Presentation on theme: "Oxidation - Reduction Reactions"— Presentation transcript:
1 Oxidation - Reduction Reactions Redox Terminology—REVIEW CHAPTER 4!!!redox reaction -- electron transfer processe.g., 2 Na + Cl2 --> 2 NaClOverall process involves two Half Reactions:oxidation -- loss of electron(s)reduction -- gain of electron(s)e.g. Na --> Na+ + e– (oxidation)Cl e– --> 2 Cl– (reduction)Related terms:oxidizing agent = the substance that gets reduced (Cl2)reducing agent = the substance that gets oxidized (Na)Oxidation and reduction always occur together so that there is no net loss or gain of electrons overall.OILRIG
2 Redox ReactionsThe transfer of electrons between species, meaning that one species is oxidized and one is reduced. The two processes will always occur together.When has a redox reaction occurred?If there is a change in the oxidation state of any element in the reaction, a redox reaction has happened.Remember that if something is oxidized, something else must be reduced! (and vice-versa)More examples:2 Ca(s) + O2(g) --> 2 CaO(s)2 Al(s) + 3 Cl2(g) --> 2 AlCl3(s)
3 HALF-REACTION METHOD for balancing Redox Equations 1. Write unbalanced ionic equations for the two half-reactions. (Look at oxidation numbers in the “skeleton equation.”)2. Balance atoms other than H and O.3. Add appropriate number of electrons.4. Balance O with H2O.5. Balance H with H+.6. If in acidic solution, then skip to step 7. If in basic solution, then add equal number of OH– to both sides to cancel all of the H+.7. Multiply balanced half-reactions by appropriate coefficients so that the number of electrons are equal.8. Rewrite the balanced half reactions.9. Add the half-reactions together.10. Cancel species that appear on both sides to get the balanced Net Ionic Equation.11. If necessary, add spectator ions to get the balanced molecular equation.Check the Final Balance (atoms and charges)!
4 Example Problem Balance the following redox equation. Ce4+(aq) + Sn2+(aq) --> Ce3+(aq) + Sn4+(aq)The reaction can be separated into a reaction involving the substance being reduced;Ce4+(aq) + e– --> Ce3+(aq)And the substance being oxidized;Sn2+(aq) --> Sn4+(aq) + 2e–We can see that the equations don’t balance; you must multiply the top equation by two, then add them to get2 Ce4+(aq) + Sn2+(aq) --> 2 Ce3+(aq) + Sn4+(aq)
5 Sample ProblemThe following redox reaction occurs in basic solution. Write complete, balanced equations for the individual half-reactions and the overall net ionic equation.MnO4–(aq) + N2H4(aq) --> MnO2(s) + NO(g)Oxidation Half Reaction:Reduction Half Reaction:Net Ionic Equation:
6 Sample ProblemThe following redox reaction occurs in basic solution. Write complete, balanced equations for the individual half-reactions and the overall net ionic equation.MnO4–(aq) + N2H4(aq) --> MnO2(s) + NO(g)Oxidation Half Reaction:8 OH–(aq) + N2H4(aq) --> 2 NO(g) H2O(l) e–Reduction Half Reaction:3 e– + 2 H2O(l) + MnO4–(aq) --> MnO2(s) OH–(aq)Net Ionic Equation:3 N2H4(aq) MnO4–(aq) --> 6 NO(g) MnO2(s) H2O(l) +8 OH–(aq)
7 Sample ProblemsBalance the following redox equations in acidic solution.Al(s) + I2(s) --> AlI3(s)KClO3(aq) + HNO2(aq) --> KCl(aq) + HNO3(aq)Cl–(aq) + MnO2(s) --> Mn2+(aq) + Cl2(g)Balance the above redox equations in basic solution.
8 Sample ProblemsBalance the following redox equations in acidic solution.Al(s) + I2(s) --> AlI3(s)2 Al(s) + 3 I2(s) --> 2 AlI3(s)KClO3(aq) + HNO2(aq) --> KCl(aq) + HNO3(aq)KClO3(aq) + 3 HNO2(aq) --> KCl(aq) + 3 HNO3(aq)Cl–(aq) + MnO2(s) --> Mn2+(aq) + Cl2(g)4H+(aq) + 2 Cl–(aq) + MnO2(s) --> Cl2(g) + Mn2+(aq) + 2 H2O(l)Balance the above redox equations in basic solution.First doesn’t change.Second: use NO2– and NO3– rather than the acids.Third: Add 4 OH- to each side,2 H2O(l) + 2 Cl–(aq) + MnO2(s) --> Cl2(g) + Mn2+(aq) + 4 OH–(aq)
9 Electrochemistry Terminology Electrochemical Cell a device that converts electrical energy into chemical energy or vice versaTwo Types:Electrolytic cellConverts electrical energy into chemical energyElectricity is used to drive a non-spontaneous reactionGalvanic (or voltaic) cellConverts chemical energy into electricity (a battery!)A spontaneous reaction produces electricityConduction:Metals: metallic (electronic) conduction free movement of electronsSolutions: electrolytic (ionic) conduction(or molten salts) -- free movement of ions
10 Galvanic Cells Galvanic Cells (batteries) -- produce electrical energy e.g. a spontaneous reaction:Cu(s) + Ag+(aq) --> Cu2+(aq) + Ag(s)(Ag metal will be deposited on a Cu wire dipped into aqueous AgNO3 solution)In a galvanic cell, the half reactions are occurring in separate compartments (half-cells)
12 Cell Potential Cell potential ~ Eºcell (an electromotive force, emf) Units of Eºcell are volts: 1 volt = 1 joule/coulombEºcell is a measure of the relative spontaneity of a cell reactionPositive (+) Eºcell --> spontaneous reactionEºcell depends on:Nature of reactantsTemperature -- superscript º means 25 ºCConcentrations -- superscript º means all conc are at M and gases are at 1.00 atmbut, Eºcell is independent of amounts of reactants
13 Standard Potentials Standard Reduction Potential: the potential of a half-cell relative to a standard reference(Pt)Sample Problem; Write the cell notation for this cell.
14 Standard Cell Potential Standard Reduction Potentials Eº (volts)F2(g) e– --> 2 F–(aq)Ag+(aq) + e– --> Ag(s)Cu2+(aq) e– --> Cu(s)2 H+(aq) e– --> H2(g)Zn2+(aq) + 2 e– --> Zn(s) – 0.76Li+(aq) + e– --> Li(s) – 3.05Ease of Reduction ~ increases with Eºe.g. F2 is easiest to reduce, Li+ is the hardestStandard Cell Potential ~ Eºcell can be determined from standard reduction potentials:Eºcell = Eºred – Eºoxid= [reduction potential of substance reduced]– [reduction potential of substance oxidized]
15 Standard Potential Example Problem Is the following a galvanic or an electrolytic cell? Write the balanced cell reaction and calculate Eºcell.Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)Cathode: Cu e– --> Cu Eº = 0.34 VAnode: Zn --> Zn e– Eº = – 0.76 VCell rxn: Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)Eºcell = Eºred – Eºox = V – (– 0.76 V)= V ∴ galvanic.
16 Another Example Problem Given the standard reduction potentials:O2(g) H+(aq) e– --> 2 H2O(l) Eº = 1.23 VCl2(g) e– --> 2 Cl–(aq) Eº = 1.36 VWrite a balanced equation and calculate Eºcell for a galvanic cell based on these half reactions. Sketch the galvanic cell.“galvanic” implies a positive value for Eºcellso, Cl2/Cl– should be the reduction half reaction,since 1.36 – 1.23 = V∴ reverse the O2 half reaction and make it the oxidationMultiply the Cl2 reaction by 2, to make e–’s cancel, hence:2 Cl2(g) + 2 H2O(l) --> 4 Cl–(aq) + O2(g) H+(aq)Note: When a half-reaction is multiplied by a coefficient, the Eº IS NOT MULTIPLIED by the coefficient.
17 Diagram of cell e– e– voltmeter Pt electrode Pt electrode cathode salt bridge(reduction)O2 gasCl2 gasK+NO3–K+K+Cl–anodeCl–K+Cl–(oxidation)NO3–K+NO3–K+supporting electrolyte
18 Sample ProblemUnder standard conditions, is the following a galvanic or an electrolytic cell? Support your conclusion with appropriate calculations and balanced chemical reactions. Ag(s), AgBr(s) | Br–(aq) || Au3+(aq) | Au(s) --From table, Au3+(aq) + 3e – Au(s) Eº = 1.50 V AgBr(s) + e – Ag(s) + Br –(aq) Eº = V
19 Sample ProblemUnder standard conditions, is the following a galvanic or an electrolytic cell? Support your conclusion with appropriate calculations and balanced chemical reactions. Ag(s), AgBr(s) | Br–(aq) || Au3+(aq) | Au(s) --From table, Au3+(aq) + 3e – Au(s) Eº = 1.50 V AgBr(s) + e – Ag(s) + Br –(aq) Eº = V 3 Ag(s) + 3 Br–(aq) + Au3+(aq) --> 3 AgBr(s) + Au(s), Eºcell = V, so galvanic.
20 Reactions of Metals with Acids Generally: metal + acid --> salt + H2(g)e.g. Zn(s) + H2SO4(aq) --> ZnSO4(aq) + H2(g)Metal is oxidized to yield the metal cationH+ from the acid is reduced to yield H2Some acids (especially HNO3) are “oxidizing acids” -- the NO3– ion is reduced to form NO2 or NO depending on concentration of the acid.e.g. reactions of Cu metal with HNO3Cu(s) + concentrated HNO3 --> NO2(g)Cu(s) + dilute HNO3 --> NO(g)
23 Cell Potential and Thermodynamics Free Energy Change (DG)DG = – Wmax (maximum work)For an electrochemical cell:W = nFEcell where n = # moles e–F = 96,500 coulombs/mole e–jouleUnits: volt =coulombcoulombsjoulejoules = (moles e–)mole e–coulomb∴ free energy is related to Ecell as follows:DG = – nFEcellunder standard conditions: use DGº and Eºcell
24 Equilibrium Constant RT nF 0.0592 n since DGº = – RTlnKc = – nFEºcell can rearrange to:Eºcell = ln Kcat 25 ºC, use values for R, T, and F, then convert to base-10 log:Eºcell = log KcRT(Use J mol-1 K-1 for R!)nF0.0592nOverall,
25 Sig Fig Review: Log, LnRules are the same for log and ln! When you report the log or ln of a number, the significant digits are the ones after the decimal point, e.g. ln (24.3) = [3 sig fig!] log (0.068) = –1.17 [2 sig fig!] Antilog is the opposite; when you convert back to natural numbers, you will “appear” to lose significant digits, e.g. Antilog (23.32) = = 2.1 x 1023 [2 sig fig!] Anti(natural log) of = e23.32 = 1.3 x 1010 [2 sig fig!]
26 Sample ProblemGiven the following standard reduction potentials, calculate the solubility product constant (Ksp) for lead sulfate, PbSO4.PbSO4(s) + 2 e– --> Pb(s) + SO42–(aq) Eº = – VPb2+(aq) + 2 e– --> Pb(s) Eº = – V
27 Sample ProblemGiven the following standard reduction potentials, calculate the solubility product constant (Ksp) for lead sulfate, PbSO4.PbSO4(s) + 2 e– --> Pb(s) + SO42–(aq) Eº = – VPb2+(aq) + 2 e– --> Pb(s) Eº = – VKc = 2 x 10–8
28 Effect of Concentration on Cell Potential For a general reaction:aA + bB cC + dDDG = DGº + RTln Qwhere Q = “concentration quotient” or “reaction quotient”= [C]c[D]d / [A]a[B]b (use given concentrations)Since DG = – nFE :– nFE = – nFEº + RTln QRearrange to get the “Nernst Equation”E = Eº – [RT/nF]ln Q(review fromCh )
29 The Nernst Equation E = Eº – [RT/nF]ln Q The Nernst Equation shows the relationship betweenthe standard cell potential (Eº) and the cell potential(E) under actual, non-standard conditions.this can be simplified at 25 ºC to:E = Eº – (0.0592/n)log QMajor use of the Nernst Equation:Determine concentrations from standard reduction potentialsUse actual concentrations (i.e. Q) to calculate Ecell
30 Sample ProblemA silver wire coated with AgCl is sensitive to the chloride ion concentration because of the following half-cell reaction.AgCl(s) + e– --> Ag(s) + Cl–(aq) Eº = VCalculate the molar concentration of Cl– when the potential of this half-cell is measured to be volts relative to a standard hydrogen electrode.
31 Sample ProblemA silver wire coated with AgCl is sensitive to the chloride ion concentration because of the following half-cell reaction.AgCl(s) + e– --> Ag(s) + Cl–(aq) Eº = VCalculate the molar concentration of Cl– when the potential of this half-cell is measured to be volts relative to a standard hydrogen electrode.3 x 10–5 M
32 Quantitative Aspects of Electrochem ~ the Faraday e.g. consider a cell with the cathode reaction:Cr3+(aq) e– --> Cr(s)Stoichiometry: 1 mole Cr3+ ~ 3 moles e– ~ 1 mole CrIf a current equivalent to 3 moles of electrons is passed through the solution of Cr3+, then 1 mole of Cr metal will be producedBy definition: 1 Faraday (F) = 1 mole electronsbut, how much electrical current is this?1 F = 9.65 x 104 coulombsand1 coulomb = 1 amp secor, 1 amp =1 coulomb1 sec
33 Batteries Dry cells (paste) Lead storage (car batteries!) Metal anode is oxidized, solid MnO2 in paste is reducedSupporting electrolyte is eitheracid (can corrode more easily, cannot recharge), oralkaline (base, lasts longer)Lead storage (car batteries!)Lead is oxidized, PbO2 is reducedAcid electrolyte (H2SO4)Rechargeable, heavy
34 Light-Weight Rechargeable Batteries NiCADCadmium is oxidized, NiO2 is reducedSupporting electrolyte is KOHNiMHHydrogen is oxidized (stored as metal hydride), NiO2 is reducedLithium ion“Concentration” cellMigration of lithium ions from graphite anode to metal oxide cathode
35 Fuel Cellslike batteries in which reactants are constantly being addedso it never runs down!Pt electrodes, KOH electrolyteSimpler diagram
36 Electrolysis-- chemical reactions that occur during electrolytic conductione.g. molten NaClreduction:Na+(l) + e– --> Na(s)Na+ ions migrate towardthe negative electrodeand are reducedoxidation:2 Cl–(l) --> Cl2(g) + 2e–Cl– ions migrate towardthe positive electrodeand are oxidized
37 Electrolysis Cell Electrodes: cathode -- where reduction occurs anode -- where oxidation occursNet Cell Reaction:add anode & cathode half-reactions so that # of electrons cancel2 Cl– --> Cl e– (anode ~ oxidation)2 [ Na+ + e– --> Na ] (cathode ~ reduction)2 Cl–(l) Na+(l) --> Cl2(g) Na(s) (Cell Reaction)Overvoltage, or overpotential — in some cases additional voltage must be applied in order to get an electrolytic reaction to occur (kinetic reasons)
38 Electrolysis of WaterIn aqueous solution electrolysis of water may occur:oxidation (H2O --> O2) -- at anode2 H2O --> O H+ + 4e– Eº = 1.23 Vreduction (H2O --> H2) -- at cathodein neutral or basic solution2 H2O + 2 e– --> H OH– Eº = Vreduction (H+ --> H2) -- at cathodein acidic solution2 H+ + 2 e– --> H2 Eº = 0.00 VHigher Eº easier to reduce!Lower Eº easier to oxidize!
39 Some Industrial Applications of Electrolysis Preparation of Aluminum from molten Al2O3(can’t use Al3+ in solution since H2O is easier to reduce)anode: 3 [O2– --> 1/2 O e– ]cathode: 2 [ Al e– --> Al ] (Eº = –1.66 V)Cell: 2 Al3+(l) O2–(l) --> 3/2 O2(g) Al(s)Electroplating reduction of metal ions (from solution) to pure metals (Ag, Cr, Cu, etc.)e.g.Cr3+(aq) e– --> Cr(s) (chrome plating)
40 Example ProblemHow long (in hours) would it take to produce 25 g of Cr from a solution of Cr3+ by a current of 2.75 amp?Cr3+(aq) e– --> Cr(s)1 mole Cr ~ 3 moles e– ~ 3 F ~ 3 x 9.65 x 104 coulombs1 mole Cr3 F25 g Cr xx= F52 g Cr1 mole Cr96,500 coulombs1.44 F x= 139,000 coulombs= 139,000 amp secF139,000 amp sec1 hrx= 14 hr2.75 amp3600 sec
41 Review Problems1. A large electrolytic cell that produces metallic aluminum from Al2O3 ore is capable of making 250 kg of aluminum in 24 hours. Determine the current (in amps) that is required for this process. Include appropriate chemical reactions.2. An aqueous solution of NaBr was electrolyzed with a current of 2.50 amps for 15.0 minutes. What volume (in mL) of M HCl would be required to neutralize the resulting solution? (Hint: H2 is produced at the cathode and Br2 at the anode.)
42 Review Problems1. A large electrolytic cell that produces metallic aluminum from Al2O3 ore is capable of making 250 kg of aluminum in 24 hours. Determine the current (in amps) that is required for this process. Include appropriate chemical reactions.3.1 x 104 amps (Al3+(aq) e– --> Al(s) is ½-reaction, overall reaction is 2 Al2O3(l) 3 O2(g) + 4 Al(s))2. An aqueous solution of NaBr was electrolyzed with a current of 2.50 amps for 15.0 minutes. What volume (in mL) of M HCl would be required to neutralize the resulting solution? (Hint: H2 is produced at the cathode and Br2 at the anode.)46.6 mL HCl soln
43 Review Problems, cont.3. Consider the following electrochemical cell in which the volume of solution in each half-cell is 100 mL.Zn(s) | Zn2+ (1.00 M) || Ag+ (1.00 M) | Ag(s)(a) Write balanced chemical equations for the anode, cathode, and overall cell reactions.(b) Determine Eºcell, DGº, and the equilibrium constant (Kc) for the cell reaction.(c) If current is drawn from this cell at a constant rate of 0.10 amp, what will the cell potential be after 15.0 hours?
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