Transcript BrCH(CO 2 H) 2 + 4 Ce +4 + 2 H 2 O HCO 2 H + 2 CO 2 + Br

Oscillation Lab Discussion
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Path A Part I:
Amber Color
• BrO3-1 + 5Br -1+ 6H+1 → 3 Br2 + 3 H2O then:
• Br2 + CH2(CO2H)2 → BrCH(CO2H)2 + Br-1 + H+1
Path A Part II:
Colorless
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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 Br
2
+ 3H2O
Start of Path A again
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Colorless
Yellow
• 2 BrO3-1 + 12H+1 + 10Ce+3 → Br2 + 6 H2O + 4 Ce+4
Yellow
Red
Colorless
Blue
• Ce+4 + Fe+2 → Ce+3 + Fe+3
• Ce+3 + Fe+3 → Fe+2 + Ce+4
Colorless
Blue
Red
Yellow
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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.
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Blue (472.2 nm)
Green (564.6 nm)
Red (635.4 nm)
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Min absorbance = 635.4 nm = red light
Time = 5.25 minutes
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
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Min absorbance = 472.2 nm = blue light
Time = 5.65 minutes
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
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Oscillation Frequency = reaction cycles per
minute = 1 reaction cycle/(2.35 min-1.25 min)
= .91 cycles per minute
Begin Cycle
End Cycle
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E = E0 – .059 log [Ce(III)]/[Ce(IV)]
E = -.056 volts
n
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
be an
area amount
of of Ce
• This should
should correspond
to a maximum
of
Ce+4 ion,
+3
ion, and a minimum amount
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?
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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?
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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
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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
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FERROIN SOLUTION USED IN EXCESS…
BrCH(CO2H)2 + 4 Ce+4 + 2 H2O  HCO2H + 2 CO2 + Br-1 + 4 Ce+3 + 5 H+1
Yellow
Red
Colorless
Blue
Ce+4 + Fe+2  Ce+3 + Fe+3
Ce+3 + Fe+3  Fe+2 + Ce+4
Colorless
Blue
Red
Yellow
• 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)
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Oscillating biological mechanisms…
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Oscillating biological mechanisms…
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Oscillating biological mechanisms…
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