14_REDOX_titrations - Both oxidation and reduction occur...

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Unformatted text preview: Both oxidation and reduction occur during a titration. The equivalence point is based on the concentration of the oxidized and reduced form of all species involved. Aox + Bred = Ared + Box Ce4+ + Fe2+ = Ce3+ + Fe3+ For a REDOX titration, the equivalence point is the point where Eforward = Ereverse. Ecell = 0 Since the E values are concentration dependent, we must rely on the Nernst equation E = Eo - 0.0592 n [A ] log [Ared] ox At the equivalence point: [A ] [B ] 0.0592 0.0592 EAo log red = EBo log red [Aox] [Box] nA nB Since Ared=Box and Bred=Aox at the equivalence point, we can reduce this to: Eeq = n E o + nBEBo Eeq = A A nA + nB nAEAo + nBEBo nA + nB Determine the Eeq for the following reaction: Fe2+ + Ce4+ = Fe3+ + Ce3+ Eo Fe3+/Fe2+ Eo Ce4+/Ce3+ = +0.771V = +1.70 V nFe = 1, nCe = 1 Eeq = 1.70V + 0.771V = 1.24 V 2 Determining Eeq becomes much more complicated for more complex systems. The inclusion of one additional species, like H+ is relatively common. Lets look at another example. 6Fe2+ + Cr2O72- + 14H+ = 6Fe3+ + 2Cr3+ + 7H2O [ Fe2+] [ Cr3+]2 7 Eeq = EoFe + 6EoCr - log [ Fe3+] [ Cr O 2-] [ H+]14 27 [ Fe2+] 0 Eeq = EoFe - .0592 log [ Fe3+] 1 o = 0.771V E Eeq = EoCr -0.0592 log 6 [ Cr3+]2 [ Cr2O72-][ H+]14 Eo = 1.33 V Eeq = EoFe + 6EoCr 7 - Eeq = 1.25V - 8.46 x10-3 At the equivalence point: [ Fe2+] [ Fe3+] First, obtain both half reactions: 2 [ Cr3+] 0.0592 log 7 [ H+]14 log So for this reaction, the equivalence point is dependent on both [ Cr3+] and [ H+]. = 6 [Cr2O72-] = 3 [Cr3+] This explains why we commonly work in 1M acid and with dilute solutions. As with acid/base titrations, we can get either to the following types of curves. If the reaction was conducted under these conditions then: Eeq = 1.25V - 8.46 x10-3 log 2 [ Cr3+] [ H+]14 2 [ Cr3+] [ 1 ]14 Eeq shows only a small dependence on Cr3+. Based on the type of reaction. over titration Ecell equivalence point buffer region percent titration Unlike acid/base titrations, we really can’t do much with this region. While some Fe3+ must be present, we can only guess what the concentration is. No Ce4+ or Ce3+ are present, so we don’t have a complete reaction. Fe3+ Fe2+ 10 20 30 40 50 60 70 80 90 90 80 70 60 50 40 30 20 10 Fe2+/Fe3+ 0.715 0.735 0.749 0.761 0.771 0.781 0.793 0.807 0.828 We can simplify our calculations by using the % titration. 2+ EFe = 0.771 - 0.0592 log %Fe3+ %Fe E 9 4 2.33 1.5 1 0.67 0.43 0.25 0.11 Between 0% and 100% titration, we can use the Nernst equation for Fe2+/Fe3+. [ Fe2+] EFe = 0.771 - 0.0592 log [ Fe3+] There is no significant level of Ce4+ to work with anyway. 0.84 0.82 0.8 There was only a change of 0.113 V from 10 to 90% titration 0.78 0.76 0.74 0.72 0.7 0 20 40 60 80 100 1.3 1.2 1.1 1.70V + 0.771V 2 1 0.9 Note the large jump in E at the equivalence point. 0.8 0.7 0.6 0 At greater than 100% titration, the predominate change is that Ce4+ is being added and diluted into a solution of Ce3+. All Fe2+ has been converted to Fe3+ and no longer figures into the calculations. We just need to keep track of the amounts of Ce3+ and Ce4+ as well as the total volume of the system. At the equivalence point: Solution volume [Ce3+] = 200 ml = 0.05M At 10% overtitration, we’ve added an additional 10ml of our Ce4+ solution so: [Ce3+] [Ce4+] = 0.0476 M = 0.00476 M 20 40 60 80 100 120 So far, we’ve not needed to worry about the actual concentrations. However, to know the volumes involved, we now need them. Lets assume that we started with 100ml of a Fe2+ solution and our titrant was 0.1M Ce4+. Now we can determine the E for overtitration. We can again use the Nernst equation to determine the E for our system. In this case, however, we’ll use the values for Ce3+ and Ce4+. 0.0476 E = 1.40 - 0.0592 log 0.00476 = 1.34 At 20% overtitration, E would be 1.36. 1.4 1.3 1.2 1.1 1 0.9 0.8 0.7 0.6 0 20 40 60 80 100 120 < 0.0592 n log or > 10 [ Indred] [ Indox] (Phen)3Fe3+ + e = (Phen)3Fe2+ N N pale blue Fe 3 Eo = 1.06V red HSO3 Eo = 0.80V irreversible reversible NH Starch-I If the goal is to reduce your analyte to a single oxidation state, you can use either a Jones or Walden Reductor. Both are columns containing a metal. You slowly wash your sample through the column with water. If the goal is to oxidize your analyte to a single form, no material is available that can be used as a column. You must have a method to remove any excess oxidizing reagents prior to titration. HCl Often, the results in the preparation step being more complicated than the rest of the method. Ag+ Methanol is added to remove the reactive C5H5N.SO3 complex. C5H5N.SO3 + CH3OH C5H5N(H)SO4CH3 This is done because the SO3 complex can react with water as well as many other species. ...
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This note was uploaded on 12/06/2011 for the course CHEM 300 taught by Professor Jameshardy during the Fall '10 term at The University of Akron.

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