Instrumental Analysis
CHEM 42 / 790
Problem Set 3
Problems 22-1, 22-2, 22-9
Problems 23-1 to 23-5, 23-14 and 23-15
Problems 24-4 (a/b), 24-4, 24-6, 24-10
Problems 25-1 to 25-6, 25-11
22-1
Calculate the electrode potentials for these half-cells using the Nernst equation
Ag(I) (0.0261M) | Ag
Since Ag(I) + 1e–  Ag(s) has a standard electrode potential of
+0.799 V vs SHE, E0 = 0.799V.
E = E0 – 2.303(RT/nF) log (aAg(s) /aAg(I))
E = 0.799 – (0.0592/n) log (1/aAg(I))
= 0.799 – (0.0592/1) log (1/0.0261M) = 0.799V – 0.094V
= 0.705V
Fe(III) (0.672 mM), Fe(II) (100 mM) | Pt
Since Fe(III) + 1e–  Fe(II) has a standard electrode potential of
+0.771V vs SHE, E0 = 0.771V.
E = E0 – 2.303(RT/nF) log (aFe(II) /aFe(III))
E = 0.771 – (0.0592/n) log (100mM/0.672 mM)
= 0.642 V
AgBr (sat’d), Br– (0.050M) | Ag
Since Ag(I) + 1e–  Ag(s) has a standard electrode potential of
+0.799 V vs SHE, E0 = 0.799V. In addition, the solubility product of AgBr is
5 x 10-13 M.
E = E0 + 2.303(RT/nF) log (Ksp) – 2.303(RT/nF) log (aBr-)
E = 0.799 + 2.303(RT/nF) log (5 x 10-13 M) – (0.0592/n) log (0.050M)
= 0.799 – 0.728 + 0.077 = 0.148 V
= 0.148 V
22-2
HCl (1.76M) | H2 (0.987 atm) | Pt
H+ + 1e–  ½ H2 has a standard electrode potential of +0.00 V vs SHE,
by definition.
E = E0 – 2.303(RT/nF) log (PH21/2 /cHCl)
E = 0.000 – (0.0592/1) log (0.9934/1.76M)
= 0.000 + 0.0147V
= 0.015V
IO3– (0.194M), I2 (2.00 x 10-4 M), H+ (3.50 x 10-3) | Pt
IO3– + 6 H+ + 5e–  ½ I2 + 3H2O has a standard electrode potential of
+1.195 V vs SHE.
E = E0 – 2.303(RT/nF) log (aI2 /aIO3-[aH+]6)
E = 1.195 – (0.0592/5) log [(2.00 x 10-4 M)1/2] /[(0.194 )(3.50 x 10-3)6])
E = 1.195 – (0.0592/5) log [(1.41 x 10-2 M)] /[3.56 x 10-16])
= 1.195 – (0.0592/5) log (3.97 x 1013)
= 1.195 – (0.0118) 13.59
= 1.195 – 0.160V
= 1.035 V
Ag2CrO4(sat’d), CrO42- (0.0520M) | Ag
Since Ag2CrO4+ 2e–  Ag(s) has a standard electrode potential of
+0.446 V vs SHE. The solubility product of Ag2CrO4
is 1.1 x 10-12 M = [Ag+]2[CrO4].
E = E0 – 2.303(RT/nF) log (aCrO42-)
E = 0.446V – 2.303(RT/2F) log (0.0520M)
E = 0.446 + 0.038 V
E = 0.484
22-9
Ni(II)(CN)42– + 2e–  Ni(s) + 4CN–
The formation constant, Kf, for Ni(CN)42– is 1.0 x 1022.
Kf = [Ni(II)(CN)42–] / [Ni(II)][CN–]4
The standard reduction potential of
Ni(II)+ 2e–  Ni(s)
is -0.250 V
E = E0 – 2.303(0.0592/n) log (aNi(s) / Kf)
E = –0.250V –2.303(0.0592/2) log (1/1.0 x 1022)
E = –0.250 V – 0.651V
E = –0.901 mV
Problems 23-1 to 23-5, 23-14 and 23-15
23-1
Nernstian behavior refers to the slope of the electrode response with respect to
concentration. 10-fold changes in concentration yield electrode changes of 59.2 mV/n,
where n is the number of electrons involved in the redox reaction.
23-2
The alkaline error observed for glass pH electrodes is due to exchange between
the hydrogen ions on the glass surface with monovalent cations in solution. The activity
of the monovalent cations is greater than that of the hydrogen ions at these low hydrogen
ion concentrations.
23-3
Ref. Electrode
0.1 M Calomel
3.5 M Calomel
Saturated Calomel
3.5 M Ag-AgCl
Sat’d Ag-AgCl
V35°C – V15°C
-0.0024
-0.008
-0.0135
-0.015
-0.020
The data above reveal that each electrode has a negative temperature coefficient
and that 0.1M Calomel has the smallest temperature coefficient of the electrodes
listed. The saturated Ag-AgCl reference electrode has the largest temperature
coefficient of the electrodes listed.
23-4
Upon immersion of a new glass pH electrode in water for the first time, the glass
hydrates. The hydration process involves dissolution of monovalent cations from the
glass, typically sodium cations, and replacement with hydrogen ions.
23-5
An electrode of the first kind is a metallic indicator electrode where the metal that
composes the electrode is used to detect that metal’s ion in solution, e.g. a Zn electrode
for detecting Zn(II). An electrode of the second kind is also a metallic inidicator
electrode, but it is used to detect analytes different from the metal that composes the
electrode. Electrodes of the second kind detect changes in potential due to complex
formation or precipitation.
23-14
CuBr(s) + 1e–  Cu(s) + Br–
The solubility product of CuBr is 5.2 x 10-9 M.
The standard potential for this reaction is:
E = E0 + 2.303(RT/nF) log (Ksp) – 2.303(RT/nF) log (aBr-)
E = 0.520 V+ 2.303(RT/nF) log (5.2 x 10-9 M) – 2.303(RT/nF) log (1 M)
= 0.520 V – 0.490 V – 0 V
= 0.030V
The schematic representation for an electrode to detect Br2 is:
SCE || CuBr(sat’d), Br- (conc M)| Cu(s)
The equation relating pBr (–log[Br2]) to the potential is:
Ecell = Eright – Eleft = E – 0.0592 log[Br–] – ESCE
E = 0.030 V – 0.0592 log[Br–] – 0.244V
cell
Ecell = 0.030 V + 0.0592 pBr – 0.244V
Ecell – 0.030 V + 0.244V = 0.0592 pBr
pBr = (Ecell – 0.030 V + 0.244V)/0.0592
pBr = (Ecell + 0.214V)/0.0592
A cell with a measured potential of –0.095V relative to SCE would have a pBr of:
pBr = (Ecell + 0.214V)/0.0592
pBr = (–0.095V + 0.214V)/0.0592
pBr = (0.119V)/0.0592
pBr = 2.0 or
[Br–] of 10-2 M (0.010 M)
23-15
Ag3AsO4(s) + 3e–  3Ag(s) + AsO4–
The solubility product of Ag3AsO4 is 1.2 x 10-22 M
Note: Ksp = [Ag+]3[ AsO4–] = 1.2 x 10-22 M
For Ag(I) + 1e–  Ag(s), E0 = 0.799V
The standard potential for this reaction is:
E = E0 – 2.303(RT/1F) log (1/[Ag+]) Note: n = 1
= E0 – 2.303(RT/3F) log (1/[Ag+]3) Note: n = 3
= E0 + 2.303(RT/3F) log (Ksp) – 2.303(RT/3F) log (aAsO4-)
E = 0.799 V+ 2.303(RT/3F) log (1.2 x 10-22 M) – 2.303(RT/3F) log (1 M)
= 0.799 V – 0.431 V – 0 V
= 0.367V
The equation relating pAsO4– (–log[AsO4–]) to the potential is:
Ecell = Eright – Eleft = E – (0.0592/3) log[AsO4–] – ESCE
E = 0.367V – (0.0592/3) log[Br–] – 0.244V
cell
Ecell – 0.367 V + 0.244V = (0.0592/3) pAsO4–
pAsO4– = (Ecell – 0.367 V + 0.244V)3/0.0592
pAsO4– = 3(Ecell – 0.123V)/0.0592
For a cell saturated in and possessing a potential of 0.247V vs. SCE, the [AsO4–] is:
pAsO4– = 3(Ecell – 0.123V)/0.0592
pAsO4– = 3(0.247 – 0.123V)/0.0592
pAsO4– = 6.3
AsO4– = 10-6.3 M , or 4.95 x 10-7 M
Problems 24-4 (a/b), 24-4, 24-6, 24-10
24-4
The solubility products of the silver(I)halides are given below.
AgCl 1.77 x 10-10
AgBr 5.35 x 10-13
AgI 8.52 x 10-17
Since the standard potential involves the Ksp, the halide with the largest will form
the Ag salt first.
E = E0 + 2.303(RT/nF) log (Ksp) – 2.303(RT/nF) log (aX-)
Cl–
E = 0.799 V + 2.303(RT/nF) log (1.77 x 10-10) – 2.303(RT/nF) log (0.025 M)
= 0.799 V– 577 V– 0.095 V
= 0.127 V
Br–
E = 0.799 V+ 2.303(RT/nF) log (5.35 x 10-13) – 2.303(RT/nF) log (0.025 M)
= 0.799 V– 0.726V– 0.095 V
= –0.022 V
I–
E = 0.799 V + 2.303(RT/nF) log (8.52 x 10-17) – 2.303(RT/nF) log (0.025 M)
= 0.799 V– 0.951V– 0.095 V
= –0.247 V
AgCl would form first, at a potential of +0.127 V vs. SHE
This is a galvanic cell, as no external potential is applied.
The potential need to lower [Br–] to 1 x 10-5 M, is
Br–
E = 0.799 V+ 2.303(RT/nF) log (5.35 x 10-13) – 2.303(RT/nF) log (1 x 10-5 M)
= 0.799 V– 0.726V– 0.296 V
= –0.220 V
at the potential of –0.220 V, the
I–
–0.220 V = 0.799 V + 2.303(RT/nF) log (8.52 x 10-17) – 2.303(RT/nF) log ([I–])
0.799 V– 0.951V + 0.220V = 2.303(RT/nF) log ([I–])
log ([I–]) = 1.14
[I–] = 6.01 x 10-2 M
so at –0.220 V, the [I–] could be as high as 6.01 x 10-2 M, since it is 0.025 M, the
separation is feasible. The potential could be set between –0.220 V and –0.247 V.
24-6
0.270g of cobalt is 4.51 x 10-3 moles of cobalt which will require 9.02 x 10-3
moles of electrons to reduce to elemental from Co(II). There are 96,485 coulombs per
mole of electrons, so 9.02 x 10-3 moles of electrons is 871 C. At a rate of 0.750 A (C/s),
it will take 1,161 seconds to electrolyze 0.270g of cobalt onto the cathode.
0.270g of Co3O4 is 1.13 x 10-3 moles. These moles contain 3 moles of cobalt,
two moles as Co(III) and one mole as Co(II). Thus, 2e– are required for each mole of
Co3O4 produced from three moles of Co(II). The reaction will require 2.25 x 10-3 moles
of electrons, or 217 C. At a rate of 0.750 A, this will take 290s.
24-10
A current of 0.0441A for 266s produces 11.73C of charge, or 1.21 x10-3 moles of
electrons. This means that the 0.0809g sample contained 1.21 x10-3 moles, or the
equivalent mass of the acid is 655 g/mol.
Problems 25-1 to 25-6, 25-11
25-1
In voltammetry, the potential is instantaneous current is measured as a function of
applied potential, thus, potential is a variable. In amperometry, the potential is
fixed and the total current is measured.
Linear scan voltammetry changes the potential in a linear fashion, pulsed methods
use different waveforms (such as a square wave) on top of the linear scan.
A rotating disk electrode is a disk electrode which is spun to allow for flow of
solution over the electrode surface. A ring disk electrode has two electrodes
arranged concentrically and analytes may be detected independently at either
electrode.
Faradaic impedence is the resistance of the electrode due to electron and mass
transfer reactions and is frequency dependent. Double-layer capacitance is due to
charging of the double-layer at the electrode surface and is not frequency
dependent.
The limiting current is the maximal current achievable in an electrochemical cell.
It includes the diffusion current (the current due to diffusion) as well as other
mass transfer currents (kinetic current, migration current etc.)
Laminar flow is flow with a smooth and regular motion. It occurs below the
Reynolds number which is related to the viscosity of the solution. At flow rates
above the Reynolds number, one observes turbulent flow which is irregular.
The standard electrode potential is the electrode potential at the conditions of the
standard state of unit activity (1M solute, 1 atm for gasses, 25°C at pH = 1 (1M
H+)) vs. the standard hydrogen electrode (SHE). The half-wave potential is the
potential at which ½ of the limiting current is passed in a voltammetry
experiment.
In normal stripping voltammetry, the analyte is concentrated onto the electrode
surface using an electrochemical process (oxidation or reduction). The analyte is
then stripped off the electrode using the reverse electrochemical process
(reduction or oxidation). In adsorptive stripping voltammetry, the electrode is
immersed in the analyte solution and the analyte absorbs to the electrode surface
without any redox reaction.
25-2
Voltammagram, the plot of the current vs. the applied potential
Hydrodynamic voltammetry, voltammetry performed on solutions in motion
relative to the working electrode surface. This may be done by rotating the
electrode or by flowing the solution over a stationary electrode by stirring or
pumping
Nernst diffusion layer, the layer of solution at the electrode surface where flow
velocity is essentially zero. The lack of convection results in concentration
gradients within the Nernst diffusion layer.
Hg film electrode, an electrode formed by the electrodeposition of Hg onto a disk
electrode surface.
Half-wave potential, the potential at which ½ of the limiting current is passed in a
voltammetry experiment.
Voltammetric sensor, a complete voltammetric cell which is used to detect an
analyte.
25-3
High concentrations of supporting electrolyte are used in electrochemical
experiments to negate the effects of analyte migration on the electrochemical
response.
25-4
The reference electrode is placed close to the working electrode so that it is in as
close of an environment as the working electrode.
25-5
It is important to buffer solutions in all types of voltammetry. Redox reactions
frequently involve the uptake or release of protons and therefore are pH sensitive.
Buffers prevent shifts in the pH that would result in changes in the potential.
25-6
Stripping methods are more sensitive than other forms of voltammetry because
they concentrate the analyte at the electrode surface prior to analysis.
25-11
The peak current in a CV experiment is described by the Randles-Sevcik
equation:
ip = 2.686x105 n3/2AcD1/2v1/2
If the peak currents for the Pb(II) and Cd(II) reductions are equal, at an identical
electrodes (the areas are identical):
ip(Cd(II)) = 2.686x105 n3/2AcD1/2v1/2 = 2.686x105 n3/2AcD1/2v1/2 = ip(Pb(II))
ip(Cd(II)) = 2.686x105 23/2A(4.38 mM) (0.72 x10-5 cm2s-1)1/2v1/2
= 2.686x105 23/2A(0.167) (0.98 x10-5 cm2s-1)1/2(2.5V/s)1/2 = ip(Pb(II))
(4.38 mM) (0.72 x10-5 cm2s-1)1/2v1/2
= (0.167) (0.98 x10-5 cm2s-1)1/2(2.5V/s)1/2
1.17x10-2 v1/2 = 8.26x10-4
v = 4.98 mV/s