2Review Oxidation reduction reactions involve a transfer of electrons. OIL- RIGOxidation Involves LossReduction Involves GainLEO-GERLose Electrons OxidationGain Electrons Reduction
3Solid lead(II) sulfide reacts with oxygen in the air at high temperatures to form lead(II) oxide and sulfur dioxide. Which substance is a reductant (reducing agent) and which is an oxidant (oxidizing agent)? PbS, reductant; O2, oxidant PbS, reductant; SO2, oxidant Pb2+, reductant; S2- oxidant PbS, reductant; no oxidant PbS, oxidant; SO2, reductant
4Applications Moving electrons is electric current. 8H++MnO4-+ 5Fe+2 +5e ® Mn+2 + 5Fe+3 +4H2OHelps to break the reactions into half reactions.8H++MnO4-+5e- ® Mn+2 +4H2O5(Fe+2 ® Fe+3 + e- )In the same mixture it happens without doing useful work, but if separate
5Connected this way the reaction starts Stops immediately because charge builds up.e-e-e-e-e-H+MnO4-Fe+2
6Galvanic CellSalt Bridge allows current to flowH+MnO4-Fe+2
7Electricity travels in a complete circuit H+MnO4-Fe+2
10Cell Potential Oxidizing agent pulls the electron. Reducing agent pushes the electron.The push or pull (“driving force”) is called the cell potential EcellAlso called the electromotive force (emf)Unit is the volt(V)= 1 joule of work/coulomb of chargeMeasured with a voltmeter
110.76H2 inCathodeAnodeH+ Cl-Zn+2 SO4-21 M ZnSO41 M HCl
12Standard Hydrogen Electrode This is the reference all other oxidations are compared toEº = 0º indicates standard states of 25ºC, 1 atm, 1 M solutions.H2 inH+ Cl-1 M HCl
13Eºcell = EºZn® Zn+2 + EºCu+2 ® Cu Cell PotentialZn(s) + Cu+2 (aq) ® Zn+2(aq) + Cu(s)The total cell potential is the sum of the potential at each electrode.Eºcell = EºZn® Zn+2 + EºCu+2 ® CuWe can look up reduction potentials in a table.One of the reactions must be reversed, so change it sign.
14Cell PotentialDetermine the cell potential for a galvanic cell based on the redox reaction.Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq)Fe+3(aq) + e-® Fe+2(aq) Eº = 0.77 VCu+2(aq)+2e- ® Cu(s) Eº = 0.34 VCu(s) ® Cu+2(aq)+2e Eº = V2Fe+3(aq) + 2e-® 2Fe+2(aq) Eº = 0.77 V
15Reduction potential More negative Eº more easily electron is added More easily reducedBetter oxidizing agentMore positive Eºmore easily electron is lostMore easily oxidizedBetter reducing agent
16Line Notation solid½Aqueous½½Aqueous½solid Anode on the left½½Cathode on the rightSingle line different phases.Double line porous disk or salt bridge.If all the substances on one side are aqueous, a platinum electrode is indicated.
17For the last reaction Fe+2 Cu2+ Cu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s)Cu2+Fe+2
18In a galvanic cell, the electrode that acts as a source of electrons to the solution is called the __________; the chemical change that occurs at this electrode is called________. a. cathode, oxidation b. anode, reduction c. anode, oxidation d. cathode, reduction
19Under standard conditions, which of the following is the net reaction that occurs in the cell? Cd|Cd2+ || Cu2+|Cu a. Cu2+ + Cd → Cu + Cd2+ b. Cu + Cd → Cu2+ + Cd2+ c. Cu2+ + Cd2+ → Cu + Cd d. Cu + Cd 2+ → Cd + Cu2+
20Galvanic CellThe reaction always runs spontaneously in the direction that produced a positive cell potential.Four things for a complete description.Cell PotentialDirection of flowDesignation of anode and cathodeNature of all the components- electrodes and ions
21PracticeCompletely describe the galvanic cell based on the following half-reactions under standard conditions.MnO H+ +5e- ® Mn+2 + 4H2O Eº=1.51 VFe+3 +3e- ® Fe(s) Eº=0.036V
22Potential, Work and DG E = work done by system / charge E = -w/q emf = potential (V) = work (J) / Charge(C)E = work done by system / chargeE = -w/qCharge is measured in coulombs.-w = q EFaraday = 96,485 C/mol e-q = nF = moles of e- x charge/mole e-w = -qE = -nFE = DG
23Potential, Work and DG DGº = -nFEº if Eº > 0, then DGº < 0 spontaneousif Eº< 0, then DGº > 0 nonspontaneousIn fact, reverse is spontaneous.Calculate DGº for the following reaction:Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq)Fe+2(aq) + e-® Fe(s) Eº = 0.44 VCu+2(aq)+2e- ® Cu(s) Eº = 0.34 V
24Cell Potential and Concentration Qualitatively - Can predict direction of change in E from LeChâtelier.2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)Predict if Ecell will be greater or less than Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 Mif [Al+3] = 1.0 M and [Mn+2] = 1.5Mif [Al+3] = 1.5 M and [Mn+2] = 1.5 M
25E = Eº - RTln(Q) nF The Nernst Equation DG = DGº +RTln(Q) -nFE = -nFEº + RTln(Q)E = Eº - RTln(Q) nF2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) Eº = 0.48 VAlways have to figure out n by balancing.If concentration can gives voltage, then from voltage we can tell concentration.
26The Nernst Equation Eº = RTln(K) nF As reactions proceed concentrations of products increase and reactants decrease.Reach equilibrium where Q = K and Ecell = 00 = Eº - RTln(K) nFEº = RTln(K) nFnF Eº = ln(K) RT
27Batteries are Galvanic Cells Car batteries are lead storage batteries.Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O
29Batteries are Galvanic Cells Alkaline Zn +MnO2 ® ZnO+ Mn2O3 (in base)
30Batteries are Galvanic Cells NiCadNiO2 + Cd + 2H2O ® Cd(OH)2 +Ni(OH)2
31Corrosion Rusting - spontaneous oxidation. Most structural metals have reduction potentials that are less positive than O2 .Fe ® Fe+2 +2e- Eº= 0.44 VO2 + 2H2O + 4e- ® 4OH- Eº= 0.40 VFe+2 + O2 + H2O ® Fe2O3 + H+Reactions happens in two places.
32Salt speeds up process by increasing conductivity WaterRustFe2+O2 + 2H2O +4e- ® 4OH-e-Iron Dissolves- Fe ® Fe+2Fe2+ + O2 + 2H2O ® Fe2O3 + 8 H+
33Preventing Corrosion Coating to keep out air and water. Galvanizing - Putting on a zinc coatHas a lower reduction potential, so it is more easily oxidized.Alloying with metals that form oxide coats.Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.
34Electrolysis Running a galvanic cell backwards. Put a voltage bigger than the potential and reverse the direction of the redox reaction.Used for electroplating.
36A battery >1.10Ve-e-ZnCu1.0 M Zn+21.0 M Cu+2CathodeAnode
37Calculating 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 metalHow long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+
38Calculating 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 if you want time to plate
39Calculate the mass of copper which can be deposited by the passage of 12.0 A for 25.0 min through a solution of copper(II) sulfate.
40How long would it take to plate 5 How long would it take to plate 5.00 g Fe from an aqueous solution of Fe(NO3)3 at a current of 2.00 A?
41Other uses Electrolysis of water. Separating mixtures of ions. More positive reduction potential means the reaction proceeds forward.We want the reverse.Most negative reduction potential is easiest to plate out of solution.
42Redox Know the table 2. Recognized by change in oxidation state. 3. “Added acid”4. Use the reduction potential table on the front cover.5. Redox can replace. (single replacement)
436. Combination Oxidizing agent of one element will react with the reducing agent of the same element to produce the free element.I- + IO3- + H+ ® I2 + H2O7. Decomposition.a) peroxides to oxidesb) Chlorates to chloridesc) Electrolysis into elements.d) carbonates to oxides
44Examples A piece of solid bismuth is heated strongly in oxygen. A strip or copper metal is added to a concentrated solution of sulfuric acid.Dilute hydrochloric acid is added to a solution of potassium carbonate.
45Hydrogen 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.
46ExamplesA 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.
47Magnesium metal is burned in nitrogen gas. Lead foil is immersed in silver nitrate solution.Magnesium turnings are added to a solution of iron (III) chloride.Pellets of lead are dropped into hot sulfuric acidPowdered Iron is added to a solution of iron(III) sulfate.
48A way to remember An Ox – anode is where oxidation occurs Red Cat – Reduction occurs at cathodeGalvanic cell- spontaneous- anode is negativeElectrolytic cell- voltage applied to make anode positive
50A 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.)
51A 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.)
52A 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.
53A 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)+ (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?