Amperometry is one of the sensitive electroanalytical methods used for quantitative determination of — Physical Chemistry — Kinetics Chemistry Question
Amperometric titration: Titration of Pb 2+ with Cr2O7 2-
Amperometry is one of the sensitive electroanalytical methods used for quantitative determination of electroactive species. A working electrode is held at a certain voltage (vs reference electrode) that is suitable for oxidation or reduction of an analyte. The analyte will be oxidized or reduced at the surface of the working electrode and the current passing through this electrode is measured. This current is directly proportional to the concentration of analyte and is used for quantitative purpose. It can be used for detection of titration end-point. In this example amperometric detection was used for monitoring a titration progress. A volume of 20 cm 3 of a lead (II) ion solution was titrated with 0.0020 mol dm -3 potassium dichromate solution. A dropping mercury electrode (DME) was used as a working electrode and the potential of -0.8 V (vs SCE, saturated calomel electrode) was applied to this electrode for enabling the reduction of lead (II) ion and dichromate ion with potassium nitrate as supporting electrolyte. The current-volume of titrant data are shown in Table 1.
Reference: Vogel’s Textbook of Quantitative Chemical Analysis, 5 th edition, John Wiley & Sons, New York, pp630.
Table 1. Titration data
Volume (cm 3 ) of 0.0020 mol dm -3 dichromate | Current (microampere)
0.00 | 9.8
2.00 | 8.0
4.00 | 6.0
6.00 | 4.0
8.00 | 2.2
10.00 | 3.5
12.00 | 5.5
14.00 | 7.6
16.00 | 9.5
Plot the titration curve and find the titration end point. (The point where the change in the slope of titration curve occurs)
Model Answer
By extrapolating the two straight lines, the titration end-point is located.
Write the titration reaction.
Model Answer
Pb 2+ (aq) + Cr2O7 2- (aq) → PbCr2O7(s)
Calculate the concentration of lead(II) cations.
Model Answer
n(Pb 2+) = n(dichromate)
0.0020 mol dm -3 × 0.008 dm 3 = c(Pb 2+) × 0.020 dm 3
Therefore, c(Pb 2+) = 0.0020 mol dm -3 × 0.008 dm 3 / 0.020 dm 3 = 8.0×10 -4 mol dm -3