Chapter 16 Redox Titrations

Chapter 16 Redox TitrationsProblems: 3, 4, 6, 8Overhead Fig 16-1 adjusted for exampleJust as calculating titration curves for acids and bases helped to tie lots of ideastogether and help to give a better understanding of acid/base reaction, let us now lookat redox titration to try to tie up some of the concepts we talked about in the Redoxchapters16-1 Theory of Redox TitrationsRather than following the book, I am going to use an example from a lab wecould do. We will use the titration of ascorbic acid (vitamins C) with iodine.Since this reaction was done in neutral water, pH 7, the most appropriate E toouse is E ' for pH 7. Thus we will use:- + oDehydroascorbate + 2e +2H Ascorbate E '=0.08 and- - - oI +2e 3I E '= .543- - +Net rxn: Ascorbate + I 3 3I + Dehydroascorbate + 2HThe physical set up looks something like Figure 16-1 in your text. We will start-with ascorbate in the beaker and have I in the buret.3Now let’s look a bit at the chemistry. Vitamin C or Ascorbic acid and dehydroascorbicAcid have the structures:

What does a titration curve for a redox reaction look like? What are we going touse for the Y axis? Well, since all redox titrations involve the flow of electrons, allredox titrations can be monitored by looking at the electrical potential of the solution.As you found in the last chapter, all you need to monitor the potential of a solution is areference electrode, and in inert electrode. So we set up an apparatus like figure 16-1and monitor E as a function of titrant volume.So a titration curve will look some thing like this:As the titration occurs, what do you have inthe solution? A mixture of Ascorbate,- -dehydroascorbate, I 3 and I . The K’s ofredox equations are usually very large (doyou remember how to calculate the K forthis reaction? Give it a try)As a result you can usually assume that thereaction goes to 100% completion so youcan calculate the [] of each of the ionsusing a simple reaction table, rather thanworrying about K eq. So where does thepotential you measure come from?Well there are two ½ reactions going on:- + oDehydroascorbate + 2e +2H Ascorbate E ’=0.08 and- - - oI +2e 3I E '= .543Figure 2 Potentiometric titration curveAnd they are going on simultaneously, so either one can be used to calculate thepotential of the cell vs the reference electrode. The trick is that only 1 will give you areasonable value, the other will make your calculator give you an error because youare either multiplying or dividing by 0.There are only three equations you need to know when working with Redoxtitration curves. One equation is applied to get the titration curve up to the equivalencepoint, a second to get the E at the equivalence point, and a third (that looks a whole lotlike the first) is used after the equivalence point.Now notice that the book uses a calomel electrode as a reference electrode, andwill occasionally throw in a factor of E=E+ -E calomel to describe the actual measuredpotential. In this example I will assume we are using a hydrogen electrode as ourreference. This means that the reference potential is 0.00 so we don’t have to put itinto any of our equation so it won’t mess us up.2

The next point is important we have a platinum wire as an inert electrode todetect the potential of the solution in the beaker. What potential will it be detecting thepotential of the Iodine ½ reaction or the potential of the ascorbic acid half reaction??The answer is both. Since these reagents are mixed together, they are inequilibrium with each other, thus E = E and any point in the titration.acorbateSince the two redox systems we are titrating together are in equilibrium, thepotential of the two pairs are equal. As a result the potential of the of the reactantmixture can be found by calculating the E for either of the redox pairs. Because of thiswe always choose the redox pair that is the easiest to calculate.Let's define concentrations and volumes for this example. I am starting with a 40-mls of a .045 M (.09N) solution of ascorbate and I am titrating it with 0.05 M (.1N) I .iodine33Location of E.P.As usual lets first determine how much titrant is need to get to the equivalencepoint.If I have 40 mL. of .045 M solution, I have 1.8 mMoles of ascorbic acid. Thereaction stoichiometry is 1:1 so I will need 1.8 mMoles of I -3. I therefore need 1.8/.05, or 36 mL. of the iodine solution to reach theequivalence point.(It isn't always this easy, will examine again in a minute)E at the initial point hooray, you can't calculate this one. To calculate an Eusing the Nernst equation, you need to have good, well defined concentrations for boththe oxidized and the reduced species in a redox pair. For the initial point, no titrant asin solution, so it can't help you. You only have the analyte, and since none of it hasbeen converted to its redox partner, the concentration for one of the two redox speciesis not well defined. This is can't be calculated.E between the initial point and the equivalence point I said earlier the E ofthe system can be determined using the E of either Redox Pair, and you choose themost convenient pair. Between the initial point and the equivalence point you havemeasurable quantities of ascorbate and dehydroascorbate. However all the Iodine(actually triiodide) you add is converted into iodide, so the concentration of Iodine is notmeasurable. Thus the ascorbate-dehydroascorbate is the system you use to get the Eof the cell. Bottom line between start and EP use analyte

The analyte is the Vitamin C/Dehydroascorbic acid pair. So we will focus on this½ reaction and ignore the other. Let’s write down this ½ reaction:- + oDehydroascorbate + 2e +2H Ascorbate E '=0.08Note: that I am writing down this reaction as the reduction reaction, eventhough in my net reaction it is actually an oxidation! Get used to this. Because ourreduction potentials are for the reduction reactions, all of our math is always set up forreduction reactions, even if the reaction is reversed and turns into an oxidation in thenet reaction!oE=E -(.059/n) log ([reduced]/[oxidized])The potential for the ascorbate couple is:4+Second Note: Although H occurs in the half reaction, I am ignoring it in this equation.In this problem we are using Formal potentials for a solution at pH 7. I am assumingthat the solution is buffered, and the pH does not change. If this is so, then it may beremoved from the cell potential equation.Let’s figure out the moles of reactants, and fill in a reaction table.Moles Ascorbate = 40 ml x .045 M = 1.8 mmoles-Moles I = 10 ml x .05M = .5 mmoles3Reaction table:- - +Ascorbate + I 3 3I + Dehydroascorbate + 2HInitial 1.8 .5 0 0 1x10 -7Reaction -.5 -.5 +.5(3/1) +.5 1x10 -7Net 1.3 0.0 1.5 .5 1x10 -7+Since our solution was buffered at pH 7, the H concentration does not change, so thiscolumn of our reaction table can be ignored[Ascorbate]=1.3 mMoles/(40+10) ml = .026 M[Dehydroascorbate]=.5 mMoles/(40+10) ml = .01M

Note: I have converted both numbers into moles by dividing by the volume. Asharp student might have figured out that I could skip this step because the volumeterm will drop out of the Nernst equation. However as you will see in when wecalculate the point after the equivalence point, this doesn’t always happen, so I don’twant you getting into a bad habit here, so you should always be sure to convert tomolarity so don’t skip this step and have it bite you in the ...... on a tricky test question.Plugging these numbers into the Nernst Equation5At the equivalence pointThe equation for the equivalence point may be fairly easily derived, if thestoichiometry of the reaction is not too difficult. For those who are interested look onpages 424 of the book. I will not ask you to derive the equation, but you do need tomemorize it. The equation is:Thus the E at the equivalence point is the mean of the to E 's weighted by theonumber of electrons in each half reaction.For this example:

Beyond the equivalence pointNow all the Ascorbate has been turned into dehydroascorbate, so only a verytiny amount remains. On the other hand measurable amount of both Iodine and Iodideexist in solution, so it is most convenient to use the E of this half reaction to determinethe E of the Cell. Let's try the point after 40 mL. of titrant have been added.Let’s figure out the moles of reactants, and fill in a reaction table.Moles Ascorbate = 40 ml x .045 M = 1.8 mmoles-Moles I = 40 ml x .05M = 2.00 mmoles3Reaction table:6- - +Ascorbate + I 3 3I + Dehydroascorbate + 2HInitial 1.8 2 0 0 1x10 -7Reaction -1.8 -1.8 +1.8 (3/1) +1.8 1x10 -7Net 0 0.2 5.4 1.8 1x10 -7+Again our solution was buffered at pH 7, the H concentration does not change, so thiscolumn of our reaction table can be ignored-[I 3 ] = .2 mmoles/(40+40) ml = .0025M-[I ] = 5.4 mM /(40+40 ml) ml = .0675 M- -And E is determined using the Nernst of the I 3 /I pair.Notice here that the numerator of the Nernst quotient has a cubed term, whilethe denominator was to the 1 power. This is a case where you had to change tomolarity. If you didn’t and used just moles instead of moles/volume, the volumes termsdon’t cancel, and you would have been up the creek without a paddle.Example 2-3With the exception of the [I ] term the above titration was fairly simple becausethe number of electrons in both half reactions was equal. Things get a little messier ifthey aren't. Let's just briefly look at the example of:3+ 2+ 2+ 4+2Fe +Sn 2Fe + Sn3+ - 2+Fe + e Fe E=0.774+ - 2+Sn + 2e Sn E=0.153+Say we are starting with 40 mL. of .196 M Fe . First where is the equivalence

72+point when you titrate with .1 M Sn ?The tricky part of this problem is keeping track of the stoichiometry. Each mole2+ 3+of Sn reacts with 2 moles of Fe , therefore if you start with 40(.196) or 7.85 mMoles3+ 2+ 2+of Fe , you only need 3.92 mMoles of Sn , or 39.2 mL. of .1M Sn . You have toremember to keep this factor of two in all the appropriate places in the calculations.For points on the curve, let’s try 10.00, 39.2, and 44.2 mls of Sn 2+10 mlStarting with moles and a reaction table40 ml of .196M Fe = 7.84 mmole Fe10 ml of .100M Sn = 1.00 mmole Sn3+ 3+2+ 2+3+2Fe +2+Sn 2+2Fe +4+SnInitial 7.84 1.00 0.00 0.00Rxn -1(2/1) -1 +1(2/1) +1Net 5.84 0.00 2.00 1.003+[Fe ] = 5.84 mmoles/(40+10) ml = .1168M2+[Fe ] = 2.00mmole/(40+10) ml = .040MUsing the Analyte ½ reaction3+ - 2+Fe + e Fe(And ignoring the actual stoichiometry in the net reaction)The Nerst Equation is:E = .77 - .059/1 log (.04/.1168)= .77 -.059log(.3425=.77+.027=.797 V39.2 ml (equivalence point)Let’s just prove this is the equivalence point using moles and the reaction table:3+ 3+40 ml of .196M Fe = 7.84 mmole Fe2+ 2+39.2 ml of .100M Sn = 3.92 mmole Sn3+2Fe +2+Sn 2+2Fe +4+SnInitial 7.84 3.92 0.00 0.00Rxn -3.92(2/1) -3.92 +3.92(2/1) +3.92Net 0.00 0.00 7.84 3.00The E at the equivalence point is:(2(.15)+.77)/3 =.357

844.2 mlStarting with moles and a reaction table3+ 3+40 ml of .196M Fe = 7.84 mmole Fe44.2 ml of .100M Sn = 4.42 mmole Sn2+ 2+3+2Fe +2+Sn 2+2Fe +4+SnInitial 7.84 4.42 0.00 0.00Rxn -7.84 -7.84(½) +7.84 +7.84(1/2)Net 0 0.5 7.84 3.922+[Sn ] = .5 mmoles/(40+44.2) ml = .005938M4+[Sn ] = 3.92mmole/(40+44.2) ml = .04656MUsing the Titrant ½ reaction4+ - 2+Sn + 2e Sn(And ignoring that this reaction is actually an oxidation in the net reaction)E = .15 - .059/2 log (.005938/.04656)= .15 + .026= .176 V16-2 Redox IndicatorsA redox indicator is a compound that changes color when if goes from itsoxidized state to its reduced state.Just as in acid/base indicators we see distinct color when we have 10:1 or 1:10ration of indicator forms. Since we want our indicator to change color as close to ouroequivalence point as possible, you choose an indicator the has an E that closelymatches the E of your equivalence point ( to +/- .06V) . A list of indicators is given inTable 16-1Many times. However indicators are not needed. Frequently the redox reagentthemselves are distinctly colored.16-3 Titrations involving Iodine.Not covered unless decide to do in lab.