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Electrochemistry. The electricity produced by chemical reactions or ….. The chemical changes brought about by electricity. Electrochemical reactions = Oxidation-Reduction Reactions. Cell = System where chemical reactions occur. Electrode = means of adding/removing electric current
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Electrochemistry The electricity produced by chemical reactions or ….. The chemical changes brought about by electricity. Electrochemical reactions = Oxidation-Reduction Reactions Cell = System where chemical reactions occur Electrode = means of adding/removing electric current to/from system. Cathode = electrode at which reduction occurs. (usually -) Anode = electrode at which oxidation occurs. (usually +) Conductivity = flow of electrons many solid metals conduct electricity electrolyte solutions (acids, bases, or soluble ionic cpds) Which of the following would produce the strongest conducting solution? a) acetic acid b) AgCl c) KCl d) glucose
Electrochemistry – Cell Types Electrolytic Cells – External source of electricity drives a nonspontaneous (DG > 0) reaction. Electrolysis Used to convert water into H2 and O2 gas. Used for electroplating – put a thin layer of one metal unto another metal at cathode. Galvanic/Voltaic Cells – A spontaneous (DG < 0) reaction is used to generate an electric current. e.g. batteries
Zn(s) Zn(s) Cu2+ Zn(s) + Cu2+→ Zn2+ + Cu(s) spontaneous DG < 0 Oxidation Zn(s)→ Zn2+ + 2e- Reduction Cu2+ + 2e- → Cu(s)
+ Zn Cu Salt bridge (KCl – 5% agar) 1M ZnSO4 1M CuSO4 voltmeter Voltaic Cell 1.100 V e- e- Zn Cu electrode electrode Cl- K+ Zn|Zn2+(1M)||Cu2+(1M)|Cu Zn2+ Cu2+ Zn(s) Zn2+ + 2e- Oxidation - anode Cu2+ + 2e- Cu(s) reduction - cathode
0.462 V e- e- + Cu Ag Salt bridge (KCl – 5% agar) Cl- K+ Cu2+ Ag+ 1M CuSO4 Cu(s) Cu2+ + 2e- Ag+ + e- Ag(s) 1M AgNO3 Voltaic Cell Cu|Cu2+(1M)||Ag+(1M)|Ag a) left b) right Which electrode is …. Reduction? …. anode?
0.763 V e- e- + Zn H2 Salt bridge (KCl – 5% agar) E (SHE) = 0 Cl- K+ E = -0.763 as reduction Zn2+ H+ 1M ZnCl2 Zn(s) Zn2+ + 2e- Oxidation - anode 2H+ + 2e- H2(g) reduction - cathode 1M HCl SHE Standard Hydrogen electrode Zn|Zn2+(1M)||H+(1M);H2(1atm)|Pt Zn
0.337 V e- e- + Cu Salt bridge (KCl – 5% agar) E (SHE) = 0 Cl- K+ E = 0.337 H+ Cu2+ 1M HCl(aq) H2(g) 2H+ + 2e- Oxidation - anode Cu2+ + 2e- Cu(s) reduction - cathode 1M CuSO4 Voltaic Cell Pt|H2(1atm);H+(1M)||Cu2+(1M)|Cu H2 Cu
See page 846 Reduction ½ rx Standard Reduction Potential E volts Strong ox. Agent more easily reduced cathode Cl2 + 2e- 2Cl- +1.360 V Ag++ e- Ag(s) +0.799 V Hg2+ + 2e- Hg(s) +0.789 V Cu2+ + 2e- Cu(s) +0.337 V 2H+ + 2e- H2(g) 0.000 V Pb2+ + 2e- Pb(s) -0.126 V Ni2+ + 2e- Ni(s) -0.25 V Fe2+ + 2e- Fe(s) -0.44 V Zn2+ + 2e- Zn(s)-0.763 V strong red. Agent oxidized anode reverse The higher up a ½ rx is on the table the more readily that element/substance is reduced.
Calculating The Standard Electrical Potential for any cell from the tabulated Standard Reduction Potentials. 1. Choose the appropriate ½ rxs from table • Write the ½ rx for the substance with the largest Eº. • Write the ½ rx for the substance with the lower Eº • as an oxidation reaction. • Write the net balanced reaction (the electrons must be balanced • but do not multiply E by the balancing coefficient!) Eº = Eºcathode - Eºanode Cl2 + 2e- 2Cl- +1.360 V Ag++ e- Ag(s) +0.799 V Hg2+ + 2e- Hg(s) +0.789 V Eº = Eºhigher - Eºlower Cu2+ + 2e- Cu(s) +0.337 V 2H+ + 2e- H2(g) 0.000 V Pb2+ + 2e- Pb(s) -0.126 V Ni2+ + 2e- Ni(s) -0.25 V Fe2+ + 2e- Fe(s) -0.44 V Zn2+ + 2e- Zn(s)-0.763 V
3Br2(g) + 2Cr(s) → 6Br-(aq) + 3Cr3+(aq) What is the voltage of a Voltaic Cell using the two Half reactions in red? +1.07 b) -1.81 c) +0.33 d) +1.81 Write the balanced equation for the reaction. Draw the cell.
1.81V e- e- + Pt + Br2 (1atm) Zn Salt bridge (KCl – 5% agar) Cl- K+ Cr3+ Br- 1M Cr3+ Cr(s) Cr3++ 3e- Oxidation - anode Br2(g) + 2e- 2Br- reduction - cathode 1M Br- 3Br2(g) + 2Cr(s) → 6Br-(aq) + 3Cr3+(aq) Cr|Cr3+(1M)||Br-(1M);Br2(1atm)|Pt Cr
Reaction Spontaneity and DG DG < 0 reaction is spontaneous – proceeds as written DG = 0 reaction is at equilibrium DG > 0 reaction proceeds in the reverse direction unless enough energy provided to drive reaction forward. Reaction Spontaneity and E E > 0 reaction is spontaneous – proceeds as written with voltage output E = 0 reaction is at equilibrium – no current flow E < 0 reaction proceeds in the reverse direction unless enough current provided to drive reaction forward. (e.g. electrolysis)
Reaction Spontaneity ― E vsDG • Represents standard conditions where P = 1atm for • each gas and the [ ]s of all reagents are 1M. Q = 1 DG = DG + RT ln Q DG = - RT ln K DG˚ = -nFE˚ or DG = -nFE F (Faraday’s Constant) = 96,485 C (J V-1 mol-1) n = the # of moles of electrons in the total balanced reaction The Nernst Equation E = E - RTlnQ nF The Nernst Equation E = E - 0.0257ln Q n The Nernst Equation E = E - 0.0592 log Q n RT/nF = 8.314 J mol-1 K-1• 298 K ÷ (96,485 V J-1 mol-1 • n mol) = 0.0257 V mol-1 log x = 2.303 ln x
The Nernst Equation E = E - 0.0257ln Q n Zn(s) + Cu2+→ Zn2+ + Cu(s) Q = [Zn2+]/[Cu2+] Zn|Zn2+(1M)||Cu2+(1M)|Cu Eº = 1.10 V Zn|Zn2+(1.5M)||Cu2+(0.5M)|Cu E = ??? V E = 1.10 - 0.0257ln(1.5/0.5) n DG = -nFE = -2 • 96,485 • 1.09 = -210 kJ/mol-1
??? V e- e- + Zn Cu Salt bridge (KCl – 5% agar) Cl- K+ Zn2+ Cu2+ 1.5 M ZnSO4 Zn(s) Zn2+ + 2e- Oxidation - anode Cu2+ + 2e- Cu(s) reduction - cathode 0.50 M CuSO4 Voltaic Cell 1.09 V Zn Cu Zn|Zn2+(1M)||Cu2+(1M)|Cu E = 1.10 - 0.0257ln(1.5/0.5) n DG = -nFE = -2 • 96,485 • 1.09 = -210 kJ/mol-1
The Nernst Equation E = E - 0.0257ln Q n Zn(s) + Cu2+→ Zn2+ + Cu(s) Q = [Zn2+]/[Cu2+] Zn|Zn2+(1M)||Cu2+(1M)|Cu Eº = 1.10 V Zn|Zn2+(1.5M)||Cu2+(0.5M)|Cu E = ??? V Note that the reaction is spontaneous (E > 0) and will proceed so that Zn is oxidized and Cu2+ ions are reduced. As the reaction proceeds E gets smaller and eventually the cell will put out no voltage when E = 0. Then …. E = + 0.0257ln K n 19.21 pg 874
19.30 b Calculate Eº, E, and DG Write balanced rx Eº = +0.760 + (-0.740) = +0.020 V E = 0.020 - 0.0257ln (0.0085/0.010) 6
anode cathode Electrolysis of Water 2H2O2H2(g) + O2(g)DGº = +474.4 kJ/mol 474,400 = - nFEº Eº = 474400/(-4 • 96485) = -1.229 V K = exp(-DGº/RT) = exp(nEº/0.0257) ~ 7.0 x 10-84 or 8.4 x 10-84 Reduction: (2H+ + 2e- H2(g)) 2 Oxidation: 2H2O 4H+ + O2 + 4e-
Friday’s Exam Topics: Thermodynamics Laws of thermodynamics enthalpy, entropy, and free energy free energy and equilibrium standard vs. non-standard states Electrochemistry oxidation states of elements in compounds recognizing redox reactions writing half-reactions Determining E° from table of reduction half-reactions Diagraming Galvanic Cells Electrolysis and electrolytic cells Stoichiometry of electroplating
Na+ Cl- Na+ + e- Na (floats) Reduction - cathode > 4.1 V 2Cl- Cl2(g) + 2e- oxidation - anode Graphite or Platinum are common inert electrodes + Molten NaCl (801˚ C) Electrolysis of molten NaCl E° = Ecath – Eanode = -2.71 – (1.36) = - 4.07 V
Electrochemistry – Cell Types Galvanic/Voltaic Cells – A spontaneous (DG < 0) reaction is used to generate an electric current. e.g. batteries Electrolytic Cells – External source of electricity drives a nonspontaneous (DG > 0) reaction. Electrolysis Used to convert water into H2 and O2 gas. Used for electroplating – put a thin layer of one metal unto another metal at cathode. Does the amount of a metal electroplated depend on the amount of electricity delivered to an electrolytic cell? a) yes b) no
Electron Stoichiometry A Coulomb (C) is the SI unit of charge 1 e- = 1.602 x 10-19 C or 1.602 x 10-19 C per e- 1 mole of e- = 96,500 C = 1 Faraday (F) Current = charge per time = C s-1 = Ampere (A) How many grams of H2 gas can be produced from water through which 1.35 x 106 C have been passed? Reduction: 2H+ + 2e- H2(g) H2OH2(g) + ½O2(g) 1.35 x 106 C •1mol e-• 1mol H2 • 2g H2 96485 C 2 mol e- 1 mol H2 = 14.0 g H2
3.4 amps for 15 seconds – How much Cu is electroplated? Cu(s)→ Cu2+ + 2e- oxidation – anode electrode ‘dissolves’ Cu2+ + 2e- → Cu(s) reduction – cathode Cu plated onto electrode 19.46 pg 876 + Cu2+ SO42- Electroplating copper 3.4C x 15s x 1 mol e- x 1 mol Cu x 63.546g Cu = 16.8 mg 1s 96,485C 2 mol e- 1 mol Cu
1 A = 1 C s-1 1 F = 96,500 C 1 C = 1 J V-1 mol-1 19.46 pg 876 Electrolysis of molten BaCl2. a) Write half reactions. The cathode product is …. ? a) Cl-. b) Ba2+. c) Cl2. d) Ba(s). Ba2+ + 2Cl-→ Ba(s) and Cl2 Ba = 137.3 g mol-1 a) How many grams of Ba(s) produced using 0.50 A for 30 minutes? a) 1.2 g b) 0.64 g c) 42 g d) 0.034 g 0.5Cx 1800s x 1 mol e- x 1 molBax 137.3 g Ba = 0.640 g 1s 96,500C 2 mol e-1 molBa
Anode = Zn Cathode = C 1.6 V Dry Cell ZnCl2; NH4Cl; MnO2 paste Zn + 2NH4+→ Zn2+ + 2NH3 + H2 Anode: (oxidation) Zn → Zn2+ + 2e- Cathode: (reduction) 2NH4+ + 2e-→ 2NH3 + H2 2MnO2 + H2→ 2MnO(OH) (removes H2 gas) Zn2+ + 4NH3 → [Zn (NH3)4]2+ (removes NH3)
Anode = Zn Cathode = C 1.5 V Alkaline Cell ZnCl2; KOH; MnO2 paste Zn + 2MnO2 + 2H2O → Zn(OH)2(s) + 2MnO(OH) Anode: (oxidation) Zn + 2OH-→ Zn(OH)2 + 2e- Cathode: (reduction) 2MnO2 + 2H2O+ 2e-→ 2MnO(OH) + 2OH- Alkaline cells have longer shelf lives
Anode = Cd Cathode = NiO2 1.4 V Ni - Cd rechargeable Ni(OH)2(s); KOH; Cd(OH)2(s) Cd + NiO2(s) + 2H2O ↔Cd(OH)2(s) + Ni(OH)2(s) + “V” Rechargeable Batteries – Nickel-Cadmium Anode: (oxidation) Cd + 2OH-→ Cd(OH)2(s) + 2e- Cathode: (reduction) NiO2 + 2H2O+ 2e-→ Ni(OH)2(s) + 2OH-
Zn(Hg) + 2OH- (aq) ZnO(s) + H2O (l) + 2e- HgO(s) + H2O (l) + 2e- Hg (l) + 2OH-(aq) Zn(Hg) + HgO (s) ZnO(s) + Hg (l) Batteries Mercury Battery Anode: Cathode:
Pb (s) + SO2- (aq) PbSO4(s) + 2e- 4 PbO2(s) + 4H+(aq) + SO2-(aq) + 2e- PbSO4(s) + 2H2O (l) 4 Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l) 4 Batteries Lead storage battery Anode: Cathode:
Batteries Solid State Lithium Battery
2H2 (g) + 4OH- (aq) 4H2O (l) + 4e- O2(g) + 2H2O (l) + 4e-4OH-(aq) 2H2 (g) + O2 (g) 2H2O (l) Batteries A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: Cathode:
Corrosion Corrosion is the term usually applied to the deterioration of metals by an electrochemical process.