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Oscillation Lab Discussion. Amber Color. Path A Part I:. BrO 3 -1 + 5Br -1 + 6H +1 → 3 Br 2 + 3 H 2 O then: Br 2 + CH 2 (CO 2 H) 2 → BrCH(CO 2 H) 2 + Br -1 + H +1. Colorless. Path A Part II:. From Path A end:. From Path B end:.
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Amber Color Path A Part I: • BrO3-1 + 5Br -1+ 6H+1→ 3 Br2 + 3 H2O then: • Br2 + CH2(CO2H)2→ BrCH(CO2H)2 + Br-1 + H+1 Colorless Path A Part II:
From Path A end: From Path B end: • BrCH(CO2H)2 + 4 Ce+4 + 2H2O → HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5H+1 High enough concentration causes: • BrO3-1 + 5 Br-1 + 6H+1 → 3 Br2 + 3H2O Start of Path A again
Colorless Yellow • 2 BrO3-1 + 12H+1 + 10Ce+3 → Br2 + 6 H2O + 4 Ce+4 Blue Yellow Red Colorless • Ce+4 + Fe+2 → Ce+3 + Fe+3 • Ce+3 + Fe+3→ Fe+2 + Ce+4 Colorless Blue Red Yellow
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.
Blue (472.2 nm) Green (564.6 nm) Red (635.4 nm)
Time = 5.25 minutes 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 = 472.2 nm = blue light Min absorbance = 564.6 nm = green light Max absorbance = 635.4 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
Oscillation Frequency = reaction cycles per minute = 1 reaction cycle/(2.35 min-1.25 min) = .91 cycles per minute Begin Cycle End Cycle
E = E0– .059 log [Ce(III)]/[Ce(IV)] n 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+1 • 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? This should be an area of
E = E0– .059 log [Ce(III)]/[Ce(IV)] n E = -.051 volts Time = 5.25 minutes BrCH(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?
BrCH(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
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
FERROIN SOLUTION USED IN EXCESS… BrCH(CO2H)2 + 4 Ce+4 + 2 H2O HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1 • 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) Blue Yellow Red Colorless Ce+4 + Fe+2 Ce+3 + Fe+3 Ce+3 + Fe+3 Fe+2 + Ce+4 Colorless Blue Red Yellow
