1Oscillation Lab DiscussionOscillation Lab Discussion

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• BrOBrO33-1-1 + 5Br + 5Br -1-1+ 6H+ 6H+1+1 → → 3 Br3 Br22 + 3 H + 3 H22O then:O then:

• BrBr22 + CH + CH22(CO(CO22H)H)22 →→ BrCH(CO BrCH(CO22H)H)22 + Br + Br-1-1 + H + H+1+1

• BrOBrO33-1-1 + 5Br + 5Br -1-1+ 6H+ 6H+1+1 → → 3 Br3 Br22 + 3 H + 3 H22O then:O then:

• BrBr22 + CH + CH22(CO(CO22H)H)22 →→ BrCH(CO BrCH(CO22H)H)22 + Br + Br-1-1 + H + H+1+1

Amber ColorAmber Color

ColorlessColorlessPath A Part II:Path A Part II:

Path A Part I:Path A Part I:

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• BrCH(COBrCH(CO22H)H)2 2 + 4 Ce+ 4 Ce+4+4 + 2H + 2H22O O →→ HCO HCO22H + 2 COH + 2 CO22 + Br + Br-1-1 + 4 Ce + 4 Ce+3+3 + 5H + 5H+1+1• BrCH(COBrCH(CO22H)H)2 2 + 4 Ce+ 4 Ce+4+4 + 2H + 2H22O O →→ HCO HCO22H + 2 COH + 2 CO22 + Br + Br-1-1 + 4 Ce + 4 Ce+3+3 + 5H + 5H+1+1

High enough concentration causes:High enough concentration causes:

• BrOBrO33-1-1 + 5 Br + 5 Br-1-1 + 6H + 6H+1 +1 →→ 3 Br 3 Br22 + 3H + 3H22O O • BrOBrO33-1-1 + 5 Br + 5 Br-1-1 + 6H + 6H+1 +1 →→ 3 Br 3 Br22 + 3H + 3H22O O

From Path A end:From Path A end: From Path B end:From Path B end:

Start of Path A againStart of Path A again

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• 2 BrO2 BrO33-1-1 + 12H + 12H+1+1 + 10Ce + 10Ce+3 +3 →→ Br Br22 + 6 H + 6 H22O + 4 CeO + 4 Ce+4+4• 2 BrO2 BrO33-1-1 + 12H + 12H+1+1 + 10Ce + 10Ce+3 +3 →→ Br Br22 + 6 H + 6 H22O + 4 CeO + 4 Ce+4+4

ColorlessColorless YellowYellow

• CeCe+4+4 + Fe + Fe+2 +2 →→ Ce Ce+3+3 + Fe + Fe+3+3• CeCe+4+4 + Fe + Fe+2 +2 →→ Ce Ce+3+3 + Fe + Fe+3+3

• CeCe+3 +3 + Fe+ Fe+3+3 →→ Fe Fe+2+2 + Ce + Ce+4+4• CeCe+3 +3 + Fe+ Fe+3+3 →→ Fe Fe+2+2 + Ce + Ce+4+4

YellowYellow RedRed ColorlessColorless BlueBlue

YellowYellowRedRedColorlessColorless BlueBlue

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• The redox reaction would take place and continue even without the addition of ferroin, as the ferroin is not part of the oscillation reaction

• The ferroin’s purpose is to simply be oxidized and reduced by the Ce+3/Ce+4 couple which changes the ferroin periodically from Fe+2, which is red, to Fe+3, which is blue, adding to the visual color change.

• The redox reaction would take place and continue even without the addition of ferroin, as the ferroin is not part of the oscillation reaction

• The ferroin’s purpose is to simply be oxidized and reduced by the Ce+3/Ce+4 couple which changes the ferroin periodically from Fe+2, which is red, to Fe+3, which is blue, adding to the visual color change.

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Blue (472.2 nm)Blue (472.2 nm)Blue (472.2 nm)Blue (472.2 nm)

Green (564.6 nm)Green (564.6 nm)Green (564.6 nm)Green (564.6 nm)

Red (635.4 nm)Red (635.4 nm)Red (635.4 nm)Red (635.4 nm)

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Min absorbance = 635.4 nm = red light

Max absorbance = 564.6 nm = yellow light

Max absorbance = 472.2 nm = blue light

Therefore,

Max concentration = Ce (IV) yellow and Fe (II) red = ORANGE

Min concentration = Ce (III) colorless and Fe (III) blue

Min absorbance = 635.4 nm = red light

Max absorbance = 564.6 nm = yellow light

Max absorbance = 472.2 nm = blue light

Therefore,

Max concentration = Ce (IV) yellow and Fe (II) red = ORANGE

Min concentration = Ce (III) colorless and Fe (III) blue

Time = 5.25 minutesTime = 5.25 minutes

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Min absorbance = 472.2 nm = blue lightMin absorbance = 472.2 nm = blue light

Min absorbance = 564.6 nm = green lightMin absorbance = 564.6 nm = green light

Max absorbance = 635.4 nm = red lightMax absorbance = 635.4 nm = red light

Therefore, Therefore,

Max concentration = Ce (III) colorless and Fe (III) blue = Max concentration = Ce (III) colorless and Fe (III) blue = BLUEBLUE

Min concentration = Ce (IV) yellow and Fe (II) redMin concentration = Ce (IV) yellow and Fe (II) red

Min absorbance = 472.2 nm = blue lightMin absorbance = 472.2 nm = blue light

Min absorbance = 564.6 nm = green lightMin absorbance = 564.6 nm = green light

Max absorbance = 635.4 nm = red lightMax absorbance = 635.4 nm = red light

Therefore, Therefore,

Max concentration = Ce (III) colorless and Fe (III) blue = Max concentration = Ce (III) colorless and Fe (III) blue = BLUEBLUE

Min concentration = Ce (IV) yellow and Fe (II) redMin concentration = Ce (IV) yellow and Fe (II) red

Time = 5.65 minutesTime = 5.65 minutes

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Oscillation Frequency = reaction cycles per minute = 1 reaction cycle/(2.35 min-1.25 min) = .91 cycles per minute

Oscillation Frequency = reaction cycles per minute = 1 reaction cycle/(2.35 min-1.25 min) = .91 cycles per minute

Begin CycleBegin CycleEnd CycleEnd Cycle

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E = -.056 volts

Time = 5.65 minutes

E = -.056 volts

Time = 5.65 minutes

BrCH(CO2H)2 + 4 Ce+4 + 2 H2O HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1BrCH(CO2H)2 + 4 Ce+4 + 2 H2O HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1

E = E0 – .059 log [Ce(III)]/[Ce(IV)]

n

E = E0 – .059 log [Ce(III)]/[Ce(IV)]

n

This should be an area of

• Time of 5.65 minutes should represent a minimum voltage - why?

• Time of 5.65 minutes, as shown earlier, represents a blue color and a maximum amount of Ce (III) and Fe (III)

• This should correspond to a maximum amount of Ce+3 ion, and a minimum amount of Ce+4 ion, according to the Nernst Equation

•This makes the log ratio large, which creates a negative value taken from E0

•The voltage reading is actually at a maximum – what is going on?

• Time of 5.65 minutes should represent a minimum voltage - why?

• Time of 5.65 minutes, as shown earlier, represents a blue color and a maximum amount of Ce (III) and Fe (III)

• This should correspond to a maximum amount of Ce+3 ion, and a minimum amount of Ce+4 ion, according to the Nernst Equation

•This makes the log ratio large, which creates a negative value taken from E0

•The voltage reading is actually at a maximum – what is going on?

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E = -.051 volts

Time = 5.25 minutes

E = -.051 volts

Time = 5.25 minutes

E = E0 – .059 log [Ce(III)]/[Ce(IV)]

n

E = E0 – .059 log [Ce(III)]/[Ce(IV)]

n

BrCH(CO2H)2 + 4 Ce+4 + 2 H2O HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1BrCH(CO2H)2 + 4 Ce+4 + 2 H2O HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1

•Time of 5.25 minutes should represent a maximum voltage - why?

•Time of 5.25 minutes, as shown earlier, represents a orange color and a maximum amount of Ce (IV) and Fe (II)

•This should correspond to a maximum amount of Ce+4 ion, and a minimum amount of Ce+3 ion, according to the Nernst Equation

•The ratio [Ce(III)]/[Ce(IV)] would be small; the log [Ce(III)]/[Ce(IV)] would be negative

• A negative value would create a positive addition to E0, and increase the E

•The voltage reading is at a minimum though – why?

•In addition, why are all of the voltages negative?

•Time of 5.25 minutes should represent a maximum voltage - why?

•Time of 5.25 minutes, as shown earlier, represents a orange color and a maximum amount of Ce (IV) and Fe (II)

•This should correspond to a maximum amount of Ce+4 ion, and a minimum amount of Ce+3 ion, according to the Nernst Equation

•The ratio [Ce(III)]/[Ce(IV)] would be small; the log [Ce(III)]/[Ce(IV)] would be negative

• A negative value would create a positive addition to E0, and increase the E

•The voltage reading is at a minimum though – why?

•In addition, why are all of the voltages negative?

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BrCH(CO2H)2 + 4 Ce+4 + 2 H2O HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1BrCH(CO2H)2 + 4 Ce+4 + 2 H2O HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1

• Ce+4 + e- Ce+3 E0 = 1.61 volts• BrCH(CO2H)2 CO2 E0 = .49 volts• Cerium is reduced, and carbon is oxidized• The Etot, then, equals Eoxid + Ered = 1.61 V + .49 V = 2.1V• When Ce+3 was at its max, the voltage was not near this value• This was not reached - why?• Resistance of the circuit? From where?•The reaction continues to oscillate between states, so there is no constant voltage• The entire experiment we read a near zero voltage•The reaction takes place completely in one container, so electrons are free to flow freely from cerium to iron, preventing a measurement of voltage - they are exchanged immediately between each other•The concentrations of products and reactants don’t vary enough to considerably change the E of the circuit; however, if this is the case, we should at least be measure the E0 of the circuit•There are also other reactions taking place that affect the voltage of the system

• Ce+4 + e- Ce+3 E0 = 1.61 volts• BrCH(CO2H)2 CO2 E0 = .49 volts• Cerium is reduced, and carbon is oxidized• The Etot, then, equals Eoxid + Ered = 1.61 V + .49 V = 2.1V• When Ce+3 was at its max, the voltage was not near this value• This was not reached - why?• Resistance of the circuit? From where?•The reaction continues to oscillate between states, so there is no constant voltage• The entire experiment we read a near zero voltage•The reaction takes place completely in one container, so electrons are free to flow freely from cerium to iron, preventing a measurement of voltage - they are exchanged immediately between each other•The concentrations of products and reactants don’t vary enough to considerably change the E of the circuit; however, if this is the case, we should at least be measure the E0 of the circuit•There are also other reactions taking place that affect the voltage of the system

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OSCILLATION FREQUENCY IS HIGHER….OSCILLATION FREQUENCY IS HIGHER….

• Why…?• Higher oscillation frequency means the overall reaction is occurring faster per unit time• This means the rate of the reaction, or kinetics, are higher• Could be an increase in temperature, or increase in concentration, or the use of a catalyst• Perhaps, instead of cerium being used as a transition catalyst, another transition metal catalyst is being used• Literature suggests that manganese can be used as an alternative to cerium

• Why…?• Higher oscillation frequency means the overall reaction is occurring faster per unit time• This means the rate of the reaction, or kinetics, are higher• Could be an increase in temperature, or increase in concentration, or the use of a catalyst• Perhaps, instead of cerium being used as a transition catalyst, another transition metal catalyst is being used• Literature suggests that manganese can be used as an alternative to cerium

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FERROIN SOLUTION USED IN EXCESS…FERROIN SOLUTION USED IN EXCESS…

• One oscillation occurred, and then the solution went red…• Why?• Ferroin contains Fe (II)• Fe (II) is oxidized by Ce (IV) to Fe (III) and Ce (III)• This will cause one oscillation• An excess of Fe (II) will use up all of the Ce (IV)• The oscillation will stop• With an excess Fe (II), the solution will appear red, as this is the color of Fe (II)

• One oscillation occurred, and then the solution went red…• Why?• Ferroin contains Fe (II)• Fe (II) is oxidized by Ce (IV) to Fe (III) and Ce (III)• This will cause one oscillation• An excess of Fe (II) will use up all of the Ce (IV)• The oscillation will stop• With an excess Fe (II), the solution will appear red, as this is the color of Fe (II)

BrCH(CO2H)2 + 4 Ce+4 + 2 H2O HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1BrCH(CO2H)2 + 4 Ce+4 + 2 H2O HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1

CeCe+4+4 + Fe + Fe+2 +2 Ce Ce+3+3 + Fe + Fe+3+3CeCe+4+4 + Fe + Fe+2 +2 Ce Ce+3+3 + Fe + Fe+3+3

CeCe+3 +3 + Fe+ Fe+3+3 Fe Fe+2+2 + Ce + Ce+4+4CeCe+3 +3 + Fe+ Fe+3+3 Fe Fe+2+2 + Ce + Ce+4+4

YellowYellow RedRed ColorlessColorless BlueBlue

YellowYellowRedRedColorlessColorless BlueBlue

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Oscillating biological mechanisms…Oscillating biological mechanisms…

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Oscillating biological mechanisms…Oscillating biological mechanisms…

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Oscillating biological mechanisms…Oscillating biological mechanisms…