agent Reduction: a gain of electrons from a reducing agent Reduction-Oxidation reaction (redox reaction) ex: Half-reactions: re: Fe3+ + e- Fe2+ ox: V2+ V3+ + e-

  3. 1. Chemistry & Electricity • Electrochemistry: the study of the interchange of chemical & electrical energy. • Electric charge (q) is measured incoulombs(C). • The magnitude of the charge of a single electron (or proton) is 1.602×10-19 C. A mole of electrons therefore has a charge of (1.602×10-19 C)(6.022×1023 /mol)= 9.649×104 C/mol, which is called the Faraday constant,F. Example at p.310

  4. 2. Electric current is proportional to the rate of a redox reaction I (ampere; A) = electric current = a flow of 1 coulomb per second = 1C/s Example at p. 310: Sn4+ + 2e- Sn2+ at a constant rate of 4.24 mmole/h. How much current flows into the solution?

  5. 3. Voltage & Electrical Work The difference in electric potential between two points measures the work that is needed (or can be done) when electrons move from one point to another.

  6. Ask yourself at p.312 • Consider the redox reaction

  7. 14.2 Galvanic Cells Chemical reaction spontaneously occurs to produce electrical energy. Ex: lead storage battery When the oxidizing agent & reducing agent are physically separated, e transfer through an external wire.  generates electricity.

  8. A cell in action Electrodes: the redox rxn occur anode: oxidation occur cathode: reduction occur Salt bridge: connect two solns. External wire

  9. Cell representation: Line Notation Example:Interpreting Line Diagrams of Cells Figure 14-4 Another galvanic cell.

  10. 14-3 Standard Potentials Cell potential ( Ecell) • The voltage difference between the electrodes.  electromotive force (emf) • can be measured by voltmeter. • emf of a cell depends on The nature of the electrodes & [ions] Temp.

  11. 14-3 Standard Potentials S.H.E. (standard hydrogen electrode ) It is impossible to measure Ecell of a half-rxn directly, need a reference rxn. standard hydrogen electrode:

  12. The standard reduction potential (E0) for each half-cell is measured by an experiment shown in idealized form in Fig.14-6.

  13. Table 14-1 & Appendix C (於1953, the 17th IUPAC meeting決定半反應以「還原反應」來表示 )

  14. Standard Reduction Potentials for reaction

  15. Standard Reduction Potentials for reaction rxn:

  16. Formal potential • AgCl (s) + e- Ag (s) + Cl- 0.222 V 0.197 V in saturated KCl (formal potentional) E0 = 0.222V: S.H.E.║ Cl- (aq, 1M) | AgCl (s) | Ag(s) E0’ (formal potential) = 0.197 V(in saturated KCl) S.H.E.║ KCl (aq, saturated) | AgCl (s) | Ag(s)

  17. Formal Potential Ex: Ce4+ + e- Ce3+ E°=1.6V with H+A- E°≠1.61V Formal potential: (E°’) • The potential for a cell containing a [reagent] ≠1M. Ex: Ce4+/Ce3+ in 1M HCl E°’=1.28V

  18. 14-4 The Nernst Equation The net driving force for a reaction is expressed by the Nernst eqn. Nernst Eqn for a Half-Reaction where E is the reduction potential at the specified concentrations n: the number of electrons involved in the half-reaction R: gas constant (8.3143 V coul deg-1mol-1) T: absolute temperature F: Faraday constant (96,487 coul eq-1) at 25°C  2.3026RT/F=0.05916

  19. Nernst equation for a half-reaction at 25ºC • E = E0 when [A] = [B] = 1M • Q (Reaction quotient ) =1 E = E0 • Where, Q = [B]b / [A]a

  20. [C] & Ecell standard conditions: [C]=1M what if [C]≠1M? (ex) • [Al3+]=2.0M, [Mn2+]=1.0M Ecell<0.48V • [Al3+]=1.0M, [Mn2+]=3.0M Ecell>0.48V

  21. Dependence of potential on pH Many redox reactions involved protons, and their potentials are influenced greatly by pH.

  22. Nernst Equation for a Complete Reaction • 1. Write reduction half-reactions for both half-cells and find E0 for each in Appendix C. • 2. Write Nernst equation for the half-reaction in the right half-cell. • 3. Write Nernst equation for the half-reaction in the left half-cell. • 4. Fine the net cell voltage by subtraction: E=E+-E-. • 5. To write a balanced net cell reaction. P.321

  23. Nernst Equation for a complete reaction Example at p. 321 Rxn: 2Ag+ (aq) + Cd (s)  Ag (s) + Cd 2+ (aq) 2Ag+ + 2e- Ag (s) E0+ = 0.799 Cd 2+ + 2e- Cd (s) E0- = -0.402

  24. Electrons Flow Toward More Positive Potential • Electrons always flow from left to right in a diagram like Figure 14-7.

  25. 14-5 E0 and the Equilibrium Constant

  26. 14-5 E0 and the Equilibrium Constant

  27. At equilibrium • E = 0 and Q = K • E0 > 0 K > 1, • E0 < 0, K < 1

  28. Ex: • One beaker contains a solution of 0.020 M KMnO4, 0.005 M MnSO4, and 0.500 M H2SO4; and a second beaker contains 0.150 M FeSO4 and 0.0015 M Fe2 (SO4)3. The 2 beakers are connected by a salt bridge and Pt electrodes are placed one in each. The electrodes are connected via a wire with a voltmeter in between. • What would be the potential of each half-cell (a) before reaction and (b) after reaction? • What would be the measured cell voltage (c) at the start of the reaction and (d) after the reaction reaches eq.? • Assume H2SO4 to be completely ionized and equal volumes in each beaker.

  29. Ans: 5Fe+2 + MnO4- + 8H+ 5Fe+3 + Mn+2 + 4H2O Pt | Fe+2(0.15 M), Fe+3(0.003 M)║MnO4-(0.02 M), Mn+2(0.005 M), H+(1.00 M) | Pt (a) EFe = EoFe(III)/Fe(II)– (0.059/1) log [Fe+2]/[Fe+3] = 0.771 – 0.059 log (0.150)/(0.0015 × 2) = 0.671 V EMn = EoMnO4-/Mn+2– (0.059/5)log [Mn+2]/[MnO4-][H+]8 = 1.51 – 0.059/5 log (0.005)/(0.02)(1.00) 8 = 1.52 V (b) At eq., EFe = EMn, 可以含鐵之半反應來看, 先找出平衡時兩個鐵離子的濃度,得 EFe = 0.771 – 0.059 log (0.05)/(0.103) = 0.790 V (c) Ecell = EMn - EFe = 1.52 – 0.671 = 0.849 V (d) At eq., EFe = EMn, 所以Ecell = 0 V

  30. Concentration Cells Determine • a) e flow direction? • b) anode? cathode? • c) E =? at 25℃

  31. (ex) Calculate Ksp for AgCl at 25℃ ε=0.58V soln: Ex: Systems involving ppt

  32. 14-6 Reference Electrodes Indicator electrode: responds to analyte concentration Reference electrode: maintains a fixed potential 

  33. Reference Electrodes • Silver-Sliver Chloride • AgCl + e- Ag(s) +Cl- • E0 = 0.222 V • E (saturated KCl) = 0.197 V • Calomel • Hg2Cl2 + 2e- 2Hg(l) +2Cl- • E0 = 0.268 V • E (saturated KCl) = 0.241 V saturated calomel electrode (S.C.E.)

  34. Voltage conversion between different reference scales • The potential of A? ?