Consider the analysis of a water sample for Cu2+ using DPASV at a hanging mercur
ID: 1075787 • Letter: C
Question
Consider the analysis of a water sample for Cu2+ using DPASV at a hanging mercury drop electrode.
a. The first step of the analysis is a “concentration” step in which a potential of -0.1 V vs. a Ag/AgCl reference is applied for several minutes. What chemical reaction takes place during this step?
b. The next step is the “stripping” step in which the applied potential is scanned from the initial -0.1 V vs. Ag/AgCl to 0.45 V vs. Ag/AgCl. What chemical reaction takes place during this step?
c. At what potential (vs. Ag/AgCl) should the copper peak appear?
d. Sketch the resulting voltammogram between the potentials of -0.1 V vs. Ag/AgCl and 0.45 V vs. Ag/AgCl.
Explanation / Answer
In the analysis of water sample for Cu2+ using DPASV at a hanging mercury drop method, the metal ion get deposited on the surface of electrode (mercury electrode) and deposited in the form of metal amalgam in the concentration step.
So the reaction occur
Cu2+ + 2e- + Hg -----------------> Cu(Hg).
In the next step which is stripping step metal amalgam again converted into its original form. So the reaction occur
Cu(Hg) --------------------> Cu2+ + 2e- + Hg
Cu2+ peak generally appear at 0.35 V when analysed against Ag/AgCl electrode.
Voltagramm can only be sketched when data is given like time, concentration of Cu2+ initially taken.