CHAPTER 20 PRINCIPLES OF REACTIVITY: ELECTRON TRANSFER REACTIONS 1-11 + all bold numbered problems 1 2 ELECTROCHEMISTRY Electric automobile 2 Why Study Electrochemistry? • Batteries • Corrosion • Industrial production of chemicals such as Cl2, NaOH, F2 and Al A rusted car. • Biological redox reactions The heme group 3 20.1 OXIDATION-REDUCTION REACTIONS • Redox reactions make up electrochemical cells. • Those that produce energy are called batteries or galvanic cells (voltaic). • Those that require energy to operate are called electrolysis cells. 4 TRANSFER REACTIONS Atom transfer HOAc + H2O ---> OAc- + H3O+ no change in oxidation state of any of the elements Electron transfer Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s) Cu looses 2 eOIL and Au+ gains 1 eRIG 5 Electron Transfer Reactions • Electron transfer reactions are oxidation-reduction or redox reactions. • Redox reactions can result in the generation of an electric current or be caused by imposing an electric current. • Therefore, this field of chemistry is often called ELECTROCHEMISTRY. 6 CHAPTER OVERVIEW • This chapter examines electron transfer reactions, redox Reduction/Oxidation Note: Oil Rig = Oxidation is Loss of an electron Reduction is Gain of an electron 7 Review of Terminology for Redox Reactions • OXIDATION—loss of electron(s) by a species; increase in oxidation number. • REDUCTION—gain of electron(s); decrease in oxidation number. • OXIDIZING AGENT—electron acceptor; species is reduced. • REDUCING AGENT—electron donor; species is oxidized. 8 Solid interacting with aqueous ions 9 OXIDATION-REDUCTION REACTIONS Direct Redox Reaction Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s) 10 Electrochemical Cells • An apparatus that allows a redox reaction to occur by transferring electrons through an external connector. • Product favored reaction ---> voltaic or galvanic cell ----> electric current • Reactant favored reaction ---> electrolytic cell ---> electric current used to cause chemical change. Batteries are voltaic cells 11 OXIDATION-REDUCTION REACTIONS Indirect Redox Reaction A battery functions by transferring electrons through an external wire from the reducing agent to the oxidizing agent. 12 CHEMICAL CHANGE --> ELECTRIC CURRENT Zn metal 2+ ions Cu Cu2+ ions With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” Zn is oxidized and is the reducing agent Zn(s) ---> Zn2+(aq) + 2eCu2+ is reduced and is the oxidizing agent Cu2+(aq) + 2e- ---> Cu(s) 13 CHEMICAL CHANGE --> ELECTRIC CURRENT Zn metal 2+ ions Cu Cu2+ ions Electrons are transferred from Zn to Cu2+, but there is no useful electric current. Oxidation: Zn(s) ---> Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e- ---> Cu(s) -------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s) 14 CHEMICAL CHANGE --> ELECTRIC CURRENT •To obtain a useful current, we separate the oxidizing w ire e le c t ro ns and reducing agents so salt Zn bridge that electron transfer occurs thru an external wire. Zn 2+ ions •This is accomplished in a GALVANIC or VOLTAIC cell. •A group of such cells is called a battery. Cu Cu2+ ions 15 •Electrons travel thru external wire. •Salt bridge allows anions and cations to move between electrode compartments. 16 •This maintains electrical neutrality. Electrons move from anode to cathode in the wire. Anions & cations move thru the salt bridge. Electrochemical Cell or Battery 17 Balancing Redox Equations OXIDATION-REDUCTION REACTIONS • To be useful, the reactions for these cells must be balanced. • The simplest method of balancing is the ion-electron or half-reaction method. • We learned this method in the first half of General Chemistry, but a quick review is in order. 18 OXIDATION-REDUCTION REACTIONS • Oxidation is the loss of electron, and reducing agents are oxidized. • Reduction is the gain of electrons, and oxidizing agents are reduced. • The simple steps in balancing a redox equation are: 19 BALANCING REDOX REACTIONS 1. Separate the reaction into half reactions. 2. Perform a mass balance for each element beginning with the element being oxidized or reduced. Next balance the oxygen atoms with water and the hydrogen atoms with hydrogen ions. 3. Perform a charge balance by adding electrons to balance the charge. 20 BALANCING REDOX REACTIONS 4. If the reaction is in basic solution, add as many hydroxide ions to both sides as there are hydrogen ions and simplify by forming water. Eliminate the excess water, if any occurs on both sides of the reaction. 5. Multiply each half-reaction by the appropriate factor so the electrons are eliminated when the half-reactions are added. 21 22 Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques. MnO4- + 5 Fe2+ + 8 H+ ---> Mn2++ 5 Fe3+ + 4 H2O 22 Balancing Equations Consider the reduction of Ag+ ions with copper metal. Cu + Ag+ -- give --> Cu2+ + Ag 23 Balancing Equations Step 1: Divide the reaction into half- reactions, one for oxidation and the other for reduction. Ox Cu ---> Cu2+ Red Ag+ ---> Ag Step 2: Balance each for mass. Already done in this case. Step 3: Balance each half-reaction for charge by adding electrons. Ox Red Cu ---> Cu2+ + 2eAg+ + e- ---> Ag 24 Balancing Equations Step 4: Multiply each half-reaction by a factor that means the reducing agent supplies as many electrons as the oxidizing agent requires. Reducing agent Cu ---> Cu2+ + 2eOxidizing agent 2 Ag+ + 2 e- ---> 2 Ag Step 5: Add half-reactions to give the overall equation. Cu + 2 Ag+ ---> Cu2+ + 2Ag The equation is now balanced for both charge and mass. 25 Balancing Equations Balance the following in acid solution— VO2+ + Zn ---> VO2+ + Zn2+ Step 1: Write the half-reactions Ox Zn ---> Zn2+ Red VO2+ ---> VO2+ Step 2: Balance each half-reaction for mass. Ox Zn ---> Zn2+ 26 Balancing Equations Balance the following in acid solution— VO2+ + Zn ---> VO2+ + Zn2+ Step 1: Write the half-reactions Ox Zn ---> Zn2+ Red VO2+ ---> VO2+ Step 2: Balance each half-reaction for mass. Ox Zn ---> Zn2+ Red VO2+ ---> VO2+ + H2O 2 H+ Add H2O on O-deficient side and add H+ on other side + for H-balance. 27 Balancing Equations Step 3: Balance half-reactions for charge. Ox Zn ---> Zn2+ + 2e- Red e- + 2 H+ + VO2+ ---> VO2+ + H2O Step 4: Multiply by an appropriate factor. Ox Zn ---> Zn2+ + 2e- Red 2e- + 4 H+ + 2 VO2+ ---> 2 VO2+ + 2 H2O Step 5: Add half-reactions Zn + 4 H+ + 2 VO2+ ---> Zn2+ + 2 VO2+ + 2 H2O 28 Tips on Balancing Equations • Never add O2, O atoms, or O2- to balance oxygen. • Never add H2 or H atoms to balance hydrogen. • Be sure to write the correct charges on all the ions. • Check your work at the end to make sure mass and charge are balanced. MORE PRACTICE!!29 20.2 CHEMICAL CHANGES LEADING TO ELECTRIC CURRENT • An electrochemical cell has two electrodes, the anode where oxidation occurs, and the cathode where reduction occurs. • The compartment containing the anode is called the anodic compartment • The compartment containing the cathode is called the cathodic compartment. • These two half-cells are frequently connected by a porous barrier or a salt bridge. 30 CHEMICAL CHANGES LEADING TO ELECTRIC CURRENT • The electrical current is carried by electrons in the metals present, and by ions in the solution. • Ions carrying charge are called electrolytes. • Electrons travel from the anode (-) to the cathode (+) 31 Reduction Oxidation Cathodic compartment Anodic compartment 32 Illustrates a general cell 33 Abbreviated Cell Notation Anode | Anodic sol’n || Cathodic sol’n | Cathode This simple notation is very efficient 2Fe3+(aq) + H2(g) 2Fe2+(aq) + 2H+(aq) Pt|H2(1atm)|H+(1M)||Fe2+(1M),Fe3+(1M)|Pt 34 Anode | Anodic sol’n || Cathodic sol’n | Cathode Cu|Cu2+(1.0 M)||Ag+(1.0 M)|Ag 35 20.3 ELECTROCHEMICAL CELLS AND POTENTIALS • The electrical potential is called the electromotive force or emf, and is given the symbol E. • The unit of charge is called the coulomb • emf is measured in joules/coulomb or volts, V • Under standard conditions, 1.0 M for solutes, 1.0 bar for gases, and 298.15 K, the emf is given the symbol Eo, and called the standard potential. 36 Eo vs. DGo • The electrical work is nFE, where F is Faraday's constant, 96, 485 coulombs/ mole electrons or J/V mole, and n is the moles of electrons in the balanced equation. 37 Eo vs. DGo This leads to the following equation between E and DG: DG = - nFE. • This can be derived from: G = H -TS = E + PDV - TDS • Since DE = q + w and w = - PDV – wext • If the process is done reversibly at constant temperature and pressure, q = TDS. DG = q - PDV – wext + PDV - TDS • This leads to DG = - wext. 38 Eo vs. DGo In the standard state, DGo = - nFEo. • Now we can use E and Eo to predict reaction spontaneity, i.e. whether a reaction is product-favored. • Calculations of DGo from Eo and visa versa are algebraic. 39 Calculating the Potential Eo of an Electrochemical Cell • All standard potentials are bases on a relative scale where the hydrogen half-cell is assigned a value to 0.00 V at all temperature, and is called the standard hydrogen electrode, SHE. • All electrode potentials are written as reductions and derive their values from direct or indirect measurements vs. the SHE. 40 Calculating the Potential Eo of an Electrochemical Cell • Demonstrates this measure for the zinc electrode potential. 41 Demonstrates this measure for the copper electrode potential. 42 CELL POTENTIAL, E Zn and Zn2+, anode 4 3 Cu and Cu2+, cathode • Electrons are “driven” from anode to cathode by an electromotive force or emf. • For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 C and when [Zn2+] and [Cu2+] = 1.0 M. 43 CELL POTENTIAL, E • For Zn/Cu cell, voltage is 1.10 V at 25 C and when [Zn2+] and [Cu2+] = 1.0 M. • This is the STANDARD CELL POTENTIAL, Eo • -- a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 C. 44 Calculating Cell Voltage • Balanced half-reactions can be added together to get overall, balanced equation. 2 I- ---> I2 + 2e- 2 H2O + 2e- ---> 2 OH- + H2 ------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2 If we know Eo for each half-reaction, we could get Eo for net reaction. 45 CELL POTENTIALS, o E • Can’t measure 1/2 reaction Eo directly. Therefore, measure it relative to a STANDARD HALF CELL, SHE. 2 H+(aq, 1 M) + 2e- --> H2(g, 1 atm) Eo = 0.0 V Typically made of Pt or Au, are inert, and important because SHE’s can act as either an anode or cathode, difficult to maintain, not used often 46 Other Reference Electrodes, know Eo, easy to make Self-contained, only make once, sealed, last a longtime ||Hg2Cl2(sat),KCl(xM)|Hg ||AgCl(sat), KCl(xM)|Ag 47 ||TlCl(sat),KCl(xM)|Tl Zn/Zn2+ half-cell hooked to a SHE. Eo for the cell = +0.76 V Remember Eo for SHE = 0.0 V Volts Zn - + Salt Bridge H2 Zn2+ Zn Zn2+ + 2eOXIDATION ANODE H+ 2 H+ + 2eH2 REDUCTION CATHODE 48 - Zn Volts Salt Bridge + H2 H+ Zn 2+ 2+ Zn Zn + 2 e- 2 H + + 2 eOXIDATION REDUCTION ANODE CATHODE Remember Eo for SHE = 0.0 V H2 Overall reaction is reduction of H+ by Zn metal. Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g) Eo = + 0.76 V Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is + 0.76 V. Reduction potential is - 0.76 V. Zn is a better reducing agent (Zn gets oxidized) than) H2. 49 Cu/Cu2+ and H2/H+ Cell Eo = +0.34 V Volts Cu + Salt Bridge H2 Cu2+ Cu2+ + 2eCu REDUCTION CATHODE H+ H2 2 H+ + 2eOXIDATION ANODE 50 Cu/Cu2+ and H2/H+ Cell Volts Cu + Salt Bridge H2 Cu 2+ Cu 2+ + 2eCu REDUCTION CATHODE • • • • H+ H2 2 H + + 2eOXIDATION ANODE Overall reaction is reduction of Cu2+ by H2 gas. Cu2+ (aq) + H2(g) ---> Cu(s) + 2 H+(aq) Measured Eo = + 0.34 V Therefore, Eo for Cu2+ + 2e- ---> Cu is + 0.34 V so Cu is a better Oxidizing agent (Cu gets reduced) the H+ 51 COMBINING Voltaic CELLS • Lets Combine the work we just did with the Zinc and Copper Cells and combine them into ONE cell and see the net effect Zn2+(aq) + 2e- ---> Zn(s) Cu2+(aq) + 2e- ---> Cu(s) Eo = - 0.76 V Eo = + 0.34 V Zn is a better reducing agent than H+ Cu is a better Oxidizing agent than H+ From this info we know Zn MUST be oxidized and that Cu MUST be reduced 52 Zn/Cu Electrochemical Cell wire elect rons Anode, negative, source of electrons (oxidation) Zn Zn 2+ ions salt bridge Cu Cu2+ ions Cathode, positive, sink for electrons (reduction) Zn(s) ---> Zn2+(aq) + 2eEo = + 0.76 V Cu2+(aq) + 2e- ---> Cu(s) Eo = + 0.34 V --------------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s) Eo (calculated) = + 1.10 V 53 Uses of o E Values These experiments show we can a) decide on relative ability of elements to act as reducing agents (or oxidizing agents) b) assign a voltage to a halfreaction that reflects this ability. wire elect rons Zn Zn 2+ ions salt bridge Cu Cu2+ ions 54 USING STANDARD POTENTIALS Eocell = Eocathodic - Eoanodic Eocell = Eoreduction + Eooxidation • The first equation uses the values right out of the table of standard electrode potentials. • The table can also give us information about which reactions will be productfavored (produce current) in the standard state since these reactions with have a positive Eocell. 55 56 57 58 USING STANDARD POTENTIALS • The following is an abbreviated table form the previous tables to give practice at using them. 59 TABLE OF STANDARD POTENTIALS Oxidizing ability of Ion Eo (V) Cu2+ + 2e- Cu +0.34 2 H+ + 2e H2 0.00 Zn2+ + 2e- Zn -0.76 reducing ability of element 60 Standard Redox Potentials, Eo oxidizing ability of ion Eo (V) Cu2+ + 2e- Cu +0.34 2 H+ + 2e- H2 0.00 Zn2+ + 2e- Zn -0.76 reducing ability of element • Any substance on the right will reduce any substance higher than it on the left. • Zn can reduce H+ and Cu2+. • H2 can reduce Cu2+ but not Zn2+ • Cu cannot reduce H+ or Zn2+. 61 Standard Redox Potentials, Eo oxidizing ability of ion Eo (V) Cu2+ + 2e- Cu +0.34 2 H+ + 2e- H2 0.00 Zn2+ + 2e- Zn -0.76 reducing ability of element • • • • • Zn(s) ---> Zn2+(aq) + 2eEo = + 0.76 V Cu2+(aq) + 2e- ---> Cu(s) Eo = + 0.34 V --------------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s) Eo (calculated) = + 1.10 V 62 Using Standard Potentials, Eo See next slide Which is the best oxidizing agent: O2, H2O2, or Cl2? _________________ • Which is the best reducing agent: Hg, Al, or Sn? ____________________ • Which is the correct direction for the following reaction to produce a favorable potential? 2Al(s) + 3 Cl2(aq) ---> AlCl3(aq) AlCl3(aq) ---> 2Al(s) + 3 Cl2(aq) 63 64 o E for a Voltaic Cell Volts Cd Fe Salt Bridge Cd2+ Cd --> Cd2+ + 2eor Cd2+ + 2e- --> Cd Fe2+ Fe --> Fe2+ + 2eor Fe2+ + 2e- --> Fe See previous slide 65 o E for a Voltaic Cell Volts Cd Fe Salt Bridge Cd2+ Fe2+ From the table, you see • Fe is a better reducing agent than Cd • Cd2+ is a better oxidizing agent than Fe2+ Overall reaction Fe + Cd2+ ---> Cd + Fe2+ Eo = + 0.04 V 66 Examples of Cells that NEED electricity (opposite of Voltaic or Galvanic cells Production of Oxygen and Hydrogen 67 Electrolysis of SnCl2(aq) Sn2+(aq) + 2Cl-(aq) Sn(s) and Cl2(g) Anode (+) Cathode (-) 68 NaI phenolphthalein More on this later, industrial applications OH- detected 69 o E and D o G Eo is related to DGo, the free energy change for the reaction. o DG = -nF o E where F = Faraday constant = 9.6485 x 104 J/V•mol and n is the number of moles of electrons transferred Michael Faraday 1791-1867 70 Michael Faraday 1791-1867 Originated the terms anode, cathode, anion, cation, electrode. Discoverer of • electrolysis • magnetic props. of matter • electromagnetic induction • benzene and other organic chemicals Was a popular lecturer. 71 o E and D o G DGo = - n F Eo For a product-favored reaction Reactants ----> Products DGo < 0 and so Eo > 0 Eo is positive, results in current For a reactant-favored reaction Reactants <---- Products DGo > 0 and so Eo < 0 Eo is negative , needs current 72 ELECTROCHEMICAL CELLS AT NON STANDARD CONDITIONS • We learned the relationship between DG and DGo in chapter 19. DG = DGo + RT ln Q. • This yields: -nFE = -nFEo + RT ln Q • At 298.15 K, we solve for E E = Eo - [0.0592 V/n] log Q or E = Eo - [0.0257 V/n] ln Q 73 20.5 Eo and the Equilibrium Constant Nernst Equations • At equilibrium for the cell reaction, E = 0, (equivalent to saying the cell is dead) Therefore: Eo = [0.0592 V/n] log K or Eo = [0.0257 V/n] ln K • These equations can be solved for K using the inverse functions. • n is the number of moles of e- 74 Calculating Ksp, Kf, and other K's from cell potentials • In the laboratory we will be calculating Ksp's and Kf 's for reactions using electrochemical cells. • Determine the Kf for AlF6-3 from the cell below: E = 2.465 V Al |10.0 mL 0.10 M Al(NO3)3,40.0 mL 1.0 M NaF ||0.10 M CuSO4|Cu 2 Al + 3 Cu+2 ===> 2 Al+3 + 3 Cu Al+3 + 6F- <===> AlF6-3 Given: [F-] = 0.68 M, [AlF6-3] = 0.020 M Note: no reaction between Al3+ or Cu2+ and SO4-2 75 2 Al + 3 Cu+2 ===> 2 Al+3 + 3 Cu Eo = 1.676 + .337 = 2.013 V Ecell = 2.465 V (given), 0.10 M Cu+2 Ecell = Eo - [0.0257 V/n] ln K 2.465 = 2.013 - [0.0257/6] Ln (Al+3)2/(.1)3 solved for Al+3, note that as soon as Al+3 forms it immediately reacts with F-, so Al+3 is NOT .1M Which gives: [Al+3] = 4.0x10-25 M Given: [F-] = 0.68 M, [AlF6-3] = 0.020 M Sub these values into Kf Kf = 5.0 x 1023 76 21.6 BATTERIES AND FUEL CELLS • There are two types of batteries: • Primary, which are not rechargeable • Secondary, or storage batteries which are easily recharged. 77 Primary Batteries • 3 types: Dry, Alkaline, Button (Hg, Li) • The dry cell produces H2 gas • The battery has a zinc anode and reduces ammonium ion to ammonia and hydrogen gas at a graphite cathode. • The net-reaction does not contain the hydrogen gas since MnO2 is added to oxidize the hydrogen gas to hydrogen ion. • This battery has a voltage of 1.5 Volts when new. 78 Dry Cell Battery Common dry cell 79 Anode (-) Zn ---> Zn2+ + 2eCathode (+) 2 NH4+ + 2e- ---> 2 NH3 + H2 1.5 Volts 79 Problem alkaline & button: gasses formed cathode 2NH + 2e 2NH3 (g ) + H2 (g ) anode Zn Zn + 4 - 2+ + 2e - The gasses NH3 (g ) + H2 (g ) are removed: Zn 2+ + 2NH 3 (g ) + 2Cl Zn(NH 3 ) 2Cl 2 ( s) - and 2MnO 2 ( s ) + H 2 (g ) Mn 2O 3 (s) + H 2O(l ) 80 Primary Batteries: No H2 Gas produced • The alkaline battery has a zinc anode and a graphite cathode, the MnO2 is reduced directly to Mn2O3 in a basic environment. • It's voltage is 1.54 Volts. • The mercury and lithium batteries are button shaped and have voltages of 1.35 V and 3.0 V respectively. • 50% stronger then a dry cell 81 Alkaline Battery Nearly same reactions as in common dry cell, but under basic conditions (1.54V) 82 Mercury Battery Anode: Zn is reducing agent under basic conditions Cathode: HgO + H2O + 2e- ---> Hg + 2 OH- Primary, Hg (1.35V, Li (3.0V) 83 Note OH-, either KOH or NaOH cathode anode Zn(s) + 2OH ZnO + 2H 2O(l ) + 2e - - 2MnO2 + H 2O(l ) + 2e- MnO2 + 2OH - Also note that Zn here is the cathode, it is the anode in a dry cell 84 At anode Primary 85 Primary: MnO replaced with-Hg/Li (1.35V, Li (3.0V)86 Secondary Batteries • The net-reaction turns lead and lead(IV)oxide, in the presence of sulfuric acid, into lead(II)sulfate and water • The nickel-cadmium battery is a rechargeable battery. Fuel Cells • Fuel cells convert hydrogen and oxygen gas into water and electrical energy without combustion. 87 88 Lead Storage Battery Anode (-) Eo = +0.36 V Pb + HSO4- ---> PbSO4 + H+ + 2eCathode (+) Eo = +1.68 V PbO2 + HSO4- + 3 H+ + 2e- ---> PbSO4 + 2 H2O 89 Ni-Cad Battery Anode (-) Cd + 2 OH- ---> Cd(OH)2 + 2eCathode (+) NiO(OH) + H2O + e- ---> Ni(OH)2 + OH- 90 CORROSION: REDOX REACTIONS IN THE ENVIRONMENT • Corrosion of iron produces iron(III)oxide and other iron oxides depending on the conditions, moisture and oxygen levels. • Other active metals also corrode and form oxides, but frequently this is in the form of a stable oxide coating on the metal surface. • The oxides of iron flake off easily, exposing fresh iron surfaces for additional oxidation. 91 CORROSION: REDOX REACTIONS IN THE ENVIRONMENT • These processes are true electrochemical processes and a cathode and an anode can be identified. • The presence of conducting ions enhances the process. (Salt water spray, salt water from ice suppression, etc.) • One protection process is called anodic inhibition and involves changing the surface of the metal. • Paint and other coatings are examples. 92 CORROSION • A more effective method is cathodic protection which places the metal in electrical contact with a more active metal like magnesium or zinc. • The latter with iron is called galvanization and the former with iron is called a sacrificial anode. 93 Cathodic Protection 94 Fe corrosion to rust Fe2O3 95 Why Aluminum does not rust 96 20.8 ELECTROLYSIS: CHEMICAL CHANGE FROM ELECTRIC ENERGY • When a reaction is driven by an external electrical energy source the process is called electrolysis. • The anode and cathode of such cells are the sites of oxidation and reduction as in a galvanic cell, but the polarity of the electrodes is reversed. • The anode is now (+) and the cathode (-) . 97 ELECTROLYSIS • For the electrolysis of molten NaCl, Na+ is reduced to Na, and Cl- is oxidized to Cl2. • In aqueous solution, H2O is reduced to H2 and OH- and Cl- is oxidized to Cl2. • These decisions are made by consulting the standard electrode potential table to determine which species are most easily oxidized and reduced. 98 ELECTROLYSIS • The oxidation and reduction of water can be a little tricky since the table is a Standard electrode potential table which means the all reactants and products are present at standard concentrations, 1M and 1 atm. • Reduction of water to form H2(g) occurs at - 0.414 V when concentration is taken into account, and oxidation of water to form O2(g) occurs at - 0.816 V. • Over-voltages for hydrogen and oxygen gases also complicate matters. 99 Electrolysis of Aqueous NaOH Electric Energy ----> Chemical Change Anode (+) Eo = - 0.40 V 4 OH- ---> O2(g) + 2 H2O + 2eCathode (-) Eo = - 0.83 V 4 H2O + 4e- ---> 2 H2 + 4 OHEo for cell = - 1.23 V 100 Electrolysis Electric Energy ---> Chemical Change electrons BATTERY + Anode Cathode Cl- Na + Electrolysis of molten NaCl. Here a battery “pumps” electrons from Cl- to Na+. 101 Electrolysis of Molten NaCl Anode (+) Eo = - 1.36 V 2 Cl- ---> Cl2(g) + 2eEo = - 2.71 V electrons Cathode (-) BATTERY Na+ + e- ---> Na + Anode Cathode Cl- Na + Eo for cell = - 4.07 V External energy needed because Eo is (-). Note that signs of electrodes are reversed from batteries. 102 Simpler process for making Na(s)? • We just saw the molten process of taking NaCl and by applying a voltage (electrolysis) we can make both Na(s) and Cl2(g) • Wouldn’t it be easier to take seawater, perform electrolysis on the “brine” to get Na(s) and Cl2(g) 103 What are the possible reactions At the anode Na+ + e- Na(s) 2H2O(l) + 2e- H2(g) + 2OH- E = -2.714 E = -0.8277 At the cathode 2Cl- Cl2 + 2e2H2O(l) O2(g) + 4H+ + 4e- E = -1.360 E = -1.229 Once a potential is applied the process with the lowest potential will start FIRST! 104 Once a potential is applied the process with the lowest potential will start FIRST! At the anode Na+ + e- Na(s) 2H2O(l) + 2e- H2(g) + 2OH- E = -2.714 E = -0.8277 At the cathode 2Cl- Cl2 + 2e2H2O(l) O2(g) + 4H+ + 4e- E = -1.360 E = -1.229 The process with the lowest potential: 4H2O(l) + 4e- 2H2(g) + 4OHE = -0.8277 2H2O(l) O2(g) + 4H+ + 4eE = -1.229 2H2O(l) O2(g) + 2H2 Ecell = -2.057V 105 Electrolysis of Aqueous NaCl Since Na(s) can not be made from seawater due to hydrogen and oxygen gas are made another technique is used on sea-water 106 “Brine Process” for making NaOH from sea-water 107 Electrolysis of Aqueous NaCl Cells like these are the source of NaOH and Cl2. In 1995 25.1 x 109 lb Cl2 26.1 x 109 lb NaOH Also the source of NaOCl for use in bleach. 108 Electrolysis of Aqueous CuCl2 Anode (+) Eo = - 1.36 V 2 Cl- ---> Cl2(g) + 2eCathode (-) Eo = + 0.34 V Cu2+ + 2e- ---> Cu electrons Eo for cell = - 1.02 V BATTERY Note that Cu+2 is more easily reduced than either H2O or Na+. The water process potential: 4H2O(l) + 4e- 2H2(g) + 4OH2H2O(l) O2(g) + 4H+ + 4e2H2O(l) O2(g) + 2H2 + Anode E = -0.8277 E = -1.229 Ecell = -2.057V Cathode Cl- Cu2+ H2O 109 THE COMMERCIAL PRODUCTION OF CHEMICALS BY ELECTROCHEMICAL METHODS • Aluminum • Chlorine, sodium hydroxide, and hydrogen • These chemicals are extremely important to the economy. • The power required for these processes can be enormous. 110 111 Producing Aluminum 2 Al2O3 + 3 C ---> 4 Al + 3 CO2 Charles Hall (1863-1914) at age of 19 developed the Al electrolysis process. Founded Alcoa! 112 COMMERCIAL PRODUCTION OF CHEMICALS • Power is measured in watts and has basic units of joules/second. • Power is calculated from the equation P = V I, (J/C) (C/S) = J/S • A kilowatt-hour is an energy unit and is equal to 3.60x106J: 3600 S/hr times 1000 W/kW 113 20.8 COUNTING ELECTRONS • Current, I, is measured in amperes and has basic units of coulombs/second. • Faraday's Law tells us that the current times the time can be used to calculate the mass of material oxidized or reduced. • F, Faraday's constant, is 9.65x104C/mol e-. • These unit factor problems change: current x time ==> coulombs ==> moles e- ==> moles oxidized or reduced species ==> mass 114 COUNTING ELECTRONS • The balanced half-reaction provides the conversion from moles electrons to moles of species sought. • Calculate the grams of aluminum formed at the cathode if a current of 4.40 A flows for 1.50 hours. –Let’s do an example first. 115 Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag+ (aq) + e- ---> Ag(s) 1 mol e---> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-? 116 Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag+ (aq) + e- ---> Ag(s) 1 mol e---> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-? charge passing Current = time I (amps) = coulombs seconds 117 Quantitative Aspects of Electrochemistry Current = charge passing time I (amps) = coulombs seconds But how is charge related to moles of electrons? Charge on 1 mol of e- = (1.60 x 10-19 C/e-)(6.02 x 1023 e-/mol) = 96,500 C/mol e- = 1 Faraday 118 Quantitative Aspects of Electrochemistry I (amps) = coulombs seconds 1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a) Calculate the charge Coulombs = amps x time = (1.5 amps)(15.0 min)(60 s/min) = 1350 C 119 Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a) Charge = 1350 C (b) Calculate moles of e- used 1 mol e1350 C • 96,500 C = 0.0140 mol e- (c) Calculate quantity of Ag 0.0140 mol e- • 1 mol Ag = 0.0140 mol Ag or 1.51 g Ag 1 mol e 120 Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eIf a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a) 454 g Pb = 2.19 mol Pb b) Calculate moles of e- c) 2 mol e2.19molPb • = 4.38 mol e1 mol Pb Calculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C 121 Quantitative Aspects of Electrochemistry a) b) c) Solution 454 g Pb = 2.19 mol Pb Mol of e- = 4.38 mol Charge = 423,000 C d) Calculate time Time(s) = Charge (C) I (amps) 423,000 C Time(s) = 282,000 s = 1.50 amp About 78 hours 122 COUNTING ELECTRONS • Remember: –The balanced half-reaction provides the conversion from moles electrons to moles of species sought. Now solve the problem: • Calculate the grams of aluminum formed at the cathode if a current of 4.40 A flows for 1.50 hours. 123 BALANCING REDOX EQUATIONS The Half-Reaction Method • Separate the equation into halfreactions. • Balance the half-reactions. • Combine the half-reactions to form a balanced equation containing no electrons. 124 Balancing Half-Reactions • First balance the element changing oxidation state. • Balance the oxygen atoms with water. • Balance the hydrogen atoms with H+. • Balance the charge with electrons. After combining the half-reactions, check for mass and charge balance. 125 Practice Problems Balance the following equation: MnO4- + H2SO3 ----> Mn+2 + SO4-2 1. Separate into half reactions: MnO4- ----> Mn+2 H2SO3 ----> SO4-2 2. Balance atoms: 8 H+ + MnO4- ----> Mn+2 + 4 H2O H2O + H2SO3 ----> SO4-2 + 4 H+ 126 Practice Problems 3. Balance charges: 5 e- + 8 H+ + MnO4-----> Mn+2 + 4 H2O H2O + H2SO3 ----> SO4-2 + 4 H+ + 2 e4. Equal electrons gained and lost: 10 16 2 2 8 2(5 e- + 8 H+ + MnO4- ----> Mn+2 + 4 H2O) 5 5 5 20 10 5(H2O + H2SO3 ----> SO4-2+ 4 H++ 2 e-) 127 Practice Problems 5. Simplify and Add: 10 16 2 2 8 3 2(5 e- + 8 H+ + MnO4- ----> Mn+2 + 4 H2O) 5 5 5 20 4 10 5(H2O + H2SO3 ----> SO4-2 + 4 H++2 e-) 2MnO4- + 5 H2SO3 ----> 2 Mn+2 + 5 SO4-2 + 4 H+ + 3 H2O 128 Practice Problems Balance the following equation: Al + NO3- ---> Al(OH)4- + NH3 1. Separate into half reactions: Al ----> Al(OH)4NO3- ----> NH3 2. Balance atoms: 4 H2O + Al ----> Al(OH)4- + 4 H+ 9 H+ + NO3- ----> NH3 + 3 H2O 129 Practice Problems 3. Balance charges: 4 H2O + Al ----> Al(OH)4- + 4 + +- 3 e+ H 8 e + 9 H + NO3 ----> NH3 + 3 H2O 4. Equal electrons gained and lost: 32 8 8 32 24 8(4 H2O + Al ----> Al(OH)4- + 4 H+ -) + 3 e 24 27 3 3 9 3(8 e- + 9 H+ + NO3- ----> NH3 + 3 H2O) 130 Practice Problems 5. Simplify and Add: 32 23 8 8(4 H2O + Al 24 27 3 H+ + 3(8 e- + 9 H2O) 23 H2O + 8 Al 32 5 24 ----> Al(OH)4- + 4 H+ + 3 e3-) 9 NO3- ----> NH3 + 3 + 3 NO3- ----> 8 Al(OH)4- + 8 5 H+ + 3 NH3 131 Practice Problems 18 6. Change to basic solution: 23 H2O + 8 Al + 3 NO3- ----> + - + 3 NH +55 H OH 3 8 Al(OH)4- + + 5 OH- 5 H2O 18 H2O + 5 OH- + 8 Al + 3 NO3- ---->8 Al(OH)4- + 3 NH3 132 Practice Problems 1. MnO2 + HBr --> Br2 + MnBr2 2e- + 4 H+ + MnO2 ----> Mn2+ + 2 H2O 2 Br- ----> Br2 + 2e4 H+ + MnO2 + 2 Br- ----> Mn2+ + 2 H2O + Br2 133 Practice Problems 2. Cl2 + NaBr --> NaCl + Br2 2e- + Cl2 --> 2 Cl2 Br --> Br2 + 2eCl2 + 2 Br- ----> 2 Cl- + Br2 134 Practice Problems 3. H2S + 3 3(H2S 6 82 HNO3 --> S 3 + 6 NO 6 ----> S + 2 H+ + 2 e-) 2 H+ + 2 NO3- 4 2(3 e- + 4 ----> NO + 2 3 HH + 2 H+ + 2 NO3- ----> 3 S + 2 NO + 4 H2O 2S2O) 135 Practice Problems 4. PbO2 6 6 + Sb --> PbO + NaSbO2 (base) 3 3 3 3(2 e- + 2H+ + PbO2 ----> PbO + 4 1 2 H2O)2 82 6 2(2 H2O + Sb ----> SbO2- + 4 + + 3 e-) H 3 PbO2 + H2O + 2 Sb --->3 PbO + 2 1 +- 2 OH- + + 2 OH 2 H2 O SbO2 + 2 H 3PbO2 + 2 OH- + 2Sb --->3PbO + - 136