Presentation on theme: "Electric Power Chemical Reactions"— Presentation transcript:
1 Electric Power Chemical Reactions Electric power conversion in electrochemistryElectrolysis / Power consumptionElectric PowerChemical ReactionsElectrochemical battery / Power generation
2 Sign or polarity of electrodes CellConstruction-+batterypowersourcee-e-conductivemedium(-)(+)vesselinertelectrodesSign or polarity of electrodes
3 Let’s examine the electrolytic cell for molten NaCl. What chemical species would be present in a vessel of molten sodium chloride, NaCl (l)?Na+Cl-Let’s examine the electrolytic cell for molten NaCl.
4 Observe the reactions at the electrodes Molten NaClObserve the reactions at the electrodes-+batteryCl2 (g) escapesNa (l)NaCl (l)Na+Cl-Na+Cl-(-)(+)electrode half-cellelectrode half-cellCl-Na+Na+ + e- Na2Cl- Cl e-
5 At the microscopic level Molten NaClAt the microscopic level-+batterye-NaCl (l)cationsmigratetoward(-)electrodeanionsmigratetoward(+)electrodeNa+Cl-Na+e-Cl-(-)(+)anodecathodeCl-Na+2Cl- Cl e-Na+ + e- Na
7 Definitions: CATHODE REDUCTION occurs at this electrode ANODE OXIDATION occurs at this electrode
8 Will the half-cell reactions be the same or different? What chemical species would be present in a vessel of aqueous sodium chloride, NaCl (aq)?Na+Cl-H2OWill the half-cell reactions be the same or different?
9 Water Complications in Electrolysis In an electrolysis, the most easily oxidized and most easily reduced reaction occurs.When water is present in an electrolysis reaction, then water (H2O) can be oxidized or reduced according to the reaction shown.Electrode Ions ... Anode Rxn Cathode Rxn E°Pt (inert) H2O H2O(l)+ 2e- g H2(g)+ 2OH-(aq) VH2O H2O(l) g 4e H+(g) + O2(g) VNet Rxn Occurring:2 H2O g 2 H2(g)+ O2 (g) E° = V
10 What could be reduced at the cathode? -+Aqueous NaClbatterypowersourcee-e-NaCl (aq)What could be reduced at the cathode?Na+Cl-(-)(+)H2Ocathodedifferent half-cellanode2Cl- Cl e-
11 Aqueous NaCl Electrolytic Cell possible cathode half-cells (-)REDUCTION Na+ + e- Na2H2O + 2e- H OH-possible anode half-cells (+)OXIDATION 2Cl- Cl e-2H2O O H e-overall cell reaction2Cl H2O H2 + Cl OH-
12 For every electron, an atom of silver is plated on the electrode. Ag e- Age-Electrical current is expressed in terms of the ampere, which is defined as that strength of current which, when passed thru a solution of AgNO3 (aq) under standard conditions, will deposit silver at the rate ofg Ag/secAg+Ag1 amp = g Ag/sec
13 Faraday’s Law Q = It 1 coulomb = 1 amp-sec = 0.001118 g Ag The mass deposited or eroded from an electrode depends on the quantity of electricity.Quantity of electricity – coulomb (Q)Q is the product of current in amps times time in secondsQ = Ittime in secondscoulombcurrent in amperes (amp)1 coulomb = 1 amp-sec = g Ag
14 molemetal depends on the half-cell reaction Ag e- Ag1.00 mole e- = 1.00 mole Ag = g Agg Ag/mole e-g Ag/coul= 96,485 coul/mole e-1 Faraday (F )mole e- = Q/Fmass = molemetal x MMmolemetal depends on the half-cell reaction
15 Examples using Faraday’s Law How many grams of Cu will be deposited in 3.00 hours by a current of 4.00 amps?Cu e- CuThe charge on a single electron is x coulomb. Calculate Avogadro’s number from the fact that 1 F = 96,487 coulombs/mole e-.
19 A series of solutions have 50,000 coulombs passed thru them, if the solutions were Au+3, Zn+2, and Ag+, and Au, Zn, and Ag were plated out respectively, calculate the amount of metal deposited at each anode.e--+battery-+-+-+e-e-e-1.0 M Au+31.0 M Zn+21.0 M Ag+Au e- AuZn e- ZnAg+ + e- Ag
20 Paper moisturized with NaCl solution Volta’s battery (1800)Alessandro VoltaPaper moisturized with NaCl solutionCuZn
21 Galvanic cells and electrodes 4/6/2017Galvanic cells and electrodesTo sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells.Relative amounts of charge can be carried by negative or positive ions (depends on their relative mobilities) through the solution.Salt bridge, consists of an intermediate compartment filled with saturated salt solution and fitted with porous barriers at each end, is used for precise measurements. The purpose of salt bridge is to minimize the natural potential difference (junction potential).Lesson 9NEEP 423Test
23 Electrodes are passive (not involved in the reaction) Olmsted WilliamsElectrodes are passive (not involved in the reaction)
24 Observe the electrodes to see what is occurring. CellConstructionSalt bridge –KCl in agarProvides conductionbetween half-cellsObserve the electrodes to see what is occurring.CuZn1.0 M CuSO41.0 M ZnSO4
25 - + What about half-cell reactions? What about the sign of the electrodes?-+Why?cathode half-cellCu e- Cuanode half-cellZn Zn e-Cuplates out or deposits on electrodeZn electrode erodesor dissolvesWhat happened at each electrode?CuZn1.0 M CuSO41.0 M ZnSO4
26 Now replace the light bulb with a volt meter. -+1.1 voltscathode half-cellCu e- Cuanode half-cellZn Zn e-CuZn1.0 M CuSO41.0 M ZnSO4
27 We need a standard electrode to make measurements against! The Standard Hydrogen Electrode (SHE)H2 input1.00 atm25oC1.00 M H+1.00 atm H2PtHalf-cell2H e- H2inertmetalEoSHE = 0.0 volts1.00 M H+
28 How do we calculate Standard Redox Potentials? We must compare the half reactions to a standardWhat is that standard?2 H3O+(aq) + 2e- H2(g) H2O(l) E°= 0.00 VThis is called the standard hydrogen electrode or SHENow that we have a standard, we can calculate standardredox potential by using the table of standard redox potentials
29 E0 is for the reaction as written The more positive E0 the greater the tendency for the substance to be reducedThe half-cell reactions are reversibleThe sign of E0 changes when the reaction is reversedChanging the stoichiometric coefficients of a half-cell reaction does not change the value of E019.3
30 Oxidizing and Reducing Agents Cell EMFOxidizing and Reducing AgentsCopyright 1999, PRENTICE HALLChapter 20
31 Now let’s combine the copper half-cell with the SHE Eo = v+0.34 vcathode half-cellCu e- Cuanode half-cellH2 2H e-H atmKCl in agarCuPt1.0 M CuSO41.0 M H+
32 Now let’s combine the zinc half-cell with the SHE Eo = v-0.76 vanode half-cellZn Zn e-cathode half-cell2H e- H2H atmKCl in agarPtZn1.0 M ZnSO41.0 M H+
33 Assigning the EoWrite a reduction half-cell, assign the voltage measured, and the sign of the electrode to the voltage.Al+3 + 3e- Al Eo = v Zn+2 + 2e- Zn Eo = v 2H+ + 2e- H2 Eo = 0.00 v Cu+2 + 2e- Cu Eo = Ag+ + e- Ag Eo = vIncreasing activity
35 2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M) What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?Cd2+ (aq) + 2e Cd (s) E0 = VCd is the stronger oxidizerCd will oxidize CrCr3+ (aq) + 3e Cr (s) E0 = VAnode (oxidation):Cr (s) Cr3+ (1 M) + 3e-x 2Cathode (reduction):2e- + Cd2+ (1 M) Cd (s)x 32Cr (s) + 3Cd2+ (1 M) Cd (s) + 2Cr3+ (1 M)E0 = Ecathode - EanodecellE0 = – (-0.74)cellE0 = 0.34 Vcell19.3
36 Calculating the cell potential, Eocell, at standard conditions H2O with O2Consider a drop of oxygenated water on an iron objectFeFe e- Fe Eo = vreverse2xFe Fe e- -Eo = vO2 (g) H2O e- 4 OH- Eo = v2Fe + O2 (g) H2O 2Fe(OH)2 (s) Eocell= vThis is corrosion or the oxidation of a metal.
37 E=? -Concentration -Related half reaction kinetic e- +2Fe3+ Fe E0 = e- +2H+ H2 E0= Fe +H+ Fe !!!!!!!!!!!!!!!!!!!!!!!! =================================================================================================== e- +Fe 3+ Fe 2+ E0= e +Fe2+ Fe E0=-0.44 Fe Fe Fe ================================================= Cu Cu Cu
38 Equilibrium Constants in Redox Reactions Whereas potential and free energy are related, and free energy and equilibrium are related, equilibrium and potential must be related to one another.DG° = –nFE°celland DG° = –RT ln Keqtherefore –RT ln Keq = –nFEocellRT ln Keq RTE°cell = ––––––––– = –––– ln KeqnF nFR and F are constant, therefore at 298 K:VE°cell = –––––––– ln Keqn
39 1) 2e +Fe2+ Fe E0=-0. 440 2) e +Fe3+ Fe2+ E0=-0 1) 2e +Fe2+ Fe E0= ) e +Fe3+ Fe2+ E0= e +Fe3+ Fe E0=??? ===============
41 Effect of Concentration on Cell EMF A voltaic cell is functional until E = 0 at which point equilibrium has been reached.The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction.The Nernst EquationThe Nernst equation relates emf to concentration usingand noting that
42 and the previous relationship: DGo = -nFEocell from thermodynamics:DGo = RT log Kand the previous relationship:DGo = -nFEocell-nFEocell = RT log Kat 25oC: Eocell = log Knwhere n is the number of electrons for the balanced reaction
43 What happens to the electrode potential if conditions are not at standard conditions? The Nernst equation adjusts for non-standard conditionsFor a reduction potential: ox + ne redin general: E = Eo – RT ln (red)nF (ox)at 25oC: E = Eo log (red)n (ox)Calculate the E for the hydrogen electrode where 0.50 M H+ and 0.95 atm H2.
44 Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s) 1) An example:Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s)According to the reduction potentials:2 e- + Ni Ni(s) V2 e- + Sn Sn(s) VOne of these needs to be reversed to get a positive voltage, and a spontaneous reaction.
45 Ni(s) + Sn2+34 Ni2+ + Sn(s) 0.090 V = Eo Ni(s) e- + Ni V2 e- + Sn Sn(s) VNi(s) + Sn2+34 Ni2+ + Sn(s) V = Eo
46 E) Calculation of the equilibrium constant 1) at equilibrium E = 0 ; Q = ____From the Nernst Equation:For the cell:Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s)
48 Sn(s) | Sn2+(1.0M)|| Pb2+(0.0010M) | Pb(s) 2) An example:Sn(s) | Sn2+(1.0M)|| Pb2+(0.0010M) | Pb(s)According to the reduction potentials:2 e- + Pb Pb(s) E0= V2 e- + Sn Sn(s) E0=-0.136VOne of these needs to be reversed to get a positive voltage, and a spontaneous reaction.!!!!!!!!!!!!!!!!!!!!!!!!!
50 Free Energy and the Cell Potential Cu Cu e- -Eo =2xAg+ + e- Ag Eo = vCu Ag+ Cu AgEocell= vDGo = -nFEocellwhere n is the number of electrons for the balanced reactionWhat is the free energy for the cell?1F = 96,500 J/v
51 Electrolysis of Copper A net reaction of zero, yet a process does take place.A concentration cell based on the Cu/Cu2+ half-reaction. A, Even though the half-reactions involve the same components, the cell operates because the half-cell concentrations are different. B, The cell operates spontaneously until the half-cell concentrations are equal. Note the change in electrodes (exaggerated here for clarity) and the equal color of solutions.
57 E = Eo pHThe scale on the pH meter is marked off so that a change of 1 pH unit equals volts or 59.2 millivolts.
58 The pH MeterIn practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas!The electrode is merely dipped into a solution, and the potential difference between the electrodes is displayed as pH.A stable reference electrode and a glass-membrane electrode are contained within a combination pH electrode.
60 Corrosion Corrosion of Iron Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen.Cathode: O2(g) + 4H+(aq) + 4e- 2H2O(l).Anode: Fe(s) Fe2+(aq) + 2e-.Dissolved oxygen in water usually causes the oxidation of iron.Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3.xH2O(s).Oxidation occurs at the site with the greatest concentration of O2.Copyright 1999, PRENTICE HALLChapter 20
65 Prevention of Corrosion Cover the Fe surface with a protective coatingPaintTinZnGalvanized iron
66 Corrosion Preventing the Corrosion of Iron Corrosion can be prevented by coating the iron with paint or another metal.Galvanized iron is coated with a thin layer of zinc.Zinc protects the iron since Zn is the anode and Fe the cathode:Zn2+(aq) +2e- Zn(s), Ered = VFe2+(aq) + 2e- Fe(s), Ered = VWith the above standard reduction potentials, Zn is easier to oxidize than Fe.Copyright 1999, PRENTICE HALLChapter 20
70 Cathodic ProtectionIn cathodic protection, an iron object to be protected is connected to a chunk of an active metal.The iron serves as the reduction electrode and remains metallic. The active metal is oxidized.Water heaters often employ a magnesium anode for cathodic protection.