Review: Balancing Redox Reactions

1 1 review : Balancing Redox Reactions Determine which species is oxidizedand which species is reduced Oxidation corresponds to an increase inthe oxidation number of an element Reduction corresponds to a decrease inthe oxidation number of an element Write half Reactions for oxidation andreduction processes Oxidation reaction will have e- s on theright side of equation Reduction reaction will have e- s on theleft side of the equationReview: Balancing Redox Reactions Balance half Reactions includingcharge balance Multiply each half reaction by a factorso that the total number of e- stransferred in each reaction are equal Add the resulting half reactionstogether to get the overall balancedredox equation2 review : Balancing Redox ReactionsExample:F2(g) + Al(s) F-(aq) + Al3+(aq)oxid. state 0 0 -1 +3 Fluorine is reduced (0 -1)Aluminum is oxidized (0 +3) Half Reactions :oxidation:Al(s) Al3+(aq) + 3 e-reduction:F2(g) + 2 e- 2 F-(aq) review : Balancing Redox ReactionsExample:F2(g) + Al(s) F-(aq) + Al3+(aq) Balance e- s transferred:multiply oxidation rxn by 22 Al(s) 2 Al3+(aq) + 6 e-multiply reduction rxn by 33 F2(g) + 6 e- 6 F-(aq) Add half reaction to get net reaction:2 Al(s) + 3 F2(g) 2 Al3+(aq) + 6 F-(aq) Check atom and charge balance:3 review : Balancing Redox ReactionsExample:Ag+(aq) + SO2(g) + H2O(l) Ag(s) + SO42-(aq) + H3O+(aq) Determine oxidation state of each elementin reaction:Ag+ + SO2 + H2O Ag + SO42- + H3O++1+4 -2-2-2-2+1+10+6 review : Balancing Redox ReactionsExample:Ag+(aq) + SO2(g) + H2O(l) Ag(s) + SO42-(aq) + H3O+(aq) Ag is reduced (+1 0) S is oxidized (+4 +6) Oxidation half reaction.

2 SO2(g) + H2O(l) SO42-(aq) + H3O+(aq) + 2 e-SO2(g) + 6 H2O(l) SO42-(aq) + 4 H3O+(aq) + 2 e-4 review : Balancing Redox ReactionsExample:Ag+(aq) + SO2(g) + H2O(l) Ag(s) + SO42-(aq) + H3O+(aq) Reduction half reaction:Ag+(aq) + e- Ag(s) Multiply reduction reaction by 2 to balancee- s transferred2 Ag+(aq) + 2 e- 2 Ag(s) review : Balancing Redox ReactionsExample:Ag+(aq) + SO2(g) + H2O(l) Ag(s) + SO42-(aq) + H3O+(aq) Add balanced half reaction to get netreactionSO2(g) + 6 H2O(l) SO42-(aq) + 4 H3O+(aq) + 2 e-2 Ag+(aq) + 2 e- 2 Ag(s)SO2(g) + 2 Ag+(aq) + 6 H2O(l) SO42-(aq) + 2 Ag(s) + 4 H3O+(aq)5 Electrochemical Cells When two half Reactions areconnected, we get an electrochemicalcell that can generate a voltagepotential and electrical currentElectrochemical CellsOxidationoccurs at theanodeReductionoccurs at thecathodeFigure Cells Each half reaction has an electricalpotential, E Electrical potential is a measure ofhow easily a species is reduced e- s added to the species to reduce itsoxidation state The emf (electromotive force) of acell is a measure of how much workthat cell can doElectrochemical Cells Work for a cell is defined as.

3 Work = charge E Work = # e- s E The potential difference ( E) ismeasured in volts Charge is measured in coulombs 1 e- has a charge of x 10-19 C 1 volt = 1 Joule/1 coulomb7 Electrochemical Cells The emf of a cell is determined bytaking the difference between thepotentials of the cathode and theanode:Ecell = Ecathode Eanode If Ecell is positive the electrochemicalreaction will proceed as written If Ecell is negative, the reversereaction will occurElectrochemical Cells Values for the potential of varioushalf Reactions can be found in tables Values are listed under standardconditions Gas phase species have a pressure of 1atm Aqueous species have a concentration of1 M Tables give as standard reductionpotentials, Eo8 Electrochemical CellsExamples: (from Appendix I)Co3+(aq) + e- Co2+(aq) Eo = VAu3+(aq) + 3 e- Au(s) Eo = VHg22+(aq) + 2 e- 2 Hg(l) Eo =.

4 789 VI2(s) + 2 e- 2 I-(aq) Eo = .535 V2 H3O+(aq) + 2 e- H2(g) + 2 H2O Eo = VPbSO4(s) + 2 e- Pb(s) + SO42-(aq) Eo = VCd2+(aq) + 2 e- Cd(s) Eo = VLi+(aq) + e- Li(s) Eo = VElectrochemical Cells Electrical potential cannot bemeasured on an absolute scale The standard hydrogen electrode(SHE) is defined as a referenceelectrode with a potential ofEo = V Potentials of all other half reactionare measured relative to the SHE9 Electrochemical CellsFigure CellsExample:Determine potential when a copper electrodein a solution of copper nitrate is connectedto a nickel electrode in a solution of nickelnitrateStep 1: write balanced half Reactions for eachelectrode (it doesn t matter yet whichelectrode you select as the anode andwhich as the cathode)Ni2+(aq) + 2 e- Ni(s) Eo = VCu(s) Cu2+(aq) + 2 e- Eo = VWhen you flip a Redox eqn.

5 , you change the sign of Eo10 Electrochemical CellsExample:Step 2: if necessary, multiply half reaction byfactor to balance e- s transferredCu2+(aq) + 2 e- Cu(s) Eo = .337 VNi(s) Ni2+(aq) + 2 e- Eo = .25 VStep 3: add half Reactions to get net reaction,and add potentials to get net cell potentialCu2+(aq) + Ni(s) Cu(s) + Ni2+(aq)Eo = .59 VBecause Eo for the cell is positive, the reactionproceeds as writtenElectrochemical CellsExample:Determine Eo for a Mg2+ solution with Ptelectrode connected to a Ag+ solution witha Ag electrodeStep 1: write balanced half reactionsMg2+(aq) + 2 e- Mg(s) Eo = VAg(s) Ag+(aq) + e- Eo = V11 Electrochemical CellsExample:Step 2: multiply anode reaction by 2 tobalance e- s2 Ag(s) 2 Ag+(aq) + 2 e- Eo = VEo is a function only of the species beingreduced or oxidized, not by how manythere areWe do not multiply the value of Eo by thesame factor used to balance the e- stransferredElectrochemical CellsExample:Step 3.

6 Add half reaction and Eo s to getresultsMg2+(aq) + 2 e- Mg(s) Eo = V2 Ag(s) 2 Ag+(aq) + 2 e- Eo = VMg2+(aq) + 2 Ag(s) Mg(s) + 2 Ag+(aq) Eo = VBecause Eo is negative, the reverse reactionoccursMg(s) + 2 Ag+(aq) Mg2+(aq) + 2 Ag(s) Eo = V12 Electrochemical Cells Shorthand notation forelectrochemical cells Phase changes are represented by asingle vertical line Salt bridges are represented by doublevertical lines Begin with anode reaction (oxidation)Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)Elect rochemical Cells Example:Write the shorthand notation for the cell:H2(g) + AgCl(s) H+(aq) + Cl-(aq) + Ag(s) 0 +1 -1 +1 -1 0 H is oxidized; Ag is reduced Notation for anode:H2(g),Pt|H+(aq) Notation for cathode:Cl-(aq),AgCl(s)|Ag(s) Overall:H2(g),Pt|H+(aq)||Cl-(aq),AgCl(s) |Ag(s)13Eo and Go The electrochemical potential, Eo, andGibb s free energy, Go, are related: Go = -nFEon = # electrons transferredF = Faraday s Constant = 96,485 C/molEo and Go Reminder: an electrochemical rxn occursspontaneously if E is positive Any rxn is spontaneous if G is negative If Eo is positive, then Go must benegative Go = -nFEo14Eo and GoExample.

7 Find Go for the reactionMg(s) + 2 Ag+(aq) Mg2+(aq) + 2 Ag(s) Eo = V2 e- s are transferred in the process Go = -(2)(96500 C/mol)( J/C) = kJ/molEo, Go, and K Since we know relation between Goand Eo and between Go and K, wecan determine equilibrium constantfor electrochemical reaction Go = -nFEo Go = -RT lnK-nFEo = -RT lnKEo = RTnF lnK = .0257 Vn lnKat T = 298 K15Eo, Go, and K If we convert from natural log tocommon log (base 10), we getEo = .0592 Vn logKat T = 298 KorK = V Concentration and Eo E at non-standard concentrations canbe determined from our knowledge of G under non-standard conditions: G = Go + RT lnQSubstituting G = -nFE gives:-nFEcell = -nFEocell + RT lnQDivide by nF:Ecell = Eocell RT/nF lnQEcell = Ecello - .0592n logQNernstEquation16 Concentration and EoExample:Find potential of the following cell:Cu(s)|Cu2+(.)

8 0037M)||Ag+(.016M)|Ag(s)Step 1: write half Reactions w/ Eo soxidation: Cu(s) Cu2+ + 2 e- Eo = Vreduction: Ag+ + e- Ag(s) Eo = .799 VStep 2: write balanced net reactionCu(s) + 2 Ag+ Cu2+ + 2 Ag(s) Eo = .462 VConcentration and EoExample:Cu(s) + 2 Ag+ Cu2+ + 2 Ag(s) Eo = .462 VStep 3: write expression for QQ = [Cu2+][Ag+]2 Step 4: solve for EE = Eo - .0592 Vn logQ = .462 V - .0592 V2 = .427 V