Ch.16. REDOX TITRATIONS

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Ch.16. REDX TITRATINS 16.1 redox titrations Chemistry useful in biology, environmental Science, material Sc 161. The Shape of a Redox Titration Curve. 16.2 Consider titration of Fe 2+ with std Ce 4+ : potentiometric detection of end point Ce 4+ + Fe 2+ Ce + Fe K ~ 10 1 in 1M HCl 4 At reference electrode : Calomel.

1 Ch.16. REDX TITRATINS 16.1 redox titrations Chemistry useful in biology, environmental Science, material Sc 161. The Shape of a Redox Titration Curve Consider titration of Fe 2+ with std Ce 4+ : potentiometric detection of end point Ce 4+ + Fe 2+ Ce + Fe K ~ 10 1 in 1M HCl 4 At reference electrode : Calomel. 2Hg (l) + 2Cl Hg 2 Cl 2(s) + 2e At Pt indicator: two reactions Fe + e Fe 2+ E o = 0.6V Ce 4+ + e Ce E o = 1.0V which of the two is reduced? higher st. red. Potential Ce 4+ Ce

2 161. The Shape of a Redox Titration Curve combined possible cell reactions: 2Fe + 2Hg (l) + 2Cl 2Fe 2+ + Hg 2 Cl 2(s) or 2Ce Hg (l) + 2Cl 2Ce + Hg 2 Cl 2(s) How the cell voltage changes as Fe 2+ is titrated with Ce 4+? Reg1). Before the Equivalence point. Before eq. excess Fe 2+ exists Need to know [Fe 2+. [Fe E = E + E = = in case of the middle point of titration, [Fe 2+ = [Fe, E= Volt similar to ph = pk a at half 161. The Shape of a Redox Titration Curve Reg2). At the eq. point [Fe produced = [Ce produced very tiny amounts of [Ce 4+, [Fe 2+ present at equilibrium [Ce 4+ = [Fe 2+, [Ce = [Fe At eq. E + is the same for both half. [ Fe E + = log [ Fe 2+ [ Ce E + = log [ Ce 4+ 2E + = log [ Fe [ Fe 2+ [ Ce [ Ce E + = E + = 1.23V 2 E = E + E SCE = = 0.99V

3 161. The Shape of a Redox Titration Curve Req 3) After Eq. all [Fe 2+ [Fe = [Ce, extra [Ce 4+ remains as it is E = E + E SCE [ Ce = ( log ) [ Ce in case of V= 2V e (V e = volume at equil. point) [Ce = [Ce 4+ E + = 1.0 before after v = 2v e equili. point v The Shape of a Redox Titration Curve ex) Suppose that we titrate 100.0mL of M Fe 2+ with 0.100M Ce 4+ using the cell in Fig 161. The equivalence point occurs when V= 50.0mL, because the Ce 4+ is twice as concentrated as the Fe 2+. Calculate the cell voltage at 36.0, 50.0, and 63.0mL.

05916 [ In( red) E = E o log n [ In( oxid) reduced form of In. comes when vice versa o 0.

4 162. Finding the End Point 16. 1) Redox Indicators : color changes upon their oxidation and reduction state In (oxid) + ne In (red) [ In( red) E = E o log n [ In( oxid) reduced form of In. comes when vice versa o E = E ± n volts [ In( red) [ In( oxid) 10 1 for ferroin E o = 1.14 thus. Color change occur in < E < volt In case of SCE as reference, 0.84< E SCE < volt 162. Finding the End Point 16.8

162. Finding the End Point 16.9 Starch Iodine complex. (blue) titrations involving Indicator. Use starch as indicator Sugar amylose Starchiodine complex 164.

50V colorless in neutral or alkaline : Mn 4 + 4H + + 3e Mn 2(s) + 2H 2 E o = 1.692V brown solid in very strongly alkaline (2M NaH) : Mn 4 + e Mn 2 4 E o = 0.

5 162. Finding the End Point 16.9 Starch Iodine complex. (blue) titrations involving Indicator. Use starch as indicator Sugar amylose Starchiodine complex 164. xidation with Potassium permanganate (KMn 4 ) KMn 4 : oxidizing agent, violet color in ph < 1, self indicator, : Mn 4 + 8H + + 5e Mn H 2 E o =1.50V colorless in neutral or alkaline : Mn 4 + 4H + + 3e Mn 2(s) + 2H 2 E o = 1.692V brown solid in very strongly alkaline (2M NaH) : Mn 4 + e Mn 2 4 E o = 0.56V 1) Preparation and Standardization. KMn 4. Has trace amount of Mn 2 standardized by Na 2 C 2 4 (sodium oxalate) Mn 4 + 5H 2 C H + 2Mn C 2 + 8H 2 or by pure Fe wire put Fe in H 2 S 4 Fe 2+

6 1) pure distinctive rxn 165. xidation with Ce Ce 4+ + e Ce in acidic sol yellow colorless. Formal potential 1.0V 1F HCl V 1F HN 3 1.4V 1F HCl 1.44V 1F H 2 S 4` 166. xidation with K 2 Cr 2 (potassium dichromate) in acidic sol. Cr 2 : powerful oxidant red Cr : Cr 2 +14H + +6e 2Cr + H 2. E o =1.36V formal p. 1.00V in 1M HCl 1.11V in 2M H 2 S 4 less powerful oxidizing agent than Mn 4, Ce 4+ in basic. Cr 2 Cr 4 (yellow) converted. Cr H 2 + 3e Cr(H) 3(s) + 5H E o = 0.12V Advantages (primary std) color change self ind. can be monitored by Pt. and SCE

7 I) ii) I) (I 1. ii) 1. iii) 16. Methods involving Iodine i) Iodimetry : I 2 (titrant). actually I 3 reducing agent titrated with I 2 I Iodometry : oxidizing I to produce I 2 2 is titrated with NaS 2 3 ) thiosulfate I 2 : slightly soluble in water. to increase solubility, I 2(aq) + I I 2 3 K = 10 iodide triiodide (complex) iodine as a titrant means I Methods involving Iodine ) Use of starch indicator starch is used for iodine. Iodimetry : starch is added at just before equivalence. (titration turn dark blue. With I 3 ) since I 3 forms starch iodine complex (some iodine bounds to starch) In Iodimetry, Reasons for keeping the solution from strongly acidic i) starch hydrolyzes in strong acid. equilibrium is affected for several red. Agents H 3 As 3 + I 2 + H 2 H 3 As 4 + 2I + 2H + strong acid, reaction goes backward. produced I tends to oxidized. by 2. 4I H + 2I 2 + 2H 2

8 16. Methods involving Iodine ) Use of sodium thiosulfate In Iodometry Cr 2 + 6I + 14H + 2Cr + 3I 2 + H 2 universal titrant for I 3 in neutral or acidic. I 3 + 2S 2 3 3I + S 4 6 I 2 + 2S 2 3 2I + S 4 6. each Cr 2 reacts with 6S 2 3 Why not titratig oxidizing agent directly with thiosulfute? strong ox, agent (Mn 4, Cr 2, I 3 ) oxidize S 2 3 S Methods involving Iodine 16.16

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

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