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.

Lecture 2711/07/05

Two things happening:

First: Titration reaction (Ce4+ + Fe2+ ⇄ Ce3+ + Fe3+)– Goes to completion– Before equiv. point: excess Fe2+

– After equiv. point: excess Ce4+

Second: Reduction of Ce4+ or Fe3+ from cell potential– Negligible effect on concentration– Produces voltage for y-axis

– 2Fe3+ +2Hg(l) +2Cl- ⇄ 2Fe2+ +Hg2Cl2 (s)

– 2Ce4+ +2Hg(l) +2Cl- ⇄ 2Ce3+ +Hg2Cl2 (s)

At the equivalence point Y-axis: [Ce3+]=[Fe3+] and [Ce4+]=[Fe2+] Ce3+ + Fe3+ ⇄ Ce4+ + Fe2+

o

ceriumoironcathode

43

32ocerium

oironcathode

4

3ocerium3

2oironcathode

4

3oceriumceriumcathode

3

2oironironcathode

EE21E

]Ce][Fe[

]Ce]Fe[log05916.0EEE2

]Ce[

]Ce[log05916.0E

]Fe[

]Fe[log05916.0EE2

]Ce[

]Ce[log

1

05916.0EEE

]Fe[

]Fe[log

1

05916.0EEE

At the equivalence point

Ecell = ½(E°Ce + E°Fe) - 0.241

Ecell = ½(1.70 + 0.767) - 0.241

Ecell = 0.99 V

Before the equivalence point Excess Fe Ecell = Ecathode – SCE X-axis: Good place to pick is ½ Ve so that [Fe3+] =

[Fe2+]

V526.0E

241.0767.0E

241.0]Fe[

]Fe[log05916.0767.0E

3

2

After the equivalence point

Excess Ce X-axis: good place to pick is 2Ve since [Ce3+] = [Ce4+]

V46.1E

241.070.1E

241.0]Ce[

]Ce[log05916.070.1E

4

3

Redox indicators

Changes color when changes oxidation state– Look at E° to find good indicator

In(ox) + ne- In(red) E° = 1.147 V – ferroin

Can see red when [In(red)] / [In(ox)] > 10/1 Can see blue when [In(red)] / [In(ox)] < 1/10

V962.0 E

244.010

1log

1

05916.0 - 147.1 E blue when

V844.0 E

244.01

10log

1

05916.0 - 147.1 red when E

Starch-Iodine complex

Not a redox reaction

I2 forms I6 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 (S2O8

2-)

– Ag+ as catalyst: S2O82- + Ag+ SO4

2- + SO4- + Ag2+

– Boiling: 2S2O82- + 2H2O 4SO4

2- + O2 +H+

AgO– Ag2+ : powerful oxidizing agent

– Boiling: 4Ag2+ + 2 H2O 4 Ag+ +O2 + 4H+

NaBiO3

– BiO3- + 4H+ + 2e- BiO+ + 2H2O

– Filter to remove excess H2O2

– Good oxidizing agent in

– Boiling: 2H2O2 H2O + O2 (gas)