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Fundamentals of Electrochemistry Chapter 13

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Fundamentals of Electrochemistry Chapter 13 Chapter 13: Fundamentals of Electrochemistry Electrochemistry is the branch of chemistry concerned with the interrelation of electrical and chemical effects. The study of chemical changes caused by passage of an electric current and production of electrical energy by chemical reactions. Electroanalytical Chemistry • Makes use of electrochemistry at the surface of electrode for the purpose of analysis • A voltage (potentiometry) or current (voltammetry) signal originating from an electrochemical cell is related to the activity or concentration of a particular species in the cell. • Excellent detection limit (10-8 ~ 10-3 M): • Inexpensive technique. • Easily miniaturized : implantable and/or portable (biosensor, biochip) Electrochemical Cells for Voltammetric Sensors Control E – measure i voltammetry – scan E amperometry – hold E Au Pt carbon 3 electrode BAS, Inc. Adapted from Sawyer, Heineman & Beebe, Chemistry Experiments for Instrumental Methods, Wiley I. Papautksy, University of Cincinnati 13-1. Basic Concepts A redox reaction: transfer of electrons from one species to another Oxidation: loss of electrons Reduction: gain of electrons Fe3+ + V2+ -->Fe2+ + V3+ ---- (13.1) electron Fe3+: oxidant V2+: reductant Chemistry and Electricity When electrons from a redox reaction flow through an electric circuit Current ∝ the rate of the electrochemical reaction Voltage ∝ the free energy change for the electrochemical reaction Electric Charge Unit: coulomb (C) The magnitude of the charge of a single electron = 1.602 x 10-19 C Charge of 1 mol of electrons = 9.649 x 104 C/mol ; Faraday constant (F) Relation between charge and moles: ------- (13.2) q = n · F n= # moles of electron transferred Example: Relating Coulombs to Quantity of Reaction Q: If 5.585 g of Fe3+ were reduced in rxn 13-1, how many coulombs of charge must have been transferred from V2+ to Fe3+ ? Sol: Atomic weight of Fe = 55.85 g/mol 5.585g of Fe3+ = 0.100 mol of Fe3+ Each Fe3+ ion requires one electron in the reaction à 0.100 mol of electrons must have been transferred (0.100 mol e-)(9.649x104 C/mol e-) = 9.649 x 103 C Charles Augustin de Coulomb (1736-1806) Michael Faraday (1791-1867) Current Current: the quantity of charge flowing each second through a circuit Unit: ampere, 1 A = 1 coulomb/sec Example: Relating Current to Rate of Reaction Q: If Sn4+ is reduced to Sn2+ at a constant (-) Power (+) rate of 4.23 mmol/h, - e Supply how much current flows ? e- Sol: (-) KCl (+) Sn4+ + 2e- Sn2+ Pt Cu Salt bridge Sn4+ reduction rate = 4.23 mmol/h K+ Cl- à Electron flows at a rate of 2 x (4.23 mmol/h ) 2e- 2e- -3 2- 2+ (8.48 mmol/h)/(3600s/h) = 2.356 x10 mmol/s SO4 Cu Cu Zn2+ Cu2+ -6 4 2+ 4+ 2+ Current = (2.356x10 mol/s)(9.649x10 C/mol) Sn Sn SO 2- 2- Cu 4 SO4 = 0.227C/s = 0.227 A Cathode Anode André-Marie Ampère (1775-1836) Voltage, Work, and Free Energy The difference in electric potential (E) between two points is a measure of the work that is needed when an electric charge moves from one point to the other. When a charge, q, moves through a potential difference, E, the work done is Work = E x q ------- (13.3) (jouls) = (volts) x (coulomb) 1 volt = 1 J/C Example: Electrical Work Q: How much work is needed to move 2.36 mmol of electrons through a potential difference of 1.05 V ? S: To get coulombs of charge, q = n x F = (2,36 x 10-3 mol)(9.649 x 104 C/mol) = 2.277 x 102 C The work required = E x q = (1.05 V) x (2,277 x102 C) = 239 J Relationship between Free Energy Difference and Electric Potential Difference At constant T, P The free energy change (ΔG) for a chemical reaction = the maximum possible electrical work that can be done by the reaction on its surroundings Work done on surroundings = - ΔG ------- (13.4) From equations (13-2, 13-3, 13-4) ΔG = - work = - E · q = - nFE ΔG = - nFE ------- (13.5) 13-2. Galvanic Cells (Voltaic Cells) A galvanic cell : uses a spontaneous chemical reaction to generate electricity To accomplish this: 1. One reagent must be oxidized 2. The other must be reduced 3. The two reagents must be physically separated à electrons are forced to flow through external circuit to go from one reagent to the other High input impedance potentiometer (voltmeter) Anode reaction Cd(s) Cd2+ (aq) + 2 e- : oxidation Cathode reaction 2 AgCl(s) + 2 e- 2Ag(s) + 2Cl-(aq) (+) : reduction Cd(s) + 2 AgCl(s) Cd2+ (aq) +2Ag(s) + 2Cl-(aq) When electrons flow from the left electrode to the right electrode : positive voltage When electrons flow from the right electrode to the left electrode : negative voltage A cell that will not work Salt Bridge: A cell that work Cd(s) Cd(NO3)2(aq) AgNO3(aq) Ag(s) Physical separation The Ag+ ions in solution can directly react at Cd(s) surface à No flow of electrons through the external circuit Positive ions increase Positive ions decrease The purpose of salt bridge is to maintain electroneutrality (no charge buildup) through the cell 13-3. Standard (reduction) Potentials (activities of all species = 1) A quantitative description of the relative driving force for a half-cell reaction. A relative quantity vs standard hydrogen electrode [assigned to zero volt. E0(SHE)=0, 25 oC] + H (aq, A = 1) + e- ½ H2 (g, A = 1) E0(SHE)=0 Reduction : spontaneous SHE Oxidation: Spontaneous 13-4. Nernst Equation (activities of all species = 1) Le Chatelier’s principle: increasing reactant concentrations drives the reacting to the right The net driving force of the reaction is expressed by the Nernst equation The Nernst equation tells us the potential of a cell whose reagents are not all unit activity Nernst Equation for a Half-Reaction R: gas constant = 8.314 J/Kmol - aA + ne bB T: temperature (K) RT Ab o B ------- (13.13) E = E - ln a nF AA ΔG = ΔGo + RT lnQ (Q; reaction quotient) -nFE = -nFEo + RT lnQ (양변을 nF 로 나누어 준다) E = Eo –(RT/nF) lnQ b o RT AB o RT E = E - ln a = E - lnQ nF AA nF Q When all activities are unity, Q = 1 and ln Q = 0, thus E = Eo Nernst Equation at 25 oC aA + ne- bB b o 0.05916V AB ------- (13.15) E = E - log( a ) n AA If Q changes 10-fold, the potential changes by (59.16/n)mV Example: Writing the Nernst Equation for a Half-Reaction Reduction of phosphorus to phosphine gas + - o 1/4 P4 (s, white) + 3H + 3e PH3 (g) E = -0.046 V 0.05916 PPH Sol: E = -0.046 - log( 3 ) 3 [H + ]3 Walter H. Nernst (1864 – 1941) : 독일 Nernst Light Source 1920년 (Infra Red) 노벨화학상 수상 Q: Find the voltage of the cell in Fig. 14-6 if the right cell contains 0.50M AgN03(aq) and the left cell contains 0.010M Cd(NO3)2(aq) (Step 1) Right electrode: 2Ag+ + 2e 2Ag(s) Eo+ = 0.799V Left electrode: Cd2+ + 2e Cd(s) Eo- = – 0.402V (Step 2) Nernst equation for right electrode: E+ = (Eo+) – (0.05916/2) log (1/[Ag+]2) = 0.799 – (0.05916/2) log(1/0.52) = 0.781 V (Step 3) Nernst equation for left electrode: E- = (Eo-) – (0.05916/2) log (1/[Cd2+]) = – 0.402 – (0.05916/2) log(1/0.010) = – 0.461 V (Step 4) Cell voltage: E = E+ – E- = 0.781 – (– 0.461) = + 1.242 V Different Description of the Same Reaction (Fig. 14-4) right electrode (1) AgCl(s) + e- Ag(s) + Cl- Eo+ = 0.222 V E+ = Eo+ - 0.05916 log [Cl-] 2+ - CdCl2 Cd + 2Cl 0.0167 M 0.0334 M E+ = 0.222- 0.05916 log (0.0334) = 0.3093 V (2) Ag+ + e- Ag(s) Eo+ = 0.799 V E+ = 0.799 - 0.05916 log(1/ [Ag+]) Why different E+ values ? AgCl Ag+ + Cl- 0.0334 1. Activity coefficient: neglected 2. Inaccurate K x x+0.0334 sp -10 + -9 x(x+0.0334) = Ksp= 1.8 x 10 , x = [Ag ] = 5.4 x 10 M E+ = 0.799- 0.05916 log (1/5.4 x 10-9) = 0.3099 V Advice for Finding Relevant Half-Reactions Your choice of reactions depend on whether concentrations of reactants is easier to figure out The Nernst Equation is Used in Measuring Standard Reduction Potentials It is nearly impossible to construct a standard electrochemical cell because we have no way to adjust the concentrations and ionic strength to give unit activities In reality, activities less than unity are used in each half-cell, and the Nernst equation is used to extract the value of Eo from the cell voltage 13-5. Eo and Equilibrium Constant • A galvanic cell produces electricity because the cell reaction is not at equilibrium • The potentiometer allows negligible current to flow àThe concentration in each half-cell remains unchanged 0.412 V At the outset [Cu2+] 증가 [Ag+] 감소 0 V At equilibrium Concentrations in the Operating Cell A high-quality pH meter (voltmeter) à Resistance = 1013 Ω If we measure 50 mV Current = E/R = 0.05V/1013 Ω = 5 x 10-15 A (5 x 10-15 C/s)/(9.649 x104 C/mol) = 5 x 10-20 mol e-/s The production rate of Cd2+ = 2.5 x 10-20 mol/s ; negligible concentration change If we replace the potentiometer with a wire, much more current would flow à Concentrations would change until the cell reach equilibrium Relationship between Eo and the Equilibrium Constant Right electrode: aA + ne- cC : Eo+ A + B C + D Left electrode: dD + ne- bB : Eo- 0.05916 Ac 0.05916 Ab E = E+ - E- = Eo+ - log( C ) - [Eo- - log( B )] n a n d AA AD 0.05916 c d 0.05916 AC AD o+ o- o ----- (13.22) E = E – E - n log( a b ) = E - n log Q AA AB Eo Q When cell is at equilibrium, nothing would be driving the reaction, and E = 0 and Q = K (DG = -nFE, DG = 0 à E = 0) o 0.05916 (14.14) becomes E = log K ------- (13.23) n Example: Using Eo to Find the Equilibrium Constant Q; Find the equilibrium constant for the reaction Cu(s) + 2Fe3+ 2Fe2+ + Cu2+ Sol; 2Fe3+ + 2e- 2Fe2+ : Eo+ = 0.771 V (-) Cu2+ + 2e- Cu(s) : Eo- = 0.339 V Eo = Eo+ - Eo- = 0.771 – 0.339 = 0.432 V 0.05916 0.05916 E o = log K = log K n 2 Therefore, K = 10(2x0.432)/0.05916 = 4 x1014 13-7.
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