Lecture 27 11/07/05

Two things happening: First: Titration reaction (Ce 4+ + Fe 2+ ⇄ Ce 3+ + Fe 3+ ) –Goes to completion –Before equiv. point: excess Fe 2+ –After equiv. point: excess Ce 4+ Second: Reduction of Ce 4+ or Fe 3+ from cell potential –Negligible effect on concentration –Produces voltage for y-axis –2Fe 3+ +2Hg(l) +2Cl - ⇄ 2Fe 2+ +Hg 2 Cl 2 (s) –2Ce 4+ +2Hg(l) +2Cl - ⇄ 2Ce 3+ +Hg 2 Cl 2 (s)

At the equivalence point Y-axis: [Ce 3+ ]=[Fe 3+ ] and [Ce 4+ ]=[Fe 2+ ] Ce 3+ + Fe 3+ ⇄ Ce 4+ + Fe 2+

At the equivalence point E cell = ½(E° Ce + E° Fe ) E cell = ½( ) E cell = 0.99 V

Before the equivalence point Excess Fe E cell = E cathode – SCE X-axis: Good place to pick is ½ V e so that [Fe 3+ ] = [Fe 2+ ]

After the equivalence point Excess Ce X-axis: good place to pick is 2V e since [Ce 3+ ] = [Ce 4+ ]

Redox indicators Changes color when changes oxidation state –Look at E° to find good indicator In(ox) + ne -  In(red) E° = V –ferroin Can see red when [In(red)] / [In(ox)] > 10/1 Can see blue when [In(red)] / [In(ox)] < 1/10

Starch-Iodine complex Not a redox reaction I 2 forms I 6 inside amylose helix and turns dark blue Two things: 1) starch is biodegradable 2) Glucose (product of starch breakdown) is reducing agent

Pre-oxidation Must be quantitative Must be able to remove/destroy oxidizing agent Peroxydisulfate or persulfate (S 2 O 8 2- ) –Ag + as catalyst: S 2 O Ag +  SO SO Ag 2+ –Boiling: 2S 2 O H 2 O  4SO O 2 +H + AgO –Ag 2+ : powerful oxidizing agent –Boiling: 4Ag H 2 O  4 Ag + +O 2 + 4H + NaBiO 3 –BiO H + + 2e -  BiO + + 2H 2 O –Filter to remove excess H 2 O 2 –Good oxidizing agent in –Boiling: 2H 2 O 2  H 2 O + O 2 (gas)