For historical, political, economic, technological as well as biological and biochemical reasons, ir — Physical Chemistry — Kinetics Chemistry Question
Iron chemistry
For historical, political, economic, technological as well as biological and biochemical reasons, iron is one of the most important elements on the periodic table. In the following task some general aspects concerning the chemistry of iron will be discussed from the viewpoint of physical chemistry.
First, let us explore the available redox states of iron in detail.
Another frequently used type of redox (and acid-base) equilibrium plot is the Pourbaix diagram. It involves individual pH-dependent redox potentials. Let us look into the problem more deeply.
For simplicity, activities will be replaced with equilibrium concentrations even though such description does not always adequately correspond to reality. Potentials are given in volts. The notation of the cationic species in the text and equations ignores the water molecules in the coordination sphere, i.e. Fe2+ means [Fe(H2O)6]2+, [Fe(OH)]+ means [Fe(H2O)5(OH)]+, etc. Therefore, e.g. [Fe(OH)3] must not be considered as solid iron(III) hydroxide but it denotes dissolved neutral [Fe(OH)3(H2O)3] complex species.
Each line in the Pourbaix diagram originates from the assumption that the activities (concentrations) of both species participating in the given equilibrium are equal.
(a) If only redox equilibria are considered, the resulting plot contains horizontal lines (Figure 1a). A typical example is the redox pair Tl+/Tl with E° = −0.34 V (in the pH range 0 to 12; in more basic solutions hydroxido complexes are formed). The mathematical expression of the given line is simple, eq. (1):
line a The potential at which activities/concentrations of Tl+ and Tl(s) are equal, i.e. a(Tl,s)/[Tl+] = 1:
(Tl+/Tl): E = E° − (0.059 / n) log(a(Tl,s) / [Tl+]) = −0.34 − 0 = −0.34 V (1)
(c) When the same approach is applied to a system in which only protolytic (e.g. hydrolytic) processes occur (with no redox contribution within the selected potential limits) leads to vertical lines (Figure 1b). For instance, the stepwise hydrolysis of Ga3+ ion occurs according to four overall stability constants of complex hydroxido species, eq. (2–5).
Ga3+ + OH− = [Ga(OH)]2+, logβ1 = [Ga(OH2+] / ([Ga3+] × [OH−]) = 11.4 (2)
Ga3+ + 2 OH− = [Ga(OH)2]+, logβ2 = [Ga(OH2+] / ([Ga3+] × [OH−]2) = 22.1 (3)
Ga3+ + 3 OH− = [Ga(OH)3], logβ3 = [Ga(OH)3] / ([Ga3+] × [OH−]3) = 31.7 (4)
Ga3+ + 4 OH− = [Ga(OH)4]−, logβ4 = [Ga(OH)4−] / ([Ga3+] × [OH−]4) = 39.4 (5)
The final expressions (6–9) can be derived and calculated as follows:
line b pH at which concentrations of Ga3+ and [Ga(OH)]2+ are equal, i.e. pH = pKa of [Ga(H2O)6]3+ leading to the formation of [Ga(H2O)5OH]2+:
(Ga3+/[Ga(OH)]2+): pH = pKw − logβ1 = 14.0 − 11.4 = 2.6; (6)
and, analogously:
line c ([Ga(OH)]2+ / [Ga(OH)2]+):
pH = pKw − logβ2 + logβ1 = 14.0 − 22.1 + 11.4 = 3.3 (7)
line d ([Ga(OH)2]+ / [Ga(OH)3]):
pH = pKw − logβ3 + logβ2 = 14.0 − 31.7 + 22.1 = 4.4 (8)
line e ([Ga(OH)3] / [Ga(OH)4]−):
pH = pKw − logβ4 + logβ3 = 14.0 − 39.4 + 31.7 = 6.3 (9)
(d) If both redox and protolytic equilibria are involved, the resulting line is sloping. One useful example is the oxygen reduction (10) and hydrated proton reduction (11), which is shown in Figure 1c. The analytical expressions of lines f (12) and g (13) have an identical slope value, i.e. both lines are parallel and they define the area of water redox stability with respect to the reduction affording H2(g) and oxidation to O2(g).
O2 + 4 H+ + 4 e− → 2 H2O, E° = 1.23 V (10)
2 H+ + 2 e− → H2, E° = 0 V (def.) (11)
line f (O2,H+/H2O):
E = E° − (0.059 / n) × log[a2(H2O,l) / (a(O2,g) × [H+]4)] = 1.23 − 0.059 × pH (12)
line g (H+/H2):
E = E° − (0.059 / n) × log[a(H2,g) / [H+]2] = −0.059 × pH (13)
[VISUAL]
Figure 1. Pourbaix diagrams of (a) Tl+/Tl system, (b) Ga3+/[Ga(OH)n](3−n+ system and (c) H2/H2O/O2 system.
Sketch the Latimer diagram for iron species (pH 0), using the following standard redox potentials: E°(FeO4 2−, H+/Fe3+) = 1.90 V, E°(Fe3+/Fe2+) = 0.77 V, E°(Fe2+/Fe) = −0.44 V. Calculate also the redox potentials for couples FeO4 2−/Fe2+, FeO4 2−/Fe and Fe3+/Fe and add them to the diagram.
Model Answer
The requested potentials are calculated as follows, (1) – (3):
E°(FeO4 2-,H+/Fe2+) = (3 × 1.90 + 1 × 0.77) / 4 V = 1.62 V (1)
E°(FeO4 2-, H+/Fe) = (3 × 1.90 + 1 × 0.77 + 2 × (−0.44)) / 6 V = 0.93 V (2)
E°(Fe3+/Fe) = (1 × 0.77 + 2 × (−0.44)) / 3 V = −0.04 V (3)
The corresponding Latimer diagram is depicted in Figure 1.
[VISUAL]
Figure 1. Latimer diagram for iron species in water (pH 0).
Determine the voltage equivalents for individual redox states of iron and plot the Frost diagram. Decide whether the mixture of FeO4 2− and Fe2+ at pH = 0 will interact spontaneously.
Model Answer
The voltage equivalent is defined as a product of the formal oxidation state N and the standard redox potential E° (4) for the reduction of the particular species to the elemental state. The Frost diagram (Figure 2) then plots voltage equivalents versus oxidation state.
Voltage equivalent = N × E°(species/element) (4)
The individual voltage equivalents are calculated from the data above (5)–(8).
Voltage equivalent (Fe) = 0 × 0 V (def.) = 0 V (5)
Voltage equivalent (Fe2+) = 2 × (−0.44) V = −0.88 V (6)
Voltage equivalent (Fe3+) = 3 × (−0.04) V = −0.12 V (7)
Voltage equivalent (FeO4 2-) = 6 × 0.93 V = 5.58 V (8)
[VISUAL]
Figure 2. Frost diagram for iron species (pH 0).
Since the imaginary line connecting both the requested oxidation states (FeO4 2− and Fe2+, dashed line in Figure 2) lies above the Fe3+ point in the diagram, a synproportionation will be favoured, (9):
FeO4 2− + 3 Fe2+ + 8 H+ → 4 Fe3+ + 4 H2O (9)
The construction of the Pourbaix diagram for all iron species is rather tedious. Thus, a rough form of the result valid for metallic iron and dissolved iron species between pH 0 and 14 is provided in Figure 2.
[VISUAL]
Figure 2. Pourbaix diagram for metallic iron and dissolved iron species in water.
a) All the lines divide the area of the diagram into many zones. Deduce which species prevail in the individual zones in Figure 2 and fill the answers into the diagram.
Model Answer
The individual zone labels follow the successive uptake of electrons (from up to down) and hydroxide anions as ligands (from left to right). The identity of any zone can be checked by comparing the appropriate borderline definitions. The answers are displayed in Figure 3.
[VISUAL]
Figure 3. Pourbaix diagram for dissolved iron species and metallic iron.
b) Using the data from 16.1 together with Table 1, derive and write down the conditions for horizontal lines 11 and 17, and vertical lines 2 and 5.
[VISUAL]
Table 1: Overall stability constants βn of ferrous / ferric hydroxido complexes.
log βn (Fe2+ + n OH−)
log βn (Fe3+ + n OH−)
n = 1: log β1 = 4.5 (ferrous), 11.8 (ferric)
n = 2: log β2 = – (ferrous), 22.3 (ferric)
n = 3: log β3 = – (ferrous), 30.0 (ferric)
n = 4: log β4 = – (ferrous), 34.4 (ferric)
Model Answer
Each borderline is drawn with the assumption that the activities of both participating species are equal. The equations which refer to lines 11 and 17, respectively, can be derived from the corresponding forms of the Nernst–Peterson equation (10) and (11) assuming the equilibrium conditions [Fe3+] = [Fe2+] and [Fe2+] = a(Fe,s). Because there is no term which depends on pH, the results are horizontal constant lines numerically equal to the standard redox potentials, see Figure 3.
line 11 (Fe3+/Fe2+): E = E° − 0.059 × log([Fe2+] / [Fe3+]), thus E = 0.77 (10)
line 17 (Fe2+/Fe): E = E° − (0.059 / 2) × log(a(Fe,s) / [Fe2+]), thus E = −0.44 (11)
The conditions for lines 2 and 5 are [Fe3+] = [Fe(OH2+] and [Fe(OH)3] = [Fe(OH)4 −], respectively, i.e. the expressions imply constant lines again, but in this case vertical ones, since there is no connection with the redox potential. The analytical expressions are represented by equations (12) and (13).
line 2 (Fe3+/[Fe(OH)]2+): pH = pKw − logβ1, thus pH = 14.0 − 11.8 = 2.2 (12)
line 5 ([Fe(OH)3]/[Fe(OH)4] −): pH = pKw + logβ3 – logβ4, thus pH = 14.0 + 30.0 − 34.4 = 9.6 (13)
d) As an example of sloping lines, derive the equation for line 6 and determine the coordinates of its intersection with lines 2 and 7.
Model Answer
The analytic expression of line 6 (14) is also derived from the Nernst–Peterson equation under the assumption [FeO4 2−] = [Fe3+].
line 6 (FeO4 2−/Fe3+): E = E° − (0.059 / 3) × log{[Fe3+] / ([FeO4 2−] × [H+]8)}, thus E = 1.90 − 0.157 × pH (14)
The first coordinate of the intersection of lines 2, 6 and 7 is obviously pH = 2.2. The second coordinate can be calculated by substituting for pH = 2.2 in equation (14), i.e. E = 1.55.
e) Ferrate(VI) anion is a stronger oxidizing agent than oxygen itself (compare line f in Figure 1c with the stability region of the ferrate ion in Figure 2), and thus is stable only in an extremely basic region, in which the potentials of O2/H2O and FeO4 2−/[Fe(OH)n](3−n+ are comparable. Thus, in general, the ferrate ion is not stable in aqueous solutions and oxidizes water to oxygen. Suggest a method for generating a ferrate ion. Write the corresponding stoichiometric equation.
Model Answer
Ferrate ion can only be produced in a very basic solution by strong oxidizing agents (stronger than elemental oxygen under these conditions), e.g. hypochlorite (32). This will overcome line 10 and produce some ferrate ions.
2 [Fe(OH)4] − + 3 ClO− + 2 OH− → 2 FeO4 2− + 3 Cl− + 5 H2O (32)
Other possibilities are for example oxidation in a mixture of melted sodium nitrate with sodium hydroxide or analogous reactions in melts.
Discuss all the species involved in the Pourbaix diagram (Figure 2) in terms ion size and surface charge density concept and explain which ligands would best match each metal-ion centre.
Model Answer
From the viewpoint of "Hard and Soft Acid-Base" (HSAB) theory, Fe2+ is an intermediary hard, Fe3+ hard and imaginary "Fe6+" would be an extremely hard acid (the hardness correlates with the ionic radii and the surface charge density). Hard acids prefer hard bases and soft acids prefer soft bases. In aqueous solutions, H2O, OH− and O2− are available (although the oxide ion is stable only when coordinated or in solid state). Thus, hard species like Fe3+ or "Fe6+" are expected to prefer more hard and more charged ligands such as OH- or O2- over H2O. This matches with the observed behavior: much higher tendency of ferric ion to undergo hydrolysis, and extreme stability of the hexavalent iron state as oxoanion FeO4 2- (ferrate).
Thermodynamic, kinetic, spectroscopic and magnetic properties are closely related to the electronic structure of the individual species. Since iron is a d-block metal, crystal field and ligand field theories help us to understand the situation. Concerning the determination of the electron configuration in the frame of split d-orbital levels, the most useful qualitative concept is the spectrochemical series.
Write down the magnetic state of the following species (high/low-spin state): [Fe(H2O)6]2+, [Fe(CN)6]4−, [Fe(H2O)6]3+, [Fe(H2O)5OH]2+, [Fe(CN)6]3−. In the approximation of Oh symmetry, calculate the LFSEs and express the results in the units of ligand field strength Δo and electron-pairing energy P.
Model Answer
The electronic, magnetic, and LFSE properties of the species are summarized in Table 1 below:
[VISUAL]
Table 1: Electronic and magnetic properties of selected iron species.
• [Fe(H2O)6]2+: d6 configuration, high-spin, t2g^4 eg^2 configuration, paramagnetic, LFSE = -0.4 Δo (with respect to free ion configuration)
• [Fe(CN)6]4-: d6 configuration, low-spin, t2g^6 eg^0 configuration, diamagnetic, LFSE = -2.4 Δo + 2P
• [Fe(H2O)6]3+: d5 configuration, high-spin, t2g^3 eg^2 configuration, paramagnetic, LFSE = 0
• [Fe(H2O)5OH]2+: d5 configuration, high-spin, (t2g^3 eg^2)* configuration (not Oh symmetry), paramagnetic, LFSE = 0
• [Fe(CN)6]3-: d5 configuration, low-spin, t2g^5 eg^0 configuration, paramagnetic, LFSE = -2.0 Δo + 2P
Figure 3 below shows the UV-Vis spectra of an orange-brown solution of FeCl3 containing a [Fe(H2O)5OH]2+ cation and of a nanosuspension of the Prussian blue (which can be approximated as Fe4[Fe(CN)6]3). The spectra are shown in an unknown order. Assign each compound to the respective spectrum.
[VISUAL]
Figure 3. Spectra of FeCl3 and Fe4[Fe(CN)6]3 (in unknown order).
Model Answer
a) FeCl3 (corresponds to spectrum (a) showing absorption primarily in the UV/blue-violet region, which leaves the solution looking orange-brown).
b) Fe4[Fe(CN)6]3 (corresponds to spectrum (b) showing strong absorption around 700 nm in the red region, which gives the pigment its characteristic deep blue colour of Prussian blue).