PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12If an environment is characterized by a certain redox reaction, it has a characteristic EhSlide 14Slide 15Slide 16Slide 17Oxidation-Reduction (Redox) ReactionsEh, Ph, and the chemistry of natural watersImportant equations and terms for redox reactions:GR = nFEGR = GR° + RTlnQEh = E = E° + (2.303RT/nF)logQ = (0.05916/n)logQE° = -(0.05916/n)logKEh = 1.23 + 0.01479log[O2] - 0.05916pHR = 1.987 x 10-3 kcal/mol°T = 298.15°KF = 23.06 kcal/Vn = number of electrons transferred in the reactionQ = reaction quotientE° = standard potential for a half-reactionEh = electromotive force (emf) generated between an electrodein any state and the H2 electrode (the referencehydrogen scale Eh) in standard state (theoretical voltage) Nernst factor at standard stateZn(s) + Cu 2+ (aq)  Zn 2+ (aq) +Cu(s)Each metal has a different relative preference for electronsElectrode PotentialSeparate the reaction into half-reactions and add the e-Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) +Cu(s)Zn(s)  Zn 2+ (aq) + 2e-Cu 2+ (aq) + 2e- Cu(s)There will be Energy gained or lost by the exchange of e-Unfortunately, such measurements are not possible (nor would these reactions occur in the natural environment:electrons are not given up except to another element or species). This requires the establishment of an arbitrary reference valueHydrogen Scale Potential Eh−+⏐ →←+eaqHgH )()(212The potential measured for the entire reaction is thenassigned to the half-cell reaction of interest:Zn(s)  Zn 2+ (aq) + 2e-becomes+++⏐ →←+ HsZngHaqZn 2)()()(22The potential for this reaction = -0.763 VSo the Eh of the reduction of Zn2+ to Zn (s) = -0.763 V(+) Eh means reaction moves left to right(-) Eh means reaction moves right to left(+) Eh means the metal ion is reduced by H2(g) to the metal(-) Eh means the metal will be oxidized to the ion and H+ is reduced+++⏐ →←+ HsZngHaqZn 2)()()(22Eh = E = E° + (2.303RT/nF)logQ = (0.05916/n)logQCalculating the Eh of Net ReactionsWe can calculate Eh values by algebraic combinations of thereactions and potentials that are listed in Table 14.3 of the text. There is, however, a “catch”Calculate the Eh for the reaction:Fe3+(aq) + 3e-  Fe(s)This reaction is the algebraic sum of two reactions listed in Table 14.3Fe3+(aq) + e-  Fe2+(aq) Fe2+(aq) + 2e-  Fe(s)Since the reactions sum, we might assume that we can simply sum the Eh values to obtain the Eh of the net reaction. Doing so, we obtain an Eh of 0.33 V. However, the true Eh of this reaction is -0.037 V.What did we do wrong?Fe3+(aq) + 3e-  Fe(s)We have neglected to take into consideration the number of electrons exchanged. In the algebraic combination of Eh values, we need to multiply the Eh for each component reaction by the number of electrons exchanged. We then divide the sum of these values by number of electrons exchanged in the net reaction to obtain the Eh of the net reaction∑=iiinetEhzznetEh1)(Fe3+(aq) + e-  Fe2+(aq) Fe2+(aq) + 2e-  Fe(s)∑=iiinetEhzznetEh1)(where the sum is over the component reactions i. By Hess’s Law, the G of the net reaction must be the simple sum of the component reaction G’s…but Eh values are obtained by multiplying G by z. GR = nFE∑=iiinetEhzznetEh1)(Eh = 1/3 (1*0.77+2*-0.44) = 1/3 (0.77-0.88) = 1/3(-0.11)=-0.037Fe3+(aq) + e-  Fe2+(aq) Fe2+(aq) + 2e-  Fe(s)nFG-Ereaction-half of potential std. EC25at volts0.05916 factor Nernst F2.303RTelectrons ofnumber n kcal/voltg 23.061 constant faraday FKelvinin temp. Tkcal/mol 0.001987 constant gasR(D)(C)(B)(A)lnnFRT EEh(volts)ofoooodcbaoΔ==========⎟⎟⎠⎞⎜⎜⎝⎛+=eqreduced oxidizeddD cC e bB aA lnFG- E][P][Hln FRT E Eh -oRo1/2Ho2+=++Δ=++=nEh = E° + (2.303RT/nF)logQ = (0.05916/n)logQFrom Garrels & Christ (1965)If an environment is characterized by a certain redox reaction, it has a characteristic EhOxidizingReducingEh = 1.23 + 0.01479log[O2] - 0.05916pHSimilar to Fig. 14.4 on p. 238[O2] = 10-83.1 atm[O2] = 1 atm[O2] conditions for natural waters1) write the equation for the reaction)()()(22212gHgOlOH +↔2) “electrode” the equation−++↔+egOHlOH 2)(2)(22122e-O in H2O gives up 2e-3)( )22/12][][log205916.0++= HOEEho4) get Eo from GrokcalOHGGeGOGHGlOHGeGOGHGGoforofofofofofofofor687.560)](([)](2)(2/1)(2[2222+=−=Δ=−==−−++=Δ++VxEo23.106.232687.56+=+=6)( )pHOEhHOEh05916.0]log[01479.023.1][][log205916.023.1222/12−+=+=+This demonstrates the Eh - pH relationship as shown in figure 14.4 p. 238 and is the equation for the lines limitingthe minimum and maximum [O2] for natural watersFigure 14.11 on p.