2Historically Historically oxidation involved reaction with O2. i.e., Rusting4 Fe(s) + 3O2 (g) Fe2O3 (s)Other exampleZn(s) + Cu2+(aq) g Zn2+(aq) + Cu(s)In this reaction:Zn(s) g Zn2+(aq) OxidationCu2+(aq) g Cu(s) ReductionIn a redox reaction, one process can’t occur without the other. Oxidation-Reduction reaction must simultaneously occurs.
3Redox BetweenIf Zn(s) and Cu2+(aq) is in the same solution, then the electron is a transferred directly between the Zn and Cu.No useful work is obtained. However if the reactants are separated and the electrons shuttle through an external path...
4Electrochemical Cells Voltaic / Galvanic Cell Apparatus which produce electricityElectrolytic Cell Apparatus which consumes electricityConsider:ZnCuInitially there is a flow of e-After some time the process stopsElectron transport stops because of charge build upBuild up of positive chargeBuild up of negative chargeThe charge separation will lead to process where it cost too much energy to transfer electron.
5Completing the Circuit Electron transfer can occur if the circuit is closedParts:Two conductorsElectrolyte solutionSalt Bridge / Porous membrane3 process must happen if e- is to flow.A. e- transport through external circuitB. In the cell, ions a must migrateC. Circuit must be closed (no charge build up)Anode (-)BlackNegative electrode generates electronOxidation OccurCathode (+)RedPositive electrode accepts electronReduction OccurACBAnode/Anion (-)Cathode/Cation(+)
6Voltaic Cell Electron transfer can occur if the circuit is closed Parts:Two conductorsElectrolyte solutionSalt Bridge / Porous membrane3 process must happen if e- is to flow.A. e- transport through external circuitB. In the cell, ions a must migrateC. Circuit must be closed (no charge build up)Anode (-)BlackNegative electrode generates electronOxidation OccurCathode (+)RedPositive electrode accepts electronReduction OccurAnode/Anion (-)Cathode/Cation(+)
7Completing the Circuit: Salt Bridge In order for electrons to move through an external wire, charge must not build up at any cell. This is done by the salt bridge in which ions migrate to different compartments neutralize any charge build up.
8Sign Convention of Voltaic Cell @ Anode: Negative Terminal (anions).Source of electron then repels electrons. Oxidation occurs.Zn(s) g Zn+2(aq) + 2e- : Electron source@ Cathode: Positive Terminal (cation)Attracts electron and then consumes electron. Reduction occurs.Electron target: 2e- + Cu+2(aq) g Cu(s)Overall:Zn(s) + Cu+2(aq) g Zn+2(aq) + Cu(s) E° = 1.10 VNote when the reaction is reverse: Cu(s) + Zn+2(aq) g Cu+2(aq) + Zn(s)Sign of E ° is also reversed E° = VOxidation: Zn(s) g Zn+2(aq) E° = 0.76 VReduction: Cu+2(aq) g Cu(s) E° = 0.34 V1.10 V = E°CELLor E°CELL = E°red (Red-cathode) - E°red (Oxid-anode)
9Other Voltaic Cell Zn(s) + 2H+ (aq) g Zn+2(aq) + H2 (g) E° = 0.76 V @ Anode: Negative Terminal (anions): Zn(s) g Zn+2(aq) + 2e- :Source of electron then repels electrons. Oxidation occurs.@ Cathode: Positive Terminal (cation): 2e H+(aq) g H2 (g)Attracts electron and then consumes electron. Reduction occurs.Net: Zn(s) + 2H+ (aq) g Zn2+ (aq) + H2 (g)
10Other Voltaic Cell Zn(s) + 2H+ (aq) g Zn+2(aq) + H2 (g) E° = 0.76 V @ Anode: Negative Terminal (anions): Zn(s) g Zn+2(aq) + 2e- :Source of electron then repels electrons. Oxidation occurs.@ Cathode: Positive Terminal (cation): 2e H+(aq) g H2 (g)Attracts electron and then consumes electron. Reduction occurs.Net: Zn(s) + 2H+ (aq) g Zn2+ (aq) + H2 (g)
11Line Notation Convention Line notation: Convenient convention for electrochemical cellSchematic Representation1. Anode g Cathode[oxidation (-) ] [reduction (+)]2. “ “ phase boundary(where potential may develop)3. “ “ Liquid junction4. Concentration of componentZn(s) ZnSO4 (aq,1.0M) CuSO4 (aq,1.0M) Cu(s)4132
12Line Notation Examples Consider : Zn(s) + Cu+2(aq) g Zn+2(aq) + Cu(sAnode: Zn g Zn e-Cathode: Cu e- g CuShorthand “Line” notationZn (s) Zn+2 (aq)(1.0M) Cu+2(aq) (1.0M) Cu(s)2nd Example : Zn(s) + 2H+ (aq) g Zn+2(aq) + H2(g)Cathode: 2H e- g H2 (g)Zn (s) Zn+2 (aq)(1.0M) H+(aq) (1.0M), H2(g, 1atm) Pt(s)
13Other Voltaic Cell & Their Line Notation Oxidation half-reactionZn(s) g Zn+2(aq) e-Oxidation half-reaction2I- (aq) g I2 (s) + 2e-Oxidation half-reactionCr(s) g Cr+3(aq) e-Reduction half-reactionMnO4-(aq) + 8H+(aq) + 5e-g Mn 2+(aq) +4H2O(l)Reduction half-reactionAg+(aq) + e- g Ag (s)Zn(s) | Zn+2 (aq)||H+(aq) , H2 (g,1atm)|PtCr(s) | Cr+3 (aq)||Ag+(aq) | Ag(s)C(s)| I-(aq) , I2 (g,1atm) || MnO4-(aq) , Mn+2 (aq)| C(s)
14Line Notation Examples Example 1: B&L 20.13Zn(s) + Ni2+(aq) g Zn+2(aq) + Ni (aq)Example 2: B&L 20.19Tl+3(aq) + 2Cr2+(aq) g Tl+(aq) + Cr+3(aq)
15Voltage of Galvanic / Voltaic Cell Transport of any object requires a net force.Consider water flowing through pipes. This occurs because of pressure gradient.Flow (Fluid Transport)Pressure (h)Pressure (i)OrSimilarly, electron are transported through wires because of the electromotive force EMF or Ecell.Object falling or transport down due to Dh(+)(-) e -
16EMF - ElectroMotive Force Potential energy of electron is higher at the anode. This is the driving force for the reaction (e- transfer)eAnode (-)D P.E. = V = Je - Ce- flow toward cathode(+) CathodeLarger the gap, the greater the potential (Voltage)
17ElectroMotive Force EMF - Electro Motive Force Potential energy difference between the two electrodesThe larger the DP.E. the larger EMF value.The magnitude of P.E. for the reaction (half reaction) is an intensive property)i.e., Size independent: r, Tbpt, Cs.Therefore EMF is also an intensive property.Analogy:Size of rock not important, only the height from ground.(Electron all have the same mass)Unit: EMF: V - Volts :1V - 1 Joule / Coulomb1 Joule of work per coulomb of charge transferred.
18Stoichiometry Relationship to E° EMF - Intensive PropertyE°cell Standard state conditions 25°C, 1atm, 1.0 ME°cell Intensive property, Size IndependentConsider:Li+ + e- g Li (s) E°Cell = Vx Li e- g 2 Li (s) E°Cell = ( V) x 2 = ??But E° = Voltage per electronE° ‘ = E° x 2 = ? g V • 2 = V2 e-\ Stoichiometry does not change E°, but reversing the reaction does change the sign of E°.
19Standard Reduction Potential Cell Potential is written as a reduction equation.M+ + e- g M E° = std red. potentialMost spontaneous <Reduction occurs> Oxidizing AgentWritten as reductionMost non-spontaneous Spontaneous in the reverse direction. <Oxidation occurs> Reducing Agent
20Zoom View of Std. Reduction Potential Cell Potential is written as a reduction equation.M+ + e- g M E° =F2 (g) e- g 2 F - (aq) VCe e- g Ce3+ (aq) V2H e- g H2 (g) VLi+(aq) + e- g Li(s) VMost spontaneous Reduction Oxidizing AgentWritten as reductionMost non-spontaneous Spontaneous in the reverse direction. Oxidation Reducing AgentAll reaction written as reduction reaction. But in electrochemistry, there can’t be just a reduction reaction. It must be coupled with an oxidation reaction.
21E°Cell Evaluation E°Cell Function of the reaction g Oxidation Process (Anode reaction)g Reduction Process (Cathode reaction)orE°Cell = E°Cathode & E°AnodeCathode (+)Anode (-) Most Negative Reduction reactionTherefore,E°Cell = E°red (Cathode) - E°red (anode)Neg Minus (Large negative)(Very Positive Value)Very Positive\ Very Spontaneous
22Standard Reduction Potential How is E°red (Cathode) and E°red (Anode) determine.E° (EMF) - State Function; there is no absolute scaleAbsolute E° value can’t be measured experimentallyThe method of establishing a scale is to measure the difference in potential between two half-cells.Consider:Zn g Zn e E°=?Can’t determine because the reaction must be coupledHow can a scale of reduction potential be determine ?Use a half reaction as reference and assign it a potential of zero.Electrochemical reaction more spontaneous than this reference will have positive E°, and those less spontaneous will have negative E°.
23Side-Bar: Relative Scale Consider a baby whose weight is to be determine but will not remain still on top of a scale. How can the parents determine the babies weight?Carry the child in arms and weight both child and parent then subtract the weight of the parent from the total to yield the baby weight.
24Reference Potential Selected half reaction is: H+ / H2 (g) couple half reaction: 2H+ (aq, 1.0M) + 2e- g H2 (g,1atm)by definition c E° = 0.0 V, the reverse is also 0.0 VH+/H2 couple - Standard Hydrogen Electrode (SHE)To determine E° for a another half reaction, the reaction of interest needs to be coupled to this SHE. The potential measured is then assigned to the half-reaction under investigation.E°Cell = V = E°red (Cat) - E°red (Anode)0.0 V (?)E°red (Anode) = V\Zn+2/Zn E° = V Reduction rxn
25Determining Other Half-Cell Potential Now consider the reaction:Zn(s)|Zn+2 (1.0 M)||Cu+2(1.0 M)|Cu(s)E°Cell = VE°Cell = E°red (Cat) - E°red (Anode)recall, E° Zn+2/Zn = VTherefore,E°Cell = E°Cu+2/Cu - E° Zn+2/Zn1.10 V = (?) ( V)E°Cu+2/Cu = V
26Example: Half-Cell Potential Example BBL20.19:For the reaction: Tl Cr2+ Tl Cr3+ E°Cell = 1.19 Vi) Write both half reaction and balanceii) Calculate the E°Cell Tl+3 Tl+iii) Sketch the voltaic cell and line notationi) Tl e- Tl+(Cr2+ 2Cr3+ + 2e- ) x E° = 0.41 Vii) E°Cell = 1.19 V = E°red (Cat) - E°red (Anode)1.19 V= E°red (Cat) Vfor Tl e- Tl+ :1.19 V = E°red (Cat) = VPtCr2+ g Cr3+Tl3+ g Tl+1.19 V
27Voltaic Vs. Electrolytic Cells Voltaic CellEnergy is released from spontaneous redox reactionElectrolytic CellEnergy is absorbed to drive nonspontaneous redox reactionGeneral characteristics of voltaic and electrolytic cells. A voltaic cell generates energy from a spontaneous reaction (DG<0), whereas an electrolytic cell requires energy to drive a nonspontaneous reaction (DG>0). In both types of cell, two external circuits provides the means or electrons to flow. Oxidation takes place all the anode, and reduction takes place at the cathode, but the relative electrode changes are opposite in the two cells.System does work on load (surroundings)Surrounding (power supply) do work on system (cell)Anode (Oxidation)Oxidation ReactionX g X+ + e-Oxidation ReactionA- g A + e-Reduction Reactione- + Y+ g YReduction Reactione- + B+ g BOverall (Cell) ReactionA- + B+ g A + B, DG> 0Overall (Cell) ReactionX + Y+ g X+ + Y, DG = 0