Chemistry 2810
Answers to Assignment #7
Topic: p-Block Elments. You are responsible for the sections of the text indicated on the course outline for this section. The
material in this chapter will be extremely relevant to the laboratory experiments you will do on the reactions of the elements
(Part II of the lab), and are useful background to further studies in several branches in chemistry. Data on redox potentials is
provided in the form of Latimer diagrams in the Appendices of SAL (or on the class Web Page).
1.
Use the Frost diagram in Fig. 8.5, page 248 SAL to rationalize all the decomposition reactions of NO3- which you perform
in the laboratory in experiment 9. This includes the reactions of various elements with nitric acid.
(a) Preparation of dinitrogen: NH4Cl + NaNO2 → NaCl + N2 + 2 H2O This is clearly a comproportionation reaction
between N(–3) and N(+3). If you draw a line between ammonium ion in acid and HNO2, then N2 lies well below the line
connecting these two points. Also, this is a complementary redox reaction, with 3 electrons transferred by each redox
reaction, which is important to reactions going at a reasonable rate.
(b) Reaction of HNO3 with iron and copper: The observations are that copper induced copious clouds of dark NO2 (the
monomer of N2O4) and produced a blue solution indicative of copper(II). Iron reacts with colorless bubbles emanating (gas
turns mildly brown above surface of liquid). The product resembles iron(III) hydroxide after basification. Thus the conclusion
is clear: 2 HNO3 + Cu + 2 H+ → 2 NO2 + Cu 2+ + 2 H2O This reaction has a cell potential of +0.46 V, and stops here
because the copper is hard to oxidize. It is also a one-electron reduction for nitrogen, while the copper is oxidized by two.
Further reduction of nitrogen does not occur under these reaction conditions (i.e. where the nitric oxides are allowed to escape
the reaction flask). However: HNO3 + Fe + 3 H+ → NO + Fe3+ + 2 H2O This reaction has a cell voltage of +1.0 V. It is
much more favorable to reduce nitrogen from +5 to +2 if the oxidized species is an appreciable reducing agent, which iron is.
Note also that this is again a complementary redox reaction, 3 electrons being transferred in each half reaction. We thus
expect this to be a kinetically fast reaction.
(c) Reaction of HNO3 with Mg: HNO3 + 4 Mg + 9 H+ → NH4+ + 4 Mg2+ + 3 H2O This is what happens when nitrate is
reacted in acidic solution with a powerful reducing agent (E° = 2.4 V). It enables the thermodynamic barrier at hydroxylamine
and hydrazine to be overcome, and nitrogen to be reduced to the most negative oxidation state.
(d) Reaction of HNO3 with S: 6 HNO3 + S → SO42– + 6 NO2 + 6 H+ Concentrated nitric acid promotes a one-electron
reduction to nitrogen dioxide. However, this is a slow reaction, because the sulfur oxidation is a 4 electron process, and the
reaction is non-complementary.
(e) Decomposition of potassium nitrate: 2 KNO3 → 2 KNO2 + O2 The nitrogen(+5) species oxidizes O(–2) to O(0)
(f) Decomposition of lead nitrate: 2 Pb(NO3)2 → 2 PbO + 4 NO2 + O2. The nitrogen(+5) species oxidizes O(–2) to O(0)
2.
Answer the following questions using the Frost diagram for chlorine in acidic and basic solutions.
a) What are the consequences of dissolving Cl2 in aqueous basic solution?
It disproportionates to Cl– and ClO–.
b) What are the consequences of dissolving Cl2 in aqueous acid?
It is stable in aqueous acid.
c) Is the failure of HClO3 to disproportionate in aqueous solution a thermodynamic or a kinetic phenomenon?
Thermodynamically stable in acid, but unstable in base (towards Cl– and ClO4–).
d) For which of the following anions is disproportionation thermodynamically favorable in acidic solution: ClO–, ClO2–,
ClO3–, and ClO4 –.
ClO– and ClO2– .
e) Which reactions in (d) are very slow at room temperature?
Both reactions are slow at room temperature, and are run industrially at about 60 °C.
3.
Use the Latimer diagram for chlorine to determine the potential for reduction of ClO4– to Cl 2. Write a balanced equation for
this half-reaction.
2 ClO4– + 16 H+ → Cl2 + 8 H2O + 14 e– E° = 1.392 V (in acid solution; you could do either in this case.)
4.
Sketch the chlor-alkali cell. Show the half-cell reactions and indicate the
direction of diffusion of the ions. Give the chemical equation for the
unwanted reaction that would occur if OH– migrated through the
membrane and into the anode compartment.
If the Cl2 and the OH– mixed, the production of hypochlorite would result
(cold mixture) or chlorate (hot mixture). These are disproportionation
reactions. For example: Cl2 + H2O → Cl– + ClO– + 2 H+
5.
Use standard potentials from Latimer diagrams to calculate the standard potential of the reaction of H3PO2 with Cu 2+. Are
H3PO32– and H3PO22– useful as oxidizing or reducing agents?
Cu 2+ + H3PO2 + H2O → H3PO3 + Cu + 2 H+ E° = 1.91 V
Both are useful reducing agents (as is obvious from the Frost diagram for the Group 15 elements, below).
6.
Rank the following species from the strongest reducing agent to the strongest oxidizing agent: a) SO42–, b) SO32–, c)
O3SO2SO32–.
Consult the Pourbaix diagram for sulfur. From this it is easy to see that: SO32– < SO42– < O3SO2SO32–
7.
Which of the following compounds presents an explosion hazard? a) NH4ClO4, b) Mg(ClO4) 2, c) NaClO4 and d)
[Fe(OH2) 6][ClO4] 2. Explain your reasoning.
a) NH4ClO4: Yes - contains an oxidizer (ClO4–) and a reducer (NH4+ ) in the same compound, and has a high tendency to release
gas molecules (dinitrogen).
b) Mg(ClO4)2: No - although the perchlorate is an oxidizer, Mg2+ cannot be oxidized further.
c) NaClO4: No - although the perchlorate is an oxidizer, Na + cannot be oxidized further.
d) [Fe(OH2)6][ClO4]2: Possibly - contains perchlorate and Fe(II), which can be oxidized to Fe(III). However, apart from steam
that could be released as a consequence of the heat of the reaction, this is not a gas-forming reaction, so it is not likely to very
explosive.
8.
Which element(s) in Group 13 are not stable in the 3+ oxidation state? Which element(s) in Group 13 are stable in the 1+
oxidation state. What is the origin of this difference?
Consider a Frost diagram for the Group 13 elements:
Here it is clear that all the Group 13 elements except the heaviest, thallium, are most stable by far in the 3+ oxidation state, but
that Tl(III) is a strong oxidizer. It is most stable in the +1 state. The origin of this difference, the formation of the so-called
d 10s2 ions, is the greater effective nuclear charge experienced by the valence orbitals of Tl as a consequence of the filling of the
4f level by 14 electrons that are very poor at shielding. The 6s orbital penetrates much better to the core than the 6p orbitals,
and hence removal of the single p electron is more favourable than the removal of the two valence s electrons.
It is also obvious from this diagram that there is a steady progression of stability for the +3 state, with Al < Ga < In < Tl.
9.
Why is it not surprising that calcium carbide reacts with water to form acetylene gas?
The crystal structures of ionic carbides, including CaC2 show unequivocally the presence of diatomic C2 ions with short C–C
bond distances. Since calcium has only one possible oxidation state in its compounds, +2, therefore these carbide ions must
bear a –2 charge. It is well known from organic chemistry that carbanions are very powerful bases. Hence the reaction of a
C22– ions with any protic source is expected to produce the protonated species, i.e. a hydrocarbon, and hence acetylene:
C22– + 2 H2O → C2H2 + 2 OH–
10. Consider the Frost diagram for the Group 14 elements in acid solution, and answer the following questions.
Frost Diagram for Group 14 in 1 M Acid
3
PbO2
2
GeH4
1
nE (V)
0
SnH4-IV
CH4
-1
CO
C
Si
-II
0
Ge
Sn
Pb
CO2
GeO
II SnO
Pb2+
GeO2
IV SnO2
C
Si
Ge
Sn
Pb
-2
-3
SiO2
-4
Oxidation State
a)
Which species among all Group 14 species is the strongest reducing agent? Explain.
Elemental silicon, since it sits at the top of the line with the steepest negative slope (producing SiO2).
b)
Which species among all Group 14 species is the strongest oxidizing agent?
PbO2 is by far the strongest oxidizer among the Group 14 compounds, as it sits atop the steepest line with positive slope.
It will be reduced to Pb 2+ during such reactions.
c)
What reaction is expected to occur when elemental tin is reacted with aqueous acid?
Tin is not a particularly good reducing agent, so that the production of hydrogen gas might not be facile. From the Frost
diagram, the disproportionation reaction to Sn(–2) and Sn(+2), or for that matter, to Sn(–4) and Sn(+4), may occur more
rapidly than hydrogen production.
d) What should happen when carbon monoxide is bubbled into an acidic solution?
From the Frost diagram, disproportionation to C and CO2 (as HCO32–) is expected to occur. In practice, this is an
exceedingly slow reaction, and carbon monoxide has very low solubility in water to start with.
e)
Write a balanced redox equation for the reaction of elemental silicon and lead dioxide.
Si + 2 PbO2 → SiO2 + 2 PbO
f)
Which Group 14 element is the most difficult to produce from its naturally occurring mineral sources?
Elemental silicon is by far the hardest to produce. Lead and tin have been in production from antiquity (e.g. the bronze
age) precisely because their ores are easily reduced to the metal. In fact, carbon in the form of crude charcoals were the
most common reducing agents for such metal production. However, silicon is of much more recent vintage, and was not
recognized as a new element till 1827 through the work of Berzelius, the renowned Swedish chemist.
11. What reaction occurs between carbon dioxide and water, and how should this reaction be classified? What are some of the
important consequences of this reaction?
The reaction is : CO2 + 2 H2O → HCO3– + H3O+ , and is best thought of as the reaction of the Lewis base OH– with the Lewis
acid, CO2. This reaction is important because it renders water open to air containing CO2 mildly acidic, typically pH = 5.5.
12. Does nitrogen show allotropy? Does phosphorus? Describe the allotropes formed.
Elemental nitrogen is only known as the diatomic molecule N2, and has no known allotropy. By contrast, phosphorus has
many allotropes. The most stable form is the molecular species P4, found in white (or yellow) phosphorus. Red phosphorus is
an amorphous compound consisting of chains and cages derived from P4 by opening one or two of the strained P–P bonds in
that tetrahedral molecule. Black phosphorus is a crystalline network solid, in which all the P–P bonds are without angle strain.
It is the least reactive form of phosphorus, although not the most thermodynamically stable form.
13. In what way do the oxides of phosphorus differ from those of carbon or sulfur?
The oxides of phosphorus, P4O6 and P4O10 are open cage structures similar to the adamantyl framework. The cage structure is
common to both oxides, with the P(V) compound having additional terminal P=O bonds. By contrast, the oxides of carbon, CO
and CO2, and the oxides of sulfur, SO2 and SO3 are simple chain compounds. Condensed chain forms of sulfur oxides are
known, however, among the anions, i.e. derivatives of sulfuric acid that have lost a molecule of water between two units.
14. How can oxygen be prepared in the laboratory?
In the laboratory, oxygen is conveniently prepared by the thermal decomposition of potassium nitrate:
KNO3 → KNO2 + ½ O2
15. Which oxygen species is the strongest oxidizing agent in acidic solution? In
basic solution? A Frost diagram is provided to help answer this question.
In acidic solution, hydrogen peroxide is a stronger oxidizing agent than is O2
itself, while in basic solution the same is true for the hydroperoxy ion, HO2–.
16. Is hydrogen peroxide thermodynamically stable? To what can you attribute the
ability to store aqueous solutions of hydrogen peroxide for long periods of
time? (Pure liquid hydrogen peroxide cannot be stored, and is a serious
explosion hazard.)
Hydrogen peroxide, in both acidic and basic solutions, and hence at any
intermediate pH, is intrinsically unstable towards disproportionation, producing
hydroxide or water and oxygen gas. However, this reaction is kinetically very
slow in the absence of a catalyst. However, when catalyzed this reaction
becomes very rapid. A common example of this catalytic effect is the employment of salts of transition metals. Dilute aqueous
hydrogen peroxide solutions (about 3% concentration) are used in medicine as a disinfectant of wounds. When blood is
present, the iron in the heme catalyzes decomposition, so that the peroxide visibly froths.
17. If you wish to prepare oxygen from water, what conditions are likely to make the reaction easier to perform? Does is
surprise you to know that basic solutions of potassium permanganate often release elemental oxygen?
From the Frost diagram, it is obvious that the oxidation of water to dioxygen is energetically less costly in basic solution (i.e.
the hydroxide ion is the less stable species compared to water itself.) Permanangate ion is a very powerful oxidizing agent,
although it is less oxidizing in basic solution than in acidic solution. However, hydroxide is easier to oxidize than water itself,
so it is not surprising to see basic permanganate solutions giving off oxygen. The reaction is slow, however, because of the
high overpotential for oxygen generation.
18. What allotropic forms of sulfur can exist in crystalline form? What allotropic forms of sulfur are amorphous? What is the
origin of the amorphous structure (i.e. solids lacking in long-range order)?
Crystalline sulfur always contains the crown-shaped S8 molecule, but there are several polymorphs, i.e. the same molecule
arranged in different crystal systems, such as rhombic and monoclinic. Of these, rhombic is the room temperature and ambient
pressure-stable form. Amorphous sulfur is most commonly seen in the form of plastic sulfur, a solid formed by rapid
quenching of molten sulfur. It contains the same broad mixture of sulfur chains that the liquid obtains, some of which can be
many dozens of atoms long. However, amorphous sulfur contains reactive radical chain ends, and even in the solid these
radicals continue to react and lead the formation eventually of stable S8 rings. Thus on standing, amorphous plastic sulfur
spontaneously converts to rhombic sulfur, which is crystalline.
19. With reference to the Pourbaix diagram for sulfur (see answer to Question 6), answer the following questions.
a) Is elemental sulfur stable when exposed to the atmosphere? Powdered sulfur is often used as a soil treatment in
agriculture where the soil is overly alkaline (as in Lethbridge). Explain.
Sulfur ought to oxidize to sulfur(VI), chiefly as the sulfate ion. This reaction can be written as:
S8 + 32 H2O → 8 SO42– + 64 H+
Thus the oxidation of sulfur releases hydronium ion to the soil, and hence sulfur is a soil acidifier. However, the reaction
is very slow, and this is the basis of its appeal to gardeners. The acid is slowly generated, and used largely to neutralize
alkali components of the soil. Thus high acid concentrations that might damage the plants are avoided. Note also that
bulk sulfur from Klaus process plants is stored in large piles open to the atmosphere. Under these conditions, the
oxidation of the sulfur is sufficiently slow that insignificant losses of material occur.
b) In some parts of the world, large deposits of elemental sulfur are found in underground deposits, from which it can be
mined by driving superheated steam into the ground and pumping out the liquefied sulfur. How can you explain the
formation and existence of such deposits?
Underground where exposure to air is prevented by the geological strata, elemental sulfur should be stable. Geologically
elemental sulfur could have been formed either by reduction of sulfates, or more likely by slow oxidation of sulfides and
hydrogen sulfide. It is known that large amounts of H2S are found in some natural gas wells, produced presumably by
similar reductive geological processes that formed the hydrocarbons geologically from decomposed plant and animal
matter.
c)
What is the most reduced form of sulfur? Where might this form of sulfur be expected to show up in the natural
environment?
The most reduced form of sulfur is H2S or its base derivatives HS– and S2–. Free H2S is a component of marsh gas, and
accounts for a large part of the unpleasant smell of bogs and marshes.
d) Is hydrogen sulfide, which is a volatile gas, expected to be stable in the atmosphere (where water is always present, as
well as oxygen)?
From the Pourbaix diagram, it is obvious that in the aqueous environment, where oxygen is present, H2S should oxidize to
sulfates and sulfuric acid, and this in fact does happen. The gas fields of Alberta constantly release low levels of
hydrogen sulfide, which dissipate into the atmosphere and are slow oxidized to the components of acid rain. Where larger
concentrations of the extremely toxic H2S are formed, it must be deliberately destroyed (e.g. by “flaring”, oxidation to SO2)
or it can be recovered in the Claus process, where it is dissolved out of the methane using ethanolamine, recovered from
the solvent using steam, then partially burnt to SO2, and finally reacted with SO2 in a comproportionation reaction:
2 H2S + SO2 → 3 S + 2 H2O
e)
Minerals such as gypsum (CaSO4) occur naturally and by weathering leads to the dissolution of calcium and sulfate
ions into the water supply. Are such solutions expected to be stable? Will they alter the acidity of the waters they
dissolve in?
Yes, SO42– is stable over most of the stability field of natural waters. Moreover, there is no acid equilibrium involving
sulfate above pH = 2, so that dissolving gypsum will have no appreciable effect on water pH. They contribute, however,
to substantially increasing the hardness of the water, and such waters must be treated with water softeners for many
applications.
f)
There are many metals whose minerals in underground deposits are present as sulfides. Explain the origin of such ores.
The metal sulfides are found for the elements known as the chalcophiles (see section 4.2.4 for the Goldschmidt
classification of ores). These soft Lewis acids have a high affinity for sulfide ion. From the Pourbaix diagram, we expect
such ores to be stable only under reducing conditions, and not on surface exposure, where oxidation to sulfates is
expected to be favourable.
g) Which common oxidation state of sulfur does not show up on the Pourbaix diagram? Explain.
Sulfur(IV) is notable by its absence on the Pourbaix diagram. In fact, SO2 is thermodynamically unstable and is oxidized to
sulfates over time. It is not thus a predominant form of sulfur, although large amounts of it can be made, e.g. by
combustion reactions.
20. What important compounds can be prepared by the industrial chlor-alkali process? What are some of the applications of
these products? What is the major cost involved in the operation of this process?
The chlor-alkali industry can directly produce many large-volume industrial chemicals: NaOH, H2, Cl2, ClO– and ClO3–. From
these primary products can also be produced ClO2. Perchloric acid, and its precursor, KClO4 are instead produced from KClO3
and the latter is better produced (on a much smaller scale) by the electrolysis of Sylvite, KCl. Although the most common
applications of these chlorine compounds is to bleaching, particularly in the paper industries, the produced chlorine is also an
important feedstock in the organic and inorganic bulk and specialty chemicals market. Sodium hydroxide has thousands of
applications including the chemical, food processing and engineering industries.
21. How are the hydrohalide acids, HX, produced? What consideration must be given to the redox properties of the halogens
during hydrohalide synthesis?
The salts of the halides, CaF2 for fluoride, NaCl for chloride, KBr or NaBr for bromides and similar salts of iodide are reacted
with a non-volatile strong acid in the absence of water, and then the gaseous HX can be distilled off, condensed, and
typically dissolved in pure water to the desired concentration. Pure HCl and HF are gases, although HF is easily liquefied
under pressure and stays liquid for a long time because it has a high enthalpy of evaporation (like water). HBr and HI are
liquids, but are commonly sold and used as concentrated aqueous solutions, as are HCl and HF.
It is very important to use non-oxidixing conditions in the manufacture of HBr and HI, and this usually necessitates the
employment of neat phosphoric acid as the medium for the reaction. The less easily oxidized halogens chloride and fluoride
can be produced from the more active liquid sulfuric acid, although phosphoric acid works for all four. The general reaction is:
MX + H3PO4 → HX + MH2PO4
22. With reference to Frost diagrams for the halogens, which element is the most difficult to prepare in its highest oxidation
state? Which group 16 element is in the same period as this halogen element? Is there evidence that that group 16 element
is also difficult to obtain in its highest oxidation state?
Bromine in the +7 oxidation state is significantly higher on the Frost diagram than all other species. It is the most difficult to
prepare, although elemental fluorine can only be prepared by electrolysis because of the large potential involved in its
formation from fluoride ion.
Selenium in Group 16 is also the hardest to oxidize to the +6 state, and SeO42– has oxidizing character. Neither sulfate nor
tellurate are strong oxidizers, by comparison.