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Chapter 14 Electrode Potentials. 14-1 Redox Chemistry & Electricity Oxidation: a loss of electrons to an oxidizing agent Reduction: a gain of electrons.

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Presentation on theme: "Chapter 14 Electrode Potentials. 14-1 Redox Chemistry & Electricity Oxidation: a loss of electrons to an oxidizing agent Reduction: a gain of electrons."— Presentation transcript:

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: Fe 3+ + e -  Fe 2+ ox: V 2+  V 3+ + e -

3 1. Chemistry & Electricity Electrochemistry: the study of the interchange of chemical & electrical energy. Electric charge (q) is measured in coulombs(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)(6.022×10 23 /mol)= 9.649×10 4 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: Sn e -  Sn 2+ at a constant rate of 4.24 mmole/h. How much current flows into the solution?

5 3. Voltage & Electrical Work wireqIE hose H2OH2OV H2O P H2O 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.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 ( E cell ) a)The voltage difference between the electrodes.  electromotive force (emf) b)can be measured by voltmeter. c)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 E cell of a half- rxn directly,  need a reference rxn. standard hydrogen electrode:

12 The standard reduction potential (E 0 ) 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 rxn: Standard Reduction Potentials for reaction

16 AgCl (s) + e -  Ag (s) + Cl V V in saturated KCl (formal potentional) E 0 = 0.222V S.H.E.║ Cl- (aq, 1M) | AgCl (s) | Ag(s) E 0’ (formal potential) = V (in saturated KCl) S.H.E.║ KCl (aq, saturated) | AgCl (s) | Ag(s) Formal potential

17 Formal Potential Ex: Ce 4+ + e -  Ce 3+ E°=1.6V with H + A - E°≠1.61V Formal potential: (E° ’ ) The potential for a cell containing a [reagent] ≠1M. Ex: Ce 4+ /Ce 3+ in 1M HCl E° ’ =1.28V

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

19 E = E 0 when [A] = [B] = 1M Q (Reaction quotient ) =1  E = E 0 Where, Q = [B] b / [A] a Nernst equation for a half-reaction at 25ºC

20 [C] & E cell standard conditions: [C]=1M what if [C]≠1M? (ex) a)[Al 3+ ]=2.0M, [Mn 2+ ]=1.0M E cell <0.48V b)[Al 3+ ]=1.0M, [Mn 2+ ]=3.0M E cell >0.48V

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

22 Equation for a Complete Reaction Nernst Equation for a Complete Reaction 1. Write reduction half-reactions for both half-cells and find E 0 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) E 0 + = Cd e -  Cd (s) E 0 - =

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

25 14-5 E 0 and the Equilibrium Constant

26

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

28 Ex: One beaker contains a solution of M KMnO 4, M MnSO 4, and M H 2 SO 4 ; and a second beaker contains M FeSO 4 and M Fe 2 (SO 4 ) 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 H 2 SO 4 to be completely ionized and equal volumes in each beaker.

29 Ans: 5Fe +2 + MnO H +  5Fe +3 + Mn H 2 O Pt | Fe +2 (0.15 M), Fe +3 (0.003 M)║MnO 4 - (0.02 M), Mn +2 (0.005 M), H + (1.00 M) | Pt (a) E Fe = E o Fe(III)/Fe(II) – (0.059/1) log [Fe +2 ]/[Fe +3 ] = – log (0.150)/( × 2) = V E Mn = E o MnO4 - / Mn +2 – (0.059/5)log [Mn +2 ]/[MnO 4 - ][H + ] 8 = 1.51 – 0.059/5 log (0.005)/(0.02)(1.00) 8 = 1.52 V (b) At eq., E Fe = E Mn, 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 E Fe = – log (0.05)/(0.103) = V (c) E cell = E Mn - E Fe = 1.52 – = V (d) At eq., E Fe = E Mn, 所以 E cell = 0 V

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

31 (ex) Calculate K sp 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 Silver-Sliver Chloride AgCl + e -  Ag(s) +Cl - E 0 = V E (saturated KCl) = V Calomel Hg 2 Cl 2 + 2e -  2Hg(l) +2Cl - E 0 = V E (saturated KCl) = V saturated calomel electrode (S.C.E.) Reference Electrodes

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


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