1

Application of O2 Activation toward Organic Pollutant Degradation

Derek F. Laine and I. Frank ChengUniversity of IdahoChemistry DepartmentMoscow, ID 83843-2343lain3267@uidaho.eduifcheng@uidaho.edu208-885-6387

The ZEA Organic Pollutant Degradation System

2

ZEA Pollutant Degradation System

Zero valent iron (ZVI) EDTA

(Ethylenediaminetetraacetic acid)

Air

Stir Plate

Stir bar and ZVI particles

Open round bottom flask

Aqueous Solution of 4-chlorophenol

3

The Search For Alternatives to the Bulk Destruction of Organic Pollutants

High temperature use of O2 Incineration

Expensive Dioxins Public reluctance

Low temperature use of O2 ZEA system

Operates at room temperature and pressure Inexpensive Common reagents Long term storage No specialized catalysts Simple Reactor Design Easily transportable Versatile (can be applied to water treatment)

4

Destruction of 4-Chlorophenol

y = 1.23E-03e-1.16E+00x

R2 = 9.66E-01

0

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

0.0014

0 0.5 1 1.5 2 2.5 3 3.5 4

Time (hours)

Con

cent

ratio

n (M

olar

ity)

Reaction Curve

Expon. (Reaction Curve)

Noradoun, Christina, et al. Ind. Eng. Chem. Res. 2003, 42, 5024-5030.

Products include low molecular weight acids and CO2.

5

Pollutants destroyed by the ZEA System Halocarbons

4-chlorophenol Pentachlorophenol

Organophosphorus Compounds (nerve agents) Malathion (vx surrogate) Malaoxon

Organics EDTA Phenol

6

Hypothesis-Oxygen Activation

Oxygen has a triplet ground state, while organic compounds have a singlet ground state.

How to overcome this kinetic barrier. Add energy in the form of heat. Addition of electrons (activation)

The ZEA system works by Reducing O2 to form reactive oxygen species

O2.-, H2O2, HO.

http://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html

7

Hypothesis-Site for O2 Activation (I) Heterogeneous

activation at the ZVI surface.

(II) Homogeneous activation by FeIIEDTA.

I Fe(0)

O2FeIIIEDTA + HO∙ + HO-

H+ H2O2

Fe2+ + EDTA → FeIIEDTA

II Fe(0)Fe2+ + EDTA FeIIEDTA

FeIIIEDTA O2

H+

H2O2

FeIIIEDTAHO∙ + HO-

8

Electrochemical Homogeneous Degradation System - Cell Design

Three electrode system:

1. Working electrode• (RVC)

2. Auxiliary electrode• Graphite rod• A salt bridge keeps

the auxiliary electrode separated from the bulk solution.

3. Reference electrode• Ag/AgCl

9

Electrochemical Pollutant Degradation System

• FeIIEDTA can reduce oxygen to form the superoxide ion (O2·- ), as well as other reactive oxygen species.

• Degradation of EDTA is measured in this system

• HPLC is used to measure the degradation of EDTA.

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH▪

+

FeIIEDTA

10

Experimental Conditions

• FeIII(NO3)3 and Na2H2EDTA were added in a 1:1 ratio to make 80 ml of a 0.5 mM FeIIIEDTA solution.

• -120 mV potential is applied to the working electrode.

• A high stir rate and large surface area working electrode is used to facilitate fast and efficient electrolysis.

• KCl is used as the supporting electrolyte.

• Oxygen is bubbled through the system.

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH▪

+

FeIIEDTA

11

HPLC Results

FeEDTA degradationAbsorption detection at 258 nm

-5

0

5

10

15

20

0 1 2 3 4

minutes

mV

t=0 hr

t=1.0 hr

t=2 hr

t=3 hr

12

Results

FeIIIEDTA degradation

0

0.1

0.2

0.3

0.4

0.5

0.6

0 2 4 6 8

time (hr)

[Fe

ED

TA

] m

M

W/O2 (-120mV)

W/N2 (-120mV)

W/O2 (+120mV)

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH▪

+

FeIIEDTA

13

Comparison of FeII/IIIEDTA degradation and pH

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1 2 3 4 5 6 7

hours

[FeE

DTA

] mM

012345678910

pH

FeEDTA

pH

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH▪

+

FeIIEDTA

14

Detection of Intermediate Oxidizing Agents (H2O2 and HO·)

Graf, Ernst; Penniston, John T. Method for Determination of Hydrogen Peroxide, with its Application illustrated by Glucose Assay. Clin. Chem. 1980, 26/5, 658-660.

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH·

+

FeIIEDTA

Electrochemical system ZEA system

15

Formation of H2O2

Starch reagents concentrated starch 40 mM HCl 0.077 mM ammonium molybdate 80 mM KI.

Add an aliquot of reaction mixture to starch reagents and analyze with UV-VIS after a 20 minute color formation period.

Any suitable oxidizing agent (such as H2O2) will oxidize the iodide to iodine.

Iodine combines with iodide to form triiodide which will then complex with starch to form a blue color.

H2O2(aq) + 3I-(aq) + 2 H+(aq) → I3-(aq) + 2 H2O(aq)

E. Graf, J.T. Penniston, Clin. Chem. 26/5 (1980) 658-660.

16

Formation of H2O2

H2O2 formation

-0.05

0

0.05

0.1

0.15

0.2

0 2 4 6 8

Time (Hrs)

[H2O

2] m

M

O2 bubbled, -120mV, 0.5 mMFeEDTAO2 bubbled, -120mV, no FeEDTA

N2 bubbled, -120mV, 0.5 mMFeEDTAO2 bubbled,+500mV, 0.5 mMFeEDTA

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH·

+

FeIIEDTA

17

Formation of HO·

•Accomplished using the spin trapping abilities of 5,5-dimethylpyrroline-N-oxide (DMPO) and electron spin resonance spectroscopy (ESR).

•The DMPO-HO· adduct has a well characterized 1:2:2:1 quartet.

Das, Kumuda C.; Misra, Hara P. Mol. Cell. Biol. 2004, 262, 127-133.

Yamazaki, Isao; Piette, Lawrence H. J. Am. Chem. Soc. 1991, 113, 7588-7593.

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH·

+

FeIIEDTA

N+

O

H

H O

N

O

H

OH+ ·

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Formation of HO·

-80000

-60000

-40000

-20000

0

20000

40000

60000

80000

3480 3490 3500 3510 3520 3530 3540 3550

G

Inte

ns

ity

•Before electrolysis, the same signal is obtained from a simple solution of FeIIIEDTA, KCl, and O2

-80000

-60000

-40000

-20000

0

20000

40000

60000

80000

3480 3490 3500 3510 3520 3530 3540 3550

G

Inte

ns

ity

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH·

+

FeIIEDTA

19

Formation of HO·

-80000

-60000

-40000

-20000

0

20000

40000

60000

80000

3480 3490 3500 3510 3520 3530 3540 3550

G

Inte

ns

ity

N+

O

H

Fe(III)

OH

HN

O

H

OH

EDTA

Fe(II)

EDTA+ -80000

-60000

-40000

-20000

0

20000

40000

60000

80000

3480 3490 3500 3510 3520 3530 3540 3550

G

Inte

ns

ity

•The two processes can be distinguished by adding methanol as a scavenger.

N+

O

H

H O

N

O

H

OH+ ·

20

Formation of HO··CH3OH H O CH2OH+ ·

N+

O

H

CH2OHN

O

H

CH2OH+ ·-80000

-60000

-40000

-20000

0

20000

40000

60000

80000

3480 3490 3500 3510 3520 3530 3540 3550

G

Inte

ns

ity

T=3 hr, 20% Methanol

T=3 hr, no Methanol

N+

O

H

Fe(III)

EDTA

CH3OHN

O

H

OMe Fe(II)

EDTA+

-80000

-60000

-40000

-20000

0

20000

40000

60000

80000

3480 3490 3500 3510 3520 3530 3540 3550

G

Inte

ns

ity

T=3 hr, no Methanol

T=0 hr, 20% Methanol

21

0

20000

40000

60000

80000

100000

120000

140000

0 100 200 300 400 500 600

Seconds

Inte

nsi

ty

A

B

Formation of HO·

A)

B) N+

O

H

Fe(III)

OH

HN

O

H

OH

EDTA

Fe(II)

EDTA+

N+

O

H

H O

N

O

H

OH+ ·

-80000

-60000

-40000

-20000

0

20000

40000

60000

80000

3480 3490 3500 3510 3520 3530 3540 3550

G

Inte

ns

ity

Growth of the quartet when adding the reaction mixture to DMPO after electrolysis.

Growth of the quartet when adding the reaction mixutre to DMPO before electrolysis

Reaction dominates after electrolysis. K = 109 M-1 S-1

Reaction dominates before electrolysis

Yamazaki, Isao; Piette, Lawrence H. J. Biol. Chem. 1990, 265, 13589-13594

22

Formation of HO·Intensity of 1:2:2:1 quartet after 3 hrs of

electrolysis and variable amounts of CH3OH scavenger

0

10000

20000

30000

40000

50000

-5 5 15 25 35 45

[Methanol] % v/v

Inte

ns

ity

23

Formation of HO·H2O2 Formation

0

0.05

0.1

0.15

0.2

0 1 2 3 4 5 6 7

Time (hr.)

[H2O

2] m

M

Intensity vs electrolysis time

0

10000

20000

30000

40000

50000

60000

0 1 2 3 4 5 6 7

Time (hr)

Inten

sity

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH·

+

FeIIEDTA

24

Cyclic voltammetry can be used to show the catalytic mechanism.

FeIIIEDTA + e- → FeIIEDTA

FeIIEDTA + O2 → FeIIIEDTA + O2·-

FeIIIEDTA

FeIIEDTA

2O2

2O2°- + 2H+ → H2O2 + O2

FeIIIEDTA + OH- + OH·

+

FeIIEDTA

25

Cyclic Voltammetry

Oxygen Activation pH=3.0

-0.20

0.20.40.60.8

11.21.41.61.8

-0.5-0.3-0.10.10.3

V

uA

FeIIIEDTA + O2

O2 only

FeIIIEDTA only

Niether FeIIIEDTA or O2

5 mV/s

26

pH Dependency

Current vs pH

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

1.2000

0 2 4 6 8 10 12

pHCu

rrent

(uA)

Zang, V; van Eldik, R. Inorg. Chem. 1990, 29, 1705-1711.

Oxygen Activation pH=3.0

-0.20

0.20.40.60.8

11.21.41.61.8

-0.5-0.3-0.10.10.3

V

uA

27

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 2 4 6 8 10 12pH

Cu

rren

t (u

A)

0.00.10.20.30.40.50.60.70.80.91.01.1

0 2 4 6 8 10 12

pH

Mo

le F

ract

ion

Free Fe(II)

FeIIEDTA(H)

FeIIEDTA

FeIIEDTA(H2)

FeIIEDTA(OH)

28

Geometrical Considerations

[FeII(EDTA)(H2O)]2- + H+ = [FeII(EDTAH)(H2O)]1-

Mizuta, T.; Wang, J.; Miyoshi, K. Bull. Chem. Soc. Jpn. 1993, 66, 2547-2551.Mizuta, T.; Wang, J.; Miyoshi, K. Inorg. Chimica Acta. 1993, 230, 119-125.

Species Bite angle on water coordinate

Bond distance from FeII to OH2

FeIIEDTA 164.0° 2.19 Å

FeIIEDTAH 172.1° 2.21 Å

29

Summary and Conclusion The ZEA system can destroy organic pollutants non-

selectively. How does the ZEA system destroy pollutants?

The ZEA system has a homogeneous reaction mechanism with activation of oxygen by FeIIEDTA followed by the Fenton reaction.

The ZEA system produces H2O2 as an intermediate. The ZEA system produces HO· which can non-selectively destroy organic

pollutants.

How can the ZEA system be made to work better? Bubble air or oxygen through the system. Optimize for pH = 3 conditions.

30

Acknowledgments

Dr. I. Frank Cheng Simon McAllister University of Idaho Dept. of Chemistry ACS Funding

NSF award number BES-0328827NIH Grant No. 1 R15 GM062777-01