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Chapter 20: Electrochemistry 1. 2. 3. 4. 5. 6. 7. 8. Balancing Redox Reactions Galvanic/Voltaic Cells EMF & Standard Cell Potentials EMF & K Nernest Eq. pH Measurements Electrolytic Cells Stoichiometry of Electrolysis Oxidation - Reduction Reactions (Redox Reactions) Involve the Transfer of Electrons Between Different Reactants Historically: Oxidation Meant the Formation of Oxides and Reduction Meant the Removal of Oxygen From the Oxide Redox Terminology Oxidation - Loss of Electrons Reduction - Gain of Electrons OIL RIG Mnemonic OIL – Oxidation Is Loss of Electron Rig – Reduction Is Gain of Electron Redox Terminology Oxidation - Loss of Electrons Reduction - Gain of Electrons Note Oxidation and Reduction Must Occur Concurrently! Oxidant - Reactant Which Oxidizes (gets Reduced) Reductant - Reactant Which Reduces (gets Oxidized) Redox Example Oxygen is a Powerful Oxidizing Agent Consider Formation of Iron(III) Oxide 4Fe + 3O2 4Fe 4Fe+3 + 12e- 3O2 + 12e4Fe +3O2 2Fe2O3 6O-2 2Fe2O3 Identification of Redox Rxns -Assign each element an Oxidation Number (State). If any element changes it’s oxidation state during a reaction, it is a redox reaction Zero Oxidation States indicate element is neutral [+] Oxidation States indicate element is “electron poor” [-] Oxidation States indicate element is “electron rich” Oxidation Numbers 1. Oxd # = 0 for pure elements 2. Oxd # = charge of monatomic ion 3. Oxd # of F = -1 in compounds with other elements 4. Oxd # of Cl, Br & I = -1 in compounds except with Oxygen & Fluorine 5. Oxd # of H= +1, except for Metal Hydrides (-1) 6. Oxd # of O = -2, except for Peroxides (-1) and Superoxides (-1/2) 7. The Sum of the oxd # ‘s of all elements in a compound = 0, and = the charge of a polyatomic ion 1/2 Reaction Method Consider the reduction of ferric iron by iodide. Fe+3 +2I- Fe+2 + I2 Is the above Equation Balanced? No! - Look at the Charge -+1 = +2 Balancing Redox Reactions 3 Techniques 1. Half Reaction Method -Good If Reaction Can Be Separated Into 2 1/2 Reactions 2. Oxidation # Method -Good If Reaction Can Not Be Separated Into 2 1/2 Reactions Half Cell Method Batteries take advantage of redox reactions where the oxidants and the reductants can be physically separated. These often occur in acidic and basic environments and so you will need to be able to balance ½ rxns in both Acidic and Basic Environments Balancing Acidic Redox Rxns 1. Split Skeletal Eq. Into 1/2 Rxns -You may need to assign oxidation #’s to all elements being oxidized or reduced 2. Balance all elements except O & H 3. Balance O by adding water 4. Balance H by adding H+ 5. Balance charge by adding e6. Multiply 1/2 rxns by appropriate integer to cancel electrons when adding ½ reactions 7. Add 1/2 reactions and check work Balancing Basic Redox Rxns 1-7. Balance as if Acidic 8. Add OH- to both sides to neutralize H+ (converting it to H2O) 1/2 Reaction Method Balance the Basic Reaction of Thiosulfate & Iodine to Form Sulfate, Iodide & Water S2O3-2 + I2 SO4-2 + I- 1/2 Reaction Method S2O3-2 + I2 SO4-2 + I- 1. We can separate the above reaction into 2 1/2 reactions Reduction 1/2 Reaction I2 I- Oxidation 1/2 Reaction S2O3-2 2O4-2 1/2 Reaction Method S2O3-2 + I2 SO4-2 + I- 2. Balance All Elements Except Oxygen and Hydrogen Reduction 1/2 Reaction I2 2I- Oxidation 1/2 Reaction S2O3-2 2SO4-2 1/2 Reaction Method S2O3-2 + I2 SO4-2 + I- 3. Balance Oxygen by Adding Water Reduction 1/2 Reaction I2 2I- Oxidation 1/2 Reaction 5H2O + S2O3-2 2SO4-2 1/2 Reaction Method S2O3-2 + I2 SO4-2 + I- 4. Balance Hydrogen by Adding H+ Reduction 1/2 Reaction I2 2I- Oxidation 1/2 Reaction 5H2O + S2O3-2 2SO4-2 + 10H+ 1/2 Reaction Method S2O3-2 + I2 SO4-2 + I- 5. Balance Charge by Adding electons Reduction 1/2 Reaction I2 + 2e- 2I- Oxidation 1/2 Reaction 5H2O + S2O3-2 2SO4-2 + 10H+ + 8e- 1/2 Reaction Method S2O3-2 + I2 SO4-2 + I- 6. Multiply ½ rxns by appropriate integers to cancel electrons Reduction 1/2 Reaction (I2 + 2e- 2I-) X 4 Oxidation 1/2 Reaction 5H2O + S2O3-2 2SO4-2 + 10H+ + 8e- 1/2 Reaction Method S2O3-2 + I2 SO4-2 + I- 7. Add ½ rxns and check work 4I2 + 8e- 8I- 5H2O + S2O3-2 2SO4-2 + 10H+ + 8e- 5H2O + S2O3-2 + 4I2 2SO4-2 + 10H+ + 8I- 1/2 Reaction Method S2O3-2 + I2 SO4-2 + I- 8. Add OH- to both sides neutralize H+ 5H2O + S2O3-2 + 4I2 +10 OH10 OH- + S2O3-2 + 4I2 2SO4-2 + 10H+ + 8I+10 OH2SO4-2 + 8I- + 5H2O Oxidation Number Method Lets Look at the following reaction in an acidic solution Cr2O7-2 + HSO3Cr(H2O)5SO4+ + SO4-2 Step 1: Write the oxidation state of all atoms which undergo a change above that atom 6 Cr2O7-2 + HSO3- Cr(H2O)5SO4+ + SO4-2 What is the Oxidation State of Chromium in Dichromate? 2( ) + 7(-2) = -2 2Cr 7O2 = Charge ( ) = (-2 +14)/2 = 6 6 4 Cr2O7-2 + HSO3- Cr(H2O)5SO4+ + SO4-2 What is the Oxidation State of Sulfur in the Hydrogen Sulfite? 6 4 Cr2O7-2 + HSO3- 3 Cr(H2O)5SO4+ + SO4-2 What is the Oxidation State of Chromium in the Cr(H2O)5SO4+ ? 6 4 Cr2O7-2 + HSO3- 3 6 6 Cr(H2O)5SO4+ + SO4-2 What is the Oxidation State of Sulfur in Sulfate? Step 2: Balance Elements in Coupled Molecules 6 4 Cr2O7-2 + HSO3- 3 6 6 2Cr(H2O)5SO4+ + SO4-2 For every Cr2O7-2 reacted, 2 Cr(H2O)5SO4+ must be Produced Step 3: Write the Change in Oxidation # of Oxidized Elements Above the Eq and Reduced Elements Below +2 6 4 Cr2O7-2 + HSO32(-3)= -6 3 6 6 2Cr(H2O)5SO4+ + SO4-2 Step 4. Balance Oxidizing and Reducing Equivalents 3(+2)= +6 6 4 Cr2O7-2 + 3HSO32(-3)= -6 3 6 6 2Cr(H2O)5SO4+ +SO4-2 Step 4: Balance Oxygen by Adding Water 6H2O + Cr2O7-2 + 3HSO3- 2Cr(H2O)5SO4+ +SO4-2 Step 5: Balance Hydrogen by Adding -add H+ in Acidic Solution -In Basic Solution, Balance as if Acidic, then Add Hydroxide to both sides to remove H+ 6H2O + 5H+ Cr2O7-2 + 3HSO3- 2Cr(H2O)5SO4+ +SO4-2 Double Check Conservation of Charge Redox Problems Balance the Following Reactions 1. Cr2O7-2 + C2O4-2 --> Cr+3 + CO2 (acidic) 2. MnO4- + NO2- --> MnO2 + NO3- (basic) 3. H2S + NO3- --> NO2 + S8 (acidic) 4. H2O2 + MnO4- --> O2 + MnO2 (basic) 1. Cr2O7-2 + C2O4-2 --> Cr+3 + CO2 (acidic) Cr2O7-2 --> 2Cr+3 Cr2O7-2 --> 2Cr+3 + 7H2O 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O 6e- + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O C2O4-2 --> 2CO2 C2O4-2 --> 2CO2 + 2e6e- + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O 3( C2O4-2 --> 2CO2 + 2e-) 3C2O4-2 + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O + 6CO2 2. MnO4- + NO2- --> MnO2 + NO3- (basic) MnO4- --> MnO2 + 2H2O 4H+ + MnO4- --> MnO2 + 2H2O 3e- + 4H+ + MnO4- --> MnO2 + 2H2O H2O + NO2- --> NO3H2O + NO2- --> NO3- + 2H+ H2O + NO2- --> NO3- + 2H+ + 2e2(3e- + 4H+ + MnO4- --> MnO2 + 2H2O) 3(H2O + NO2- --> NO3- + 2H+ + 2e-) (3NO2- + 2H+ + 2MnO4- --> 2MnO2 + H2O + 3NO3-) +2OH+2OH(3NO2- + H2O + 2MnO4- --> 2MnO2 + 2OH- + 3NO3-) 3. H2S + NO3- --> NO2 + S8 (acidic) 8H2S --> S8 8H2S --> S8 + 16H+ 8H2S --> S8 + 16H+ + 16e- NO3- --> NO2 + H2O 2H+ + NO3- --> NO2 + H2O e- + 2H+ + NO3- --> NO2 + H2O 8H2S --> S8 + 16H+ + 16e16(e- + 2H+ + NO3- --> NO2 + H2O) 16H+ + 16NO3- + 8H2S --> S8 + 16NO2 + 16H2O 4. H2O2 + MnO4- --> O2 + MnO2 (basic) MnO4- --> MnO2 + 2H2O 4H+ + MnO4- --> MnO2 + 2H2O 3e- + 4H+ + MnO4- --> MnO2 + 2H2O H2O2 --> O2 + 2H+ H2O2 --> O2 + 2H+ + 2e2(3e- + 4H+ + MnO4- --> MnO2 + 2H2O) 3(H2O2 --> O2 + 2H+ + 2e-) 3 H2O2 + 2H+ + 2MnO4- --> 2MnO2 + 4H2O + 3O2 +2OH- +2OH- 3 H2O2 + 2MnO4- --> 2MnO2 + 2H2O + 3O2 + 2OH- Electrochemical Cells 2 Types of Electrochemical Cells 1. Galvanic Cells (batteries)- spontaneous redox reactions, produce energy 2. Electrolytic Cells - nonspontaneous redox reactions - require energy, used in commercial syntheisis (electroplating…) Electrochemical Cells Note: We call galvanic cells "batteries" which is actually a misnomer. A battery is actually a series of cells (the term comes from a battery of cannons) and may be constituted of galvanic or electrolytic cells Voltaic (Galvanic) Cells - “Classical Batteries”, Will Release Energy Is DG Negative or Positive? DG < 0 -Galvanic Cells Spontaneously Release Energy Activity Series The Activity Series of Single Replacement Reactions Describes Spontaneity Zn(s) + CuCl2 ZnCl2 + Cu(s) Zn(s) + CuCl2 ZnCl2 + Cu(s) What is the net ionic equation? Zn(s) + Cu+2 Cu(s) + Zn+2 This can be Split into 2 1/2 reactions Zn(s) Zn+2 + 2e- Cu+2 + 2e- Cu(s) Zn(s) + CuCl2 Zn(s) ZnCl2 + Cu(s) Zn+2 + 2e- Cu+2 + 2e- Cu(s) Electrochemical Cells physically separate half reactions while allowing charge transfer Zn(s) + CuCl2 Zn(s) ZnCl2 + Cu(s) Zn+2 + 2e- Zinc: Reductant or reducing agent (gets oxidized while reducing copper) Cu+2 + 2e- Cu(s) Copper: Oxidant or oxidizing agent (gets reduced while oxidizing copper) Zn(s) + CuCl2 Oxidation Half Cell ZnCl2 + Cu(s) Reduction Half Cell Salt Bridge: Allows Flow of Ions While Keeping Oxidant & Reductant Separated Net Cell Reaction: Zn(s) + Cu+ 2(aq) --> Zn+2(aq) + Cu(s) Shorthand Notation 2 2 Zn( s) Zn (aq) Cu (aq) Cu( s) Anode 1/2 rxn Cathode 1/2 rxn Salt Bridge - + Oxidation Occurs at the Anode Negative Electrode Reduction Occurs at the Cathode Positive Electrode Electric Potentials & Units Work (Electrical) = (Charge)(Potential Differnce) W = qDV C = Coulomb - Charge of 6.2 x 1018 electrons F = Faraday = charge of 1 mole of e1F = 96,500 coulombs J 1V C The Flow of electricity is measured in amps 1 amp = 1 C/sec EMF (Ecell) EMF: Electromotive Force A galvanic cell’s maximum DV -Open Circuit Potential (the cell potential drops when current flows) Wmax = -nFEcell DGrxn = -nFEcell Eo & DGo Wmax = o -nFE Wmax = o DG o DG rxn = rxn o -nFE F = Faraday’s Constant -the charge of 1 mole of e- 96,500 J/(V x mol) Reference Potential What is 0 Potential? q1q2 V k r V = 0 when two charged objects are separated by infinity We are interested in DV between Anode and Cathode Electric Potentials Electrons Spontaneously Flow From Anode (oxidation) to Cathode (reduction) Zn Anode V Zn+2 + 2e- Zn(s) e- flows to the lower potential Cu Cathode Cu+2 + 2e- Cu(s) (Oxidation of Anode & Reduction of Cathode) Standard Potentials Reduction 1/2 Reaction F2(g) + 2 eCl2(g) + 2 eBr2(g) + 2 eFe+3(aq) + eCu+2(aq) + 2e2H+(aq) + 2 eFe+2(aq) + 2eZn+2(aq) + 2eNa+(aq) + eK+(aq) + eLi+(aq) + e- Eo (V) 2 F-(aq) 2 Cl-(aq) 2 Br-(aq) Fe+2(aq) Cu (s) 2 H2(g) Fe (s) Zn (s) Na(s) K(s) Li(s) +2.87 +1.36 +1.06 +0.54 +0.34 0.000 -0.44 - 0.76 -2.71 -2.93 -3.05 Standard Electrode Potentials Eocell - Standard Electrode Potential - the cell emf when all reactants are at standard state conditions Refers to Potential Eocell = Eored(red) + Eoox(ox) Refers to Half Reaction At Standard State Conditions, all solutes are at 1M concentrations and all gasses are at 1 atm pressure Standard Electrode Potentials How Would You Write the Standard Cell for the Reaction of Zinc & Hydrogen? 2 Zn Zn (1M ) H (1M ) H 2 (1atm), Pt (s) How Can We Set up a Scale to Measure Electrode Potentials? Zn Zn2 (1M ) H (1M ) H 2 (1atm), Pt (s) video Standard Potentials Reduction 1/2 Reaction F2(g) + 2 eCl2(g) + 2 eBr2(g) + 2 eFe+3(aq) + eCu+2(aq) + 2e2H+(aq) + 2 eFe+2(aq) + 2eZn+2(aq) + 2eNa+(aq) + eK+(aq) + eLi+(aq) + e- Eo (V) 2 F-(aq) 2 Cl-(aq) 2 Br-(aq) Fe+2(aq) Cu (s) 2 H2(g) Fe (s) Zn (s) Na(s) K(s) Li(s) +2.87 +1.36 +1.06 +0.54 +0.34 0.000 -0.44 - 0.76 -2.71 -2.93 -3.05 o E cell Determination of from Reduction Potentials Refers to Potential Eocell = Eored(red) + Eoox(ox) Refers to Half Reaction Oxidation is the opposite of reduction, so the oxidation potential is the negative of the reduction potential Eocell = Eored(red) - Eored(ox) Don’t get Confused, the Concept is Simple: The greater the “reduction potential”, the more something wants to get reduced (gain electrons). The greater the “oxidation potential”, the more something wants to get oxidized (give electrons). You simply change the sign to go from a reduction potential to an oxidation potential because they are opposite processes. What is Eocell for the following cell? Zn(s) Zn2 (aq) Cu 2 (aq) Cu(s) (The Zinc gets Oxidized and the Cu+2 gets Reduced) Cu+2(aq) Zn (s) + 2e- Cu (s) Zn+2(aq) + 2e- Eocell = 0.34 - (-.76) = 1.10V Eored = 0.34 Eored =- .76 What is Eocell for the following cell? Fe(s) Fe2 (aq) Zn2 (aq) Zn(s) (The Iron gets Oxidized and the Zn gets Reduced) But Nothing Happens Fe+2(aq) Fe (s) Zn+2(aq) + 2e- + 2e- Zn (s) Eocell = -.76 – (-.44) = -.32V Eored = -.44 Eored = -0.76 o E o DG & Spontaneity rxn = o -nFE Positive Eocell – Spontaneous Negative Eocell - Nonspontaneous Use Electrode Potentials to Determine Spontaneity for the following reaction. . -(aq) Cl2(g) + 2I 2Cl-(aq)+ I2(s) Eocell = Eored + Eoox Cl2(g) + 2I-(aq) 2Cl-(aq)+ I2(s) Identify the 1/2 Reactions Cl2(g) + 2e2I-(s) 2Cl-(aq) I2(s) + 2e- Eored = 1.36 Eored = .54 Eocell = 1.36 + (-.54) = .82V Spontaneous Class Problem If Eo Is 0.82 V, Determine the Standard Free Energy and Equilibrium Constant Cl2(g) + 2I- 2Cl- + I2(s) DG o nFE o (2mol)(96500 molJV )(.82V ) 158kJ K e Go DRT e ( 158KJ ) ( 8.314J )( 298K ) K 4.9 x1027 Nernst Eq What Is the Effect of Concentration on Cell EMF? DG = DGo + RTlnQ DG = -nFE -nFE = -nFEo +RTlnQ E = Eo - (RT/nF)lnQ E = Eo - (0.0592/n)logQ At 298 K Class Problem Using Standard Reduction Potentials Calculate DG, K & E at 298 K for. 2Br-(aq) + Cl2(g) Br2(l) + 2Cl-(aq) Given: [Br-] = 0.1M, [Cl-] = 0.01M & PCl2 = 0.50 atm 2Br-(aq) + Cl2(g) Br2(l) + 2Cl-(aq) Given: [Br-] = 0.1M, [Cl-] = 0.01M & PCl2 = 0.50 atm E = Eo - (RT/nF)lnQ Cl2(g) + 2 eBr2(g) + 2 e- 2 Cl-(aq) 1.36V 2 Br-(aq) 1.06V Eo = 1.36 - 1.06 = .3V (8.314)(298) [.01]2 E .3V ln 2(96500) .5[.1]2 E = .35V 2Br-(aq) + Cl2(g) Br2(l) + 2Cl-(aq) DG = -nFEcell What is n? 2Br-(aq) Br2(l) + 2eCl2(g) +2e-(aq) 2Cl-(aq) J DG (2mol)(96500 V mol )(.35V ) DG = -67.55kJ What is K? K e o D G RT e nFEo RT K = 1.4 x e 10 10 J )(. 3V ) 2 ( 96500mol V 8.314molJK ( 298K ) Nernst Equation and Concentration of an unknown What is the concentration of Zn+2 in the following cell if the potential is 1.13V? Zn(s) Zn2 (aq) Cu 2 (1.80M ) Cu( s) Nernst Equation and Concentration of an unknown What is the concentration of Zn+2 in the following cell if the cell Potential is 1.13V? Zn(s) Zn2 (aq) Cu 2 (1.80M ) Cu( s) From Standard Potentials: Zn (s) Cu+2(aq) + 2e- Zn+2(aq) + 2e- Eored = -.76 Cu (s) Eored = 0.34 Eocell = -(-0.76) + 0.34 = 1.10V Nernst Equation and Concentration of an unknown What is the concentration of Zn+2 in the following cell if the cell Potential is 1.13V? Zn(s) Zn2 (1.0) Cu 2 (1.80M ) Cu(s) Zn( s ) Cu 2 Zn 2 Cu ( s ) RT RT Zn 2 0 EE ln Q E ln 2 nF nF Cu Zn 2 nF 0 ln 2 E E Cu RT 0 2 nF 0 E E 2 RT Zn Cu e 1.8Me 2(96,500)(1.11.13) 8.314(298) 0.17 M Classical “Batteries” Galvanic (Voltaic) Cells Which Can Store Electrochemical Energy and Release it for Work During Discharge Lead Storage Batteries E0 Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) +2e- 0.296V PbO2(s) +3H+(aq)+ HSO4-(aq) + 2e- ---> PbSO4(s) + 2H2O(l) 1.628V Pb(s) + PbO2(s) +2H+(aq)+ HSO4-(aq) ---> 2PbSO4(s) + 2H2O(l) 1.92V How do we get a 12 Volt Car Battery? By “stacking” six 2 volt cells in series Edison’s Alkaline Battery KOH Electrolyte (5 Cells) NiO2(s) + 2H2O(l) + 2e- Ni(OH)2(s) + 2OH- Eored= .49V Fe(OH)2(s) + 2e Fe(s) +2OHEored=-.877V Edison’s Alkaline Battery KOH Electrolyte (5 Cells) Nickel Oxide Cathode Iron Anode NiO2(s) + 2H2O(l) + Fe(s) +2OH- 2e- Ni(OH)2(s) + Fe(OH)2(s) + 2e- 2OH- Eored= .49V Eored=+.877V NiO2(s)+2H2O(l)+Fe(s) Ni(OH)2(s)+Fe(OH)2(s) Eocell= 1.37V Alkaline Dry Cell Zn(s) + 2OH- --> ZnO(s) +H2O(l) + 2e2 MnO2(s) +2H2O(aq) + 2e- ---> Mn2O3(s) +2OH-(aq) Fuel Cells 2H2(g) + 4OH-(aq) --> 4H2O(l) + 4eO2(g) + 2H2O(l) + 4e- --> 4 OH-(aq) Fuel Cells PEM Electrolyzer/Fuel Cell PEM Electrolyzer/Fuel Cell Corrosion Spontaneous Redox Reaction Resulting in Formation of Oxides From Pure Metals Only Noble Metals do Not Corrode Corrosion 4Fe+2(aq) + O2(g) + 4H+(aq) --> 4Fe+3(aq) + 2H2O 2Fe+3 + 4 H2O(l) --> Fe2O3.H2O + 6H+ Why does Aluminum not Rust? It does form a layer of aluminum oxide, but this is impervious to oxygen, and functions as a protective coat. Cathodic Protection Iron can be protected by hooking up another metal with a lower reduction potential, which functions as a Sacrificial Anode Cathodic Protection Iron can be protected by hooking up another metal with a lower reduction potential, which functions as a Sacrificial Anode Will Copper Work as a Sacrificial Anode? Will Zinc Work as a Sacrificial Anode? Will Magnesium Work as a Sacrificial Anode? Galvanic & Electrolytic Cells Galvanic (Voltaic) Cells Positive Potentials Spontaneous Provide Energy for Work (DG<0) Electrolytic Cells Negative Potentials Non Spontaneous Require Energy to Work (DG>0 unless coupled to an external voltage source) Galvanic & Electrolytic Cells Sodium and Chlorine Spontaneously Form Sodium Chloride in a Galvanic Cell Ecell = 1.36V -(-2.71V) = 4.07V Where the Sodium is being oxidized and the Chlorine is being reduced An Electrolytic Cell Is Formed by Applying a Potential Greater Than 4.07 V in the Opposite Direction, Making the Nonspontaneous Reaction Occur Electrolysis of Molten NaCl Driving Non Spontaneous Reactions + - 2Na+(aq) + 2Cl-(aq) --> 2Na(s) + Cl2(g) Electrolysis of aqueous NaF What are the possible reduction 1/2 rxns? Na+(aq) e- + 2H2O(l) + 2e- Na(s) H2(g) + 2OH-(aq) E0red= -2.71V E0red= -0.83V What are the possible oxidation 1/2 rxns? 2F-(aq) 2H2O(l) 4OH-(aq) 2e- F2(g) + O2(g) + 4H+(aq) + 4eO2(g) + 2H2O(l) + 4e- E0red= 2.87V E0red= 1.23V E0red= 0.40V What species gets reduced? H 2O What species gets oxidized? OH- Electrolysis of aqueous NaF The net cell reaction? cathode: 2[2H2O(l) + 2eanode: 4OH-(aq) cell: 2H2O(l) H2(g) + 2OH-(aq)] E0red= -083V O2(g) + 2H2O(l) + 4eE0red= 0.40V 2H2(g) + O2(g) The Electrolysis of Water E0cell= -1.23V Electroplating & Active Electrodes When metallic cations are being reduced, they can bind to the cathode Active Electrodes Stoichiometry & Electrolysis How do we measure the rate at which an electric current carries charge? 1 Amp = 1C/sec How many moles of electrons does this represtent? 1F = 96,500 C = charge of 1 mole of e1 Amp = 1.04x10-5 mole e-/ sec Stoichiometry & Electrolysis The current across an electrolytic cell provides a stoichiometric relation for the number of electrons transferred to the number of species oxidized and reduced 1F = 96,500 C = charge of 1 mole of e1 Amp = 1C/sec 2 Types of Problems: Stoichiometry & Electrolysis 1. Measure Current & time, determine quantity of species oxidized and reduced. 2. Measure quantity of any species oxidized or reduced over a given time span, and you can determine the current Stoichiometry & Electrolysis Lithium is a very active metal, and can be formed through the electrolysis of LiCl. What mass of Lithium and chlorine can be formed through the electrolysis of molten LiCl if 4300.0 amps passes through a cell operating for 24.0 hours? m Li ( 4300C sec mCl2 ( 4300C sec (24h)( 60min h (24h)( 60min h )( 60sec min )( )( 60sec min mol e 96500C )( mole )( mol e 96500C Cl2 71g Cl2 )( 2mol )( mol ) 137kg Cl 2 mol e )( mol Li 6.9 g Li mol ) 26.6kg Li Electroplating How many minutes will it take to plate out 125 g Cr+3 with a constant current of 200 amps? molCr 3 3mol e 96500C sec min 125gCr molCr 3 mol e 200C 60sec 58.0 min 52 g 3