Redox Titrations. -the oxidation/reduction reaction between analyte and titrant. -the equivalence point is based upon:

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1 Redox Titrations -the oxidation/reduction reaction between analyte and titrant -titrants are commonly oxidizing agents, although reducing titrants can be used -the equivalence point is based upon: A ox + B red A red + B ox Rx n goes to completion after each addition of titrant Potentiometric Titration:

2 Titration reaction: Reference half-reaction: Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ (1) 2Hg(l) + 2Cl- Hg 2 Cl 2 (s) + 2e- At the Pt indicator electrode (Indicator half-reaction) Fe 3 + e- Fe 2+ E 0 = V (2) Ce 4+ + e- Ce 3+ E 0 = 1.70V (3) Cell reactions (in 1 M HClO 4 ): 2Fe Hg(l) + 2Cl - 2Fe 2+ + Hg 2 Cl 2 (s) (4) 2Ce Hg(l) + 2Cl - 2Ce 3+ + Hg 2 Cl 2 (s) (5) Relationships - Cell reactions are not the same as the titration reaction - May describe the cell voltage with either (4) or (5) or both

3 Balance: -atoms -# of electrons transferred Example: Balancing Redox Reactions Cr(s) + Ag+ Cr 3+ + Ag(s) 1. Write the half reactions: 2. Balance the electrons: 3. Recombine

4 Equilibrium constants for oxidation-reduction reactions Cu(s) + 2Ag Cu Ag(s) Galvanic cell: [Cu ] [Ag ] 2+ K eq = + 2 E cell = E cathode E anode = E Ag+ - E Cu2+ Under equilibrium conditions, the potential of the cell becomes zero, thus can write: E cell = O = E cathode E anode = E Ag+ - E Cu2+ or E cathode = E anode = E Ag+ = E Cu2+ -Also when in equilibrium, electrode potentials of all systems are identical: E Ox1 = E Ox2 = E Ox3 = E Ox4 Where E Ox1..are electrode potentials for the four halfreactions

5 Calculating Equilibrium Constants Cu(s) + 2Ag Cu Ag(s) E 0 Ag [Ag log = ] + 2 E 0 Cu log 2 1 [Cu2 + ] E 0 Ag+ - E 0 Cu2+ = 2( log 2 [ Ag+ ] ) log 2 [Cu 2 2+ [Cu ] [Ag+] E Ag + E Cu2+ = = log = log K 2 eq Ex: Calculate the equilibrium constant: ] 2+ [Cu ] 2( ) log K eq = log = 2 [Ag+] = 15.6 K eq = antilog 15.6 = 4.1 x = 4 x Redox Titration Curves Fe 2+ + Ce 4+ Fe 3+ + Ce 3+ E Ce 4+ = E Fe3+ = E system E In = E Ce4+ = E Fe3+ = E system

6 Equivalence Point Potentials Fe 3 + e- Fe 2+ Ce 4+ + e- Ce [Ce 3+ ] E 0 eq = E log Ce4+ 1 [Ce 4 + ] [Fe 2 + ] E 0 eq = E log Fe3+ 1 [Fe 3+ ] [Ce ][Fe ] 2 E 0 0 eq = E + E (1) Ce4 + Fe [Ce ][Fe ] Definition of e.p. requires that: [Fe 3+ ] = [Ce 3+ ] [Fe 2+ ] = [Ce 4+ ] [Ce ][Ce ] 2 E 0 0 eq = E + E log = Ce4 + Fe E 0 0 Ce4 + + E Fe 3 + [Ce ][Ce ] E eq = E0 0 3 Ce E Fe (2) 2

7 The Derivation of Titration Curves Titration of ml of M Fe 2+ with M Ce 4+ in a solution that is 1.0 M in H 2 SO 4 at all times. Ce 4+ + e - Ce 3+ E f = 1.44V Fe 3+ + e- Fe 2+ E f = 0.68V 1. Initial potential Ce and Fe 3+ only present in very small amounts. 2. Potential after addition of 5.00 ml of Ce 4+ [Fe 3+ ]= 5.00 x [Ce 4 + ] [Fe 2+ ] = x x [Ce ] Substitution into Nernst equation: E = log = 0.64V system / / E.P. potential E eq f f E E Ce 4 + Fe = + + = = V

8 3. Potential after addition of ml of Ce 4+ [Fe 2+ ] = amt of Ce 4+ left unreacted, therefore added to C Ce4+ calculated from the volumes of the two solutions and subtracted from C Ce3+ Conc of two cerium ion species: [Ce 3+ ]= x [ Fe2 + ] [Ce ] E = log = log [Ce ] 1 = V Effect of system variables on redox titration curves 2.500/ / Concentration independent of analyte and reagent concentrations. Exception: Electrode potentials dependent upon dilution I 3 +2e - 3 I [I ] E = E 0 log 2 [I - ] num-mol/l 3, denom-mol/l 3 Completeness of reaction the change in E system in the e.p. region becomes larger as the reaction becomes more concentrated.

9 Redox indicators a. specific indicators react with one of the participants in the titration to produce a color, e.g. thiocyanate b. Oxidation-reduction indicators- respond to the potential of the system rather than to the appearance or disappearance of some species during the course of the titration, e.g. methylene blue Color changes will occur over the range: E = ( E 0 ± ) Volts n where n= # of electrons in the indicator half-reaction -larger diff in std potential between titrant and analyte, the sharper the break in the titration curve at the e.p. 0.2 V, best detected potentiometrically

10 Gran plot - more accurate way to use potentiometric data - uses data well before e.p. (V e ) to locate V e For the oxidation of Fe 2+ to Fe 3+, the potential prior to V e is: 2 + ' [Fe ] E = [ E log( ) E 3 + [Fe ] ref where, E 0' = formal potential for Fe 3+ Fe 2+ and Eref is the potential of the reference electrode. If vol of analyte = V 0 and the vol of titrant = V, and if reaction goes to completion with each addition of titrant: [Fe2+] / [Fe3+] = (V e -V) V 10 -ne/ = V e 10 -n(eref E0 )/ V 10 -n(eref E0 )/ y b x m

11 Adjustment of Analyte Oxidation State -before titration, e.g. Mn 2+ preoxidized to MnO 4 - -excess preadjustement reagent must be destroyed so that it will not interfere in subsequent titration Preoxidation -powerful oxidants can be removed after preoxidation, e.g. peroxydisulfate (S 2 O 8 2- ) requires Ag+ as a catalyst. S 2 O Excess reagent destroyed: Ag + SO2 - + SO - + Ag S O2 - boiling + 2 H O 4SO2 - + O + 4H Prereduction -Stannous chloride (SnCl 2 ) will reduce Fe 3+ to Fe 2+ in hot HCl Excess reductant is then destroyed: Sn HgCl 2 Sn 4+ + HgCl Cl -

12 Oxidation with Potassium Permanganate -strong oxidant, violet color In strongly acidic solutions, reduced to colorless Mn 2+ : MnO H + + 5e - Mn H 2 0 In neutral or alkaline solution, the product is the brown solid, MnO 2 : MnO H + + 3e - MnO 2 (s) + 2H 2 O In strongly alkaline solution (2 M NaOH), green manganate is produced: MnO e- MnO 4 2- Tales see below

13 Note: permanganate solutions are unstable, therefore not a primary standard. 4MnO H 2 O > 4MnO 2 + 4OH - + 3O 2 (MnO 2 catalyses this reaction) Permanganate must be standardized for example with oxalate; H 2 C 2 O 4 > 2H + + CO 2 + 2e - Overall:. 2MnO H 2 C 2 O H + > 2Mn CO 2 + 8H 2 O Initially the reaction is slow but is catalyzed by Mn 2+ so becomes more rapid. Can also standardize with arsenic (III) oxide As(III) > As(V) + 2e- The reaction of As (III) with permanganate ion takes place without complications in acidic medium if a trace of an iodine compound (for example potassium iodate) is added as a catalyst. The reaction generally carried out in HCl rather than H 2 SO 4... in the latter a brown green coloration occurs due to formation of a manganese arsenate compound

14 KMnO4 can serve as own indicator, since product Mn 2+ is colorless.

15 Cerium(IV) Strong oxidant > Ce(III) Ce 4+ [Yellow ]+ e - > Ce 3+ [Colorless] Note however that the color change not good enough for it to act as self indicator. Ce(IV) not found in acid solution as simple aqua ion.. forms complexes. Dichromate reactions Dichromate ion is an oxidizing agent Cr 2 O H + + 6e - > 2Cr H 2 O E = +1.33V Dichromate has replace permanganate in many analyses... notably iron (II)... it can be prepared as a standard solution and so avoids the need to standardize as is the case with permanganate. Iodine Methods I 2 + 2e - > 2I - E = +0.54V Value for E is intermediate can therefore be reduced or oxidized...iodine can be reduced to iodide by for example As(III), Sn(II) whilst iodide can be oxidized to iodine by for example permanganate. Use of iodide as titrant..practical problems..so add excess potassium iodide and titrate the liberated iodine with for example standard thiosulphate solution

16 Miscellaneous Oxidizing agents Sodium bismuthate and lead (IV) oxide are strong oxidizing agents. NaBiO 3 + 6H + + 2e - > Na + + Bi H 2 O E = +1.6V PbO 2 + 4H + + 2e - > Pb H 2 O E = +1.5V Hydrogen peroxide... strong oxidant even in alkaline conditions. H 2 O 2 + 2H + + 2e - > 2H 2 O E = +1.77V Excess peroxide may be removed by boiling...decomposes

17 Ex: Derive a titration curve for the titration of ml of M U 4+ with M Ce 4+