Presentation on theme: "Chapter 20 Electrochemistry (modified for our needs)"— Presentation transcript:
1Chapter 20 Electrochemistry (modified for our needs) Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. BurstenChapter 20 Electrochemistry (modified for our needs)John D. BookstaverSt. Charles Community CollegeSt. Peters, MO 2006, Prentice Hall, Inc.
2Resources and Activities Ch. 20ElectrochemistryOxidation StatesBalancing Oxidation-Reduction EquationsVoltaic cellsCell EMFSpontaneity of redox reactionsThe effect of concentration on EMFElectrolysisResources and ActivitiesStudents to review chapter 5 notes on redox equations and assigning oxidation numbersTextbook - chapter 20 & ppt fileOnline practice quizChemtour videos from NortonGalvanic cell (Glencoe) animation
3Activities and Problem set for chapter 20 (due date_______) Independent work - students to view animations & interactive activities (5 in total from Norton) and write summary notes on each. These summaries are to be included in your portfolio. Some of these may be previewed in class.Norton Animations :Zinc-copper cell, free energy, cell potential, alkaline battery, fuel cellOnline practice quiz ch 20 due by_____Chapter 20 practice problems packet
4Electrochemical Reactions In electrochemical reactions, electrons are transferred from one species to another. In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.
5Examples of reactions that are redox reactions (write equations) A piece of solid bismuth is heated strongly in oxygen.A strip or copper metal is added to a concentrated solution of sulfuric acid.Magnesium turnings are added to a solution of iron (III) chloride.A stream of chlorine gas is passed through a solution of cold, dilute sodium hydroxide.A solution of tin ( II ) chloride is added to an acidified solution of potassium permanganateA solution of potassium iodide is added to an acidified solution of potassium dichromate.
6Hydrogen peroxide solution is added to a solution of iron (II) sulfate. Propanol is burned completely in air.A piece of lithium metal is dropped into a container of nitrogen gas.Chlorine gas is bubbled into a solution of potassium iodide.Magnesium metal is burned in nitrogen gas.Lead foil is immersed in silver nitrate solution.Pellets of lead are dropped into hot sulfuric acidPowdered Iron is added to a solution of iron(III) sulfate.
7Combination: Oxidizing agent of one element will react with the reducing agent of the same element to produce the free element. I- + IO3- + H+ ® I2 + H2O Decomposition. a) peroxides to oxides b) Chlorates to chlorides c) Electrolysis into elements. d) carbonates to oxides
8Oxidation and Reduction A species is oxidized when it loses electrons.Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.
9Oxidation and Reduction A species is reduced when it gains electrons.Here, each of the H+ gains an electron and they combine to form H2.
10Oxidation and Reduction What is reduced is the oxidizing agent.H+ oxidizes Zn by taking electrons from it.What is oxidized is the reducing agent.Zn reduces H+ by giving it electrons.
11Assigning Oxidation Numbers Elements in their elemental form have an oxidation number of 0.The oxidation number of a monatomic ion is the same as its charge.Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.
12Assigning Oxidation Numbers Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.Fluorine always has an oxidation number of −1.The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.The sum of the oxidation numbers in a neutral compound is 0.The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
13Balancing Oxidation-Reduction Equations Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.
14Half-Reaction MethodAssign oxidation numbers to determine what is oxidized and what is reduced.Write the oxidation and reduction half-reactions.Balance each half-reaction.Balance elements other than H and O.Balance O by adding H2O.Balance H by adding H+.Balance charge by adding electrons.Multiply the half-reactions by integers so that the electrons gained and lost are the same.
15Half-Reaction MethodAdd the half-reactions, subtracting things that appear on both sides.Make sure the equation is balanced according to mass.Make sure the equation is balanced according to charge.
16MnO4−(aq) + C2O42−(aq) Mn2+(aq) + CO2(aq) Half-Reaction MethodConsider the reaction between MnO4− and C2O42− :MnO4−(aq) + C2O42−(aq) Mn2+(aq) + CO2(aq)
17Half-Reaction Method MnO4− + C2O42- Mn2+ + CO2 First, we assign oxidation numbers.MnO4− + C2O42- Mn2+ + CO2+7+3+4+2Since the manganese goes from +7 to +2, it is reduced.Since the carbon goes from +3 to +4, it is oxidized.
18Oxidation Half-Reaction C2O42− CO2To balance the carbon, we add a coefficient of 2:C2O42− 2 CO2The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.C2O42− 2 CO2 + 2 e−
19Reduction Half-Reaction MnO4− Mn2+The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.MnO4− Mn H2OTo balance the hydrogen, we add 8 H+ to the left side.8 H+ + MnO4− Mn H2O
20Reduction Half-Reaction 8 H+ + MnO4− Mn H2OTo balance the charge, we add 5 e− to the left side.5 e− + 8 H+ + MnO4− Mn H2O
21Combining the Half-Reactions Now we evaluate the two half-reactions together:C2O42− 2 CO2 + 2 e−5 e− + 8 H+ + MnO4− Mn H2OTo attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.
22Combining the Half-Reactions 5 C2O42− 10 CO e−10 e− + 16 H+ + 2 MnO4− 2 Mn H2OWhen we add these together, we get:10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 2 Mn H2O + 10 CO2 +10 e−
23Combining the Half-Reactions 10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 2 Mn H2O + 10 CO2 +10 e−The only thing that appears on both sides are the electrons. Subtracting them, we are left with:16 H+ + 2 MnO4− + 5 C2O42− 2 Mn H2O + 10 CO2
24Balancing in Basic Solution If a reaction occurs in basic solution, one can balance it as if it occurred in acid.Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.If this produces water on both sides, you might have to subtract water from each side.(Practice Problems)
25Voltaic CellsIn spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.
26Voltaic CellsWe can use that energy to do work if we make the electrons flow through an external device.We call such a setup a voltaic cell.See ANIMATIONS
27Voltaic Cells A typical cell looks like this. The oxidation occurs at the anode.The reduction occurs at the cathode.
28Voltaic CellsOnce even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.
29Voltaic CellsTherefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.Cations move toward the cathode.Anions move toward the anode.
30Voltaic CellsIn the cell, then, electrons leave the anode and flow through the wire to the cathode.As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.
31Voltaic CellsAs the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode.The electrons are taken by the cation, and the neutral metal is deposited on the cathode.
32Electromotive Force (emf) Water only spontaneously flows one way in a waterfall.Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.
33Electromotive Force (emf) The potential difference between the anode and cathode in a cell is called the electromotive force (emf).It is also called the cell potential, and is designated Ecell.Cell potential is measured in volts (V).1 V = 1JC
34Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated.
35Standard Hydrogen Electrode Their values are referenced to a standard hydrogen electrode (SHE).By definition, the reduction potential for hydrogen is 0 V:2 H+ (aq, 1M) + 2 e− H2 (g, 1 atm)
36Standard Cell Potentials The cell potential at standard conditions can be found through this equation:Ecell= Ered (cathode) − Ered (anode)Because cell potential is based on the potential energy per unit of charge, it is an intensive property.
37Cell Potentials Ered = −0.76 V Ered = +0.34 V Ecell = Ered For the oxidation in this cell,For the reduction,Ered = −0.76 VEred = VEcell=Ered(cathode) −(anode)= V − (−0.76 V)= V
38Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials.The strongest reducers have the most negative reduction potentials.
39Oxidizing and Reducing Agents The greater the difference between the two, the greater the voltage of the cell.
40Free Energy G for a redox reaction can be found by using the equation G = −nFEwhere n is the number of moles of electrons transferred, and F is a constant, the Faraday.1 F = 96,485 C/mol = 96,485 J/V-molUnder standard conditions,G = −nFE
41Nernst Equation Remember that G = G + RT ln Q This means −nFE = −nFE + RT ln Q
42Nernst EquationDividing both sides by −nF, we get the Nernst equation:E = E −RTnFln Qor, using base-10 logarithms,E = E −2.303 RTnFlog Q
43Nernst Equation At room temperature (298 K), 2.303 RT F = 0.0592 V Thus (when T = 298 K) the equation becomesE = E −0.0592nlog Q
44Concentration Cells Ecell Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.For such a cell,would be 0, but Q would not.EcellTherefore, as long as the concentrations are different, E will not be 0.
45Applications of Oxidation-Reduction Reactions Batteries
49Electrolysis (animations) Using electrical energy to drive a reaction in a non-spontaneous direction
50ElectrolysisUsing electrical energy to drive a reaction in a nonspontaneous directionUsed for electroplating, electrolysis of water, separation of a mixture of ions, etc. (Most negative reduction potential is easiest to plate out of solution.)
51Calculating plating Have to count charge. Measure current I (in amperes)1 amp = 1 coulomb of charge per secondq = I x tq/nF = moles of metalMass of plated metalFaraday Constant (F)(96,480 C/mol e-) gives the amount of charge (in coulombs that exist in 1 mole of electrons passing through a circuit.1volt = 1joule/coulomb
52Calculating plating Current x time = charge Charge ∕Faraday = mole of e-Mol of e- to mole of element or compoundMole to grams of compoundor the reverse these steps if you want to find the time to plateHow many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 2.00A is run through a solution of CuSO4 for a period of 20min?How many hours would it take to produce 75.0g ofmetallic chromium by the electrolytic reduction of Cr3+ with a current of 2.25 A?
53How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 2.00A is run through a solution of CuSO4 for a period of 20min?Answer:Cu2+(aq) + 2e- →Cu(s)2.00A = 2.00C/s and 20min (60s/min) = 1200sCoulombs of e- = (2.00C/s)(1200s) = 2400Cmol e- = (2400C)(1mol/96,480C) = .025mol(.025mol e-)(1mol Cu/2mol e-) = .0125mol Cug Cu = (.0125mol Cu)(63.55g/mol) = .79g
54How many hours would it take to produce 75.0g of metallic chromium by the electrolytic reduction of Cr3+ with a current of 2.25 A?Answer:75.0g Cr/(52.0g/mol) = 1.44mol Crmol e- = (1.44mol Cr)(3mol e-/1mol Cr) = 4.32mol e-Coulombs = (4.32mol e-)(96,480C/mol) = 416,793.6 C (4.17x105C)Seconds = (4.17x105C)/(2.25C/s) = 1.85x105 sHours = (1.85x105 s)(1hr/3600 s) = 51.5 hours
55AP ProblemA student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be volt.(a) Draw a diagram of this cell.(b) Describe what is happening at the cathode (Include any equations that may be useful.)
56AP ProblemA student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be volt.(c) Describe what is happening at the anode. (Include any equations that may be useful.)
57AP ProblemA student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be volt.(d) Write the balanced overall cell equation.(e) Write the standard cell notation.
58AP ProblemA student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be volt. (f) The student adds 4 M ammonia to the copper sulfate solution, producing the complex ion Cu(NH3)4+2 (aq). The student remeasures the cell potential and discovers the voltage to be 0.88 volt. What is the Cu2+ (aq) concentration in the cell after the ammonia has been added?