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:
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 -
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
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 4+ + 2e - Sn 2+ at a constant rate of 4.24 mmole/h. How much current flows into the solution?
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.
Ask yourself at p.312 Consider the redox reaction
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
A cell in action Electrodes: the redox rxn occur anode: oxidation occur cathode: reduction occur Salt bridge: connect two solns. External wire
Cell representation: Line Notation ： Example ： Interpreting Line Diagrams of Cells Figure 14-4 Another galvanic cell.
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.
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:
The standard reduction potential (E 0 ) for each half-cell is measured by an experiment shown in idealized form in Fig.14-6.
Table 14-1 & Appendix C ( 於 1953, the 17th IUPAC meeting 決定半反應以「還原反應」來表示 )
rxn: Standard Reduction Potentials for reaction
AgCl (s) + e - Ag (s) + Cl - 0.222 V 0.197 V in saturated KCl (formal potentional) E 0 = 0.222V S.H.E.║ Cl- (aq, 1M) | AgCl (s) | Ag(s) E 0’ (formal potential) = 0.197 V (in saturated KCl) S.H.E.║ KCl (aq, saturated) | AgCl (s) | Ag(s) Formal potential
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
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 -1 mol -1 ) T: absolute temperature F: Faraday constant (96,487 coul eq -1 ) at 25°C 2.3026RT/F=0.05916 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 = 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
[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
Dependence of potential on pH Many redox reactions involved protons, and their potentials are influenced greatly by pH.
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
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 + = 0.799 Cd 2+ + 2e - Cd (s) E 0 - = -0.402
Electrons Flow Toward More Positive Potential Electrons always flow from left to right in a diagram like Figure 14-7.
E = 0 and Q = K E 0 > 0 K > 1, E 0 < 0, K < 1 At equilibrium
Ex: One beaker contains a solution of 0.020 M KMnO 4, 0.005 M MnSO 4, and 0.500 M H 2 SO 4 ; and a second beaker contains 0.150 M FeSO 4 and 0.0015 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.
Ans: 5Fe +2 + MnO 4 - + 8H + 5Fe +3 + Mn +2 + 4H 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 ] = 0.771 – 0.059 log (0.150)/(0.0015 × 2) = 0.671 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 = 0.771 – 0.059 log (0.05)/(0.103) = 0.790 V (c) E cell = E Mn - E Fe = 1.52 – 0.671 = 0.849 V (d) At eq., E Fe = E Mn, 所以 E cell = 0 V
Determine a) e - flow direction? b) anode? cathode? c) E =? at 25 ℃ Concentration Cells
(ex) Calculate K sp for AgCl at 25 ℃ ε=0.58V soln: Ex: Systems involving ppt
14-6 Reference Electrodes Indicator electrode: responds to analyte concentration Reference electrode: maintains a fixed potential
Silver-Sliver Chloride AgCl + e - Ag(s) +Cl - E 0 = 0.222 V E (saturated KCl) = 0.197 V Calomel Hg 2 Cl 2 + 2e - 2Hg(l) +2Cl - E 0 = 0.268 V E (saturated KCl) = 0.241 V saturated calomel electrode (S.C.E.) Reference Electrodes
Voltage conversion between different reference scales The potential of A ? ?