Transcript Tutorial 9

Instrumental Analysis
Polarography and Voltammetry
Tutorial 9
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Learning Outcomes
By the end of the tutorial, student should be able to:
1. Analyze voltamogram and polarogram results.
2. Apply standard addition technique to
voltametry and polarography.
3. Define cyclic voltammetry principles.
4. Analyze cyclic voltamogram
5. Apply CV principles on drug analysis.
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Example IIn 1 M NH3/1 M NH4Cl solution, Cu2+ is reduced to Cu+ near –0.3 V
(versus SCE), and Cu+ is reduced to Cu (on Hg) near –0.6 V.
a) Sketch a qualitative sampled current polarogram for a
solution of Cu+.
b) Sketch a polarogram for a solution of Cu 2+.
c) Suppose that Pt, instead of Hg, were used as the working
electrode. Which, if any, reduction potential would you
expect to change?
Solution
(c) The potential for the reaction Cu(I)
Cu(Hg) will
change if Pt is used, since the product obviously cannot
be copper amalgam and the potential at which reduction of
Cu+ starts definitely depends on the type of electrode
upon which redox reaction occurs.
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Example IISuppose that the diffusion current in a polarogram for reduction
of Cd2+ at a mercury electrode is 14 A. if the solution contains
25 mL of 0.50 M Cd2+, what percentage of Cd2+ is reduced in the
3.4 min required to scan from 0.6 to 1.2 V? comment!
• Solution: Number of moles of electrons flowing in 3.4 min=
(14  10 6 C / s)(3.4 min)(60 s / min)
 2.96  10  8 mol
96485 C / mol
• For the reaction:
moles of Cd2+ =
1/2
Cd2+ + 2e-
Cd(Hg)
moles of e- = 1.48 x 10-8 mol (forming amalgam)
• moles of Cd2+ in 25 mL of 0.5 M solution = 25 x 0.50 x 10-3
= 1.25 x 10-2 mol
(starting concentration)
• percentage Cd2+ of reduced =
1.48  10 8
 100  1.2  10  4 %
2
1.25  10
• Comment: In polarography, it is clear that a very small amount of Cd+2 (1.2 x
10-4%)has been reduced during the full time of the experiment(3.4 min). This is
in contrast to electrogravimetry and coulometry in which all the analyte
should be quantitatively reduced on the electrode surface. This little amount
is representative of the concentration in builidng up a calibration curve. 4
Example IIIThe polarogram of 3.00 mL of solution containing the
antibiotic tetracycline in 0.1 M acetate, pH 4, gives a
maximum current of 152 nA at a potential of –1.05 V (versus
S.C.E.). When 0.500 mL containing 2.65 mg/mL of tetracycline
was added, the current increased to 206 nA. Calculate the
concentration of tetracycline in the original solution.
Solution
In this case, both std. solution and unknown solution
are mutually dilute each other.
cu
i

cs ( dil .)  cu ( dil .)
i
cu (mg / mL )
152 nA

206 nA
 0.500 
 3.00 
2.65 
  cu 

 3.50 
 3.50 
cu  0.760 mg/mL
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Example IV- (Self Study)
The drug Librium gives a polarographic wave with E1/2 = 0.265 V
(versus SCE) in 0.05 M H2SO4. A 50.0 mL sample containing Librium
gave a wave height of 0.37 A. When 2.00 mL of 3.00 mM Librium in
0.05 M H2SO4 were added to the sample, the wave height increased
to 0.80 A. Find the molarity of Librium in the unknown.
Answer: 0.096 mM
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Cyclic Voltammetry (CV)
final potential
Reversible
reaction
1st cycle
2nd cycle
Initial potential
Criteria for reversible reactions:
Irreversible
reaction
 The ratio of the cathodic peak current (Ipc) to the anodic peak
current (Ipa) should be 1.
 The half-wave potential (E1/2) is the same
for the cathodic and anodic waves.
 The separation between the cathodic peak potential
(Epc) and the anodic peak potential (Epa) should be close to 59/n mV
Ep = |Epa - Epc| = 59/n mV
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Example VI-
Consider the cyclic voltammogram of the Co 3+ compound Co(B9C2H11)2 .
Suggest a chemical reaction to account for each wave. Are the
reaction reversible? How many electrons are involved in each
step? Sketch the sampled current polarogram expected for this
compound.
First wave
Second wave
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Solution
•
We see two consecutive reactions.
•
From the value of
Epa – Epc, we find that one electron is
involved in each reduction (Ep = |Epa - Epc| = 59/n mV )
•
•
A possible sequence of reaction is:
Co3+
+ e-
Co2+
+ e-
Co2+
Co+
The equality of the anodic and cathodic heights suggests that
the reactions are reversible (This is confirmed by the same
value of half wave potential E1/2 in the cathodic and anodic
•
The expected sampled current
polarogram:
current
directions).
E
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voltage
Example VII- A bonus idea
The cyclic voltammogram of the antibiotic chloramphenicol
(abbreviated RNO2) is shown in the figure below. The scan was
started at 0 V, and the potential was swept in the negative
voltage. The first cathodic wave, A, is from the reaction
RNO2
+ 4e- + 4H+
RNHOH + H2O.
Explain what happens at peaks B and C using the reaction
RNO + 2e- +2H+
RNHOH
Why was peak C not seen in the initial scan?
Answer
• Peak B is due to the anodic reaction:
RNHOH
RNO + 2e- +2H+
• Peak C is due to the cathodic reaction:
RNO + 2e- +2H+
RNHOH
• There was no peak C in the first run because
no RNO intermediate B present before the
initial scan.
Try to solve problems 17-22, 17-26, and 17-27
(Harris text book, p403-404)
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