1 Oscillation Lab Discussion

2 BrO Br H +1 → 3 Br H 2 O then:BrO Br H +1 → 3 Br H 2 O then: Br 2 + CH 2 (CO 2 H) 2 → BrCH(CO 2 H) 2 + Br -1 + H +1Br 2 + CH 2 (CO 2 H) 2 → BrCH(CO 2 H) 2 + Br -1 + H +1 BrO Br H +1 → 3 Br H 2 O then:BrO Br H +1 → 3 Br H 2 O then: Br 2 + CH 2 (CO 2 H) 2 → BrCH(CO 2 H) 2 + Br -1 + H +1Br 2 + CH 2 (CO 2 H) 2 → BrCH(CO 2 H) 2 + Br -1 + H +1 Amber Color Colorless Path A Part II: Path A Part I:

3 BrCH(CO 2 H) Ce H 2 O → HCO 2 H + 2 CO 2 + Br Ce H +1BrCH(CO 2 H) Ce H 2 O → HCO 2 H + 2 CO 2 + Br Ce H +1 High enough concentration causes: BrO Br H +1 → 3 Br 2 + 3H 2 OBrO Br H +1 → 3 Br 2 + 3H 2 O From Path A end: From Path B end: Start of Path A again

4 2 BrO H Ce +3 → Br H 2 O + 4 Ce +42 BrO H Ce +3 → Br H 2 O + 4 Ce +4 Colorless Yellow Ce +4 + Fe +2 → Ce +3 + Fe +3Ce +4 + Fe +2 → Ce +3 + Fe +3 Ce +3 + Fe +3 → Fe +2 + Ce +4Ce +3 + Fe +3 → Fe +2 + Ce +4 Yellow Red Colorless Blue Yellow Red Colorless Blue

5 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.

6 Blue (472.2 nm) Green (564.6 nm) Red (635.4 nm)

7 Min absorbance = nm = red light Max absorbance = nm = yellow light Max absorbance = 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 = nm = red light Max absorbance = nm = yellow light Max absorbance = 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 minutes

8 Min absorbance = nm = blue light Min absorbance = nm = green light Max absorbance = nm = red light Therefore, Max concentration = Ce (III) colorless and Fe (III) blue = BLUE Min concentration = Ce (IV) yellow and Fe (II) red Min absorbance = nm = blue light Min absorbance = nm = green light Max absorbance = nm = red light Therefore, Max concentration = Ce (III) colorless and Fe (III) blue = BLUE Min concentration = Ce (IV) yellow and Fe (II) red Time = 5.65 minutes

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

10 E = volts Time = 5.65 minutes E = volts Time = 5.65 minutes BrCH(CO 2 H) Ce H 2 O  HCO 2 H + 2 CO 2 + Br Ce H +1 E = E 0 –.059 log [Ce(III)]/[Ce(IV)] n E = E 0 –.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 E 0 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 E 0 The voltage reading is actually at a maximum – what is going on?

11 E = volts Time = 5.25 minutes E = volts Time = 5.25 minutes E = E 0 –.059 log [Ce(III)]/[Ce(IV)] n E = E 0 –.059 log [Ce(III)]/[Ce(IV)] n BrCH(CO 2 H) Ce H 2 O  HCO 2 H + 2 CO 2 + Br Ce 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 E 0, 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 E 0, and increase the E The voltage reading is at a minimum though – why? In addition, why are all of the voltages negative?

12 BrCH(CO 2 H) Ce H 2 O  HCO 2 H + 2 CO 2 + Br Ce H +1 Ce +4 + e -  Ce +3 E 0 = 1.61 volts BrCH(CO 2 H) 2  CO 2 E 0 =.49 volts Cerium is reduced, and carbon is oxidized The E tot, then, equals E oxid + E red = 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 E 0 of the circuit There are also other reactions taking place that affect the voltage of the system Ce +4 + e -  Ce +3 E 0 = 1.61 volts BrCH(CO 2 H) 2  CO 2 E 0 =.49 volts Cerium is reduced, and carbon is oxidized The E tot, then, equals E oxid + E red = 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 E 0 of the circuit There are also other reactions taking place that affect the voltage of the system

13 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

14 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(CO 2 H) Ce H 2 O  HCO 2 H + 2 CO 2 + Br Ce H +1 Ce +4 + Fe +2 Ce +3 + Fe +3 Ce +4 + Fe +2  Ce +3 + Fe +3 Ce +3 + Fe +3 Fe +2 + Ce +4 Ce +3 + Fe +3  Fe +2 + Ce +4 Yellow Red Colorless Blue Yellow Red Colorless Blue

15 Oscillating biological mechanisms…

16 Oscillating biological mechanisms…

17 Oscillating biological mechanisms…