Sept/Oct 2002 Syllabus changes

This section outlines the omissions that can be made from Conquering Chemistry as a result of the changes that were made to the syllabus in September/October 2002. The revised syllabus can be obtained from the Board of Studies web site, www.boardofstudies.nsw.edu.au.

The first of the two sub-sections below is for students: it tells you what you can omit from CCHSC. The second sub-section is primarily for teachers: it gives more detail about what the changes are and how they affect CCHSC.

In both sub-sections LC, MC and RC mean left, middle and right column respectively and DP means dot point. The references are to the original syllabus (1999 or March/April 2001)

1. For students

As a result of these syllabus changes you should omit the following:

2. For teachers

The changes to the syllabus and their consequences for Conquering Chemistry are set out in the following table.

Choosing an option

Two major considerations for teachers In selecting the one option (of the five available) for their students to study, are first their students' likely interest in the topic and secondly their ability to cope with the amount and complexity of the material. Other considerations are likely to be how well the material integrates with and reinforces core material and how straight-forward and/or predictable are likely examination questions on the material.

Judging by the number of pages I needed to cover the material in each of the three options treated in Conquering Chemistry, the Shipwrecks and salvage option appears to be the shortest, and the material does to a significant extent reinforce and extend core material, though some of it can be difficult to understand for weaker students. In some ways the material in Industrial chemistry may be more straight forward, but apart from chemical equilibrium it does not flow as directly from core material. Probably there is less scope for problem-solving type exam questions and more reliance upon factual recall: these latter facts could be seen as advantages to certain groups of students. There appears to be much more material in Forensic chemistry and much of it is quite complex for students whose background in organic chemistry is just the core material. While a good option for students with a strong biological interest, it could be daunting for weaker students.

Another factor (for teachers, but not for their students) is that Shipwrecks and salvage is closely related to the Oxidation and reduction elective of the previous syllabus.

In writing Conquering Chemistry I thought that the other two options, Biochemistry of movement and Chemistry of art, contained too much material of too diverse a nature for the seven weeks available for the study of the option. Of course these options will particularly appeal to certain groups of students – those strongly interested in sport and dance and those with a strong artistic bent – and strong motivation can easily overcome barriers that may seem daunting to others.

Some teaching points

It was surprising that the original syllabus omitted Volta. No account of the development of electricity could be complete without Volta's contribution. Maybe it was Galvani's 'silly' experiment of sticking wires into frogs' legs that inspired Volta to develop his more productive experiments and to show that the generation of electricity was chemical and not physiological, though Volta still got the interpretation wrong. It was Davy who first recognised that it was chemical reaction at the interface between electrode surface and aqueous solution (and not contact voltages between two different metals) that was producing the electric current. Fortunately the Sept/Oct 2002 syllabus changes put Volta back into the history. Volta had been fully treated in CCHSC right from the start.
 

2. Cell voltages and electrode potentials

Cell voltages and electrode potentials are no longer mentioned in the syllabus for this option. However I guess there is nothing to stop examiners asking a question on them, though I think this is most unlikely while examiners are in their current 'regurgitate facts' mode. The extra worked examples that used to be here have been moved to Module 1 (though I feel that there is less need for them now than when I first wrote these pages).

All of the above are with inert electrodes – electrodes that do not undergo chemical change during the electrolysis. Common cases in which an electrode reacts are electrolysis of aqueous solutions of copper and silver salts using a copper or silver anode respectively (Cu ® Cu²⁺ and Ag ® Ag⁺).

Most electrolyses are not particularly sensitive to the concentration of the ions present, the exception being chloride where concentrated solutions produce chlorine gas  while dilute ones form hydroxide (page 393).

Correlation with electrode potentials
What happens during electrolysis of aqueous solutions can be roughly correlated with electrode potentials (pages 393–4). Basically, the higher the electrode potential the more readily the reduction half reaction occurs and the less readily the oxidation (reverse) half reaction occurs. Hence Ag⁺ and Cu²⁺ are more easily reduced than is H⁺, but H⁺ is reduced in preference to Al³⁺ or Na⁺ (E^os are +0.80, +0.34, 0.00, –1.66 and –2.71 V respectively). Iodide is more easily oxidised than bromide which in turn is more easily oxidised than water; however water is more easily oxidised than is fluoride (E^os are +0.54, +1.09, +1.23, and +2.87 V respectively). Water and chlorine are close together (1.36 and 1.23 V) so changing the concentration of chloride can change which is oxidised preferentially. (Remember E^os refer to reduction half reactions.)

4. Bacterial corrosion of shipwrecks

2. Acid sulfate soils (and acid mine drainage)

On page 418 it is explained how certain bacteria can reduce sulfate (to sulfide) and so bring about corrosion of iron at ocean depths too great for there to be sufficient dissolved oxygen to corrode the iron. This may bring to mind an environmental problem that often occurs in Australia, namely acid sulfate soils (which is not a very informative name!).

Some soils when they are disturbed – dug up for farming or excavated for sub-divisions and buildings – produce acid which drains into waterways and so has a detrimental effect upon aquatic life. The problem here is oxidation of sulfide to sulfate – in some ways the opposite to the bacterial oxidation of shipwrecks.

The soils that produce acidic run-offs are ones that contain iron pyrites, FeS₂. This unexpected formula arises because the two S atoms are covalently bonded together to form the S₂^2– ion: ^–S–S^–. The peroxide anion O₂^2– in Na₂O₂ or BaO₂ is similar: compare with the structure of hydrogen peroxide H–O–O–H .

On exposure to air (oxygen) iron pyrites oxidises to Fe(II) sulfate and sulfuric acid:
2FeS₂(s) + 7O₂(g) + 2H₂O(l) ® 2Fe²⁺(aq) + 4SO₄^2–(aq) + 4H⁺(aq)
(the products could be written 2FeSO₄ + 2H₂SO₄)
This overall equation is made up of the two half equations:
FeS₂ + 8H₂O ® Fe²⁺ + 2SO₄^2– + 16H⁺ + 14e^–
O₂ + 4H⁺ + 4e^– ® 2H₂O

The Fe²⁺ is then oxidised to Fe³⁺ (at the low pH involved, generally by bacteria such as Thiobaccillus ferrooxidans) and precipitates out as a yellow-brown gelatinous Fe(OH)₃. This leads to further acid production:
4Fe²⁺(aq) + O₂(g) + 4H⁺(aq) ® 4Fe³⁺(aq) + 2H₂O(l)
4Fe³⁺(aq) + 12H₂O(l) ® 4Fe(OH)₃(s) + 12H⁺(aq)

Water draining from coal mines is often quite acidic. The cause is this same series of reactions: mining exposes iron pyrites to air and so oxidation occurs. This is called acid mine drainage and is a serious problem for most coal mines.

 
3. Acidic and alkaline forms of equations

Sometimes equations, particularly half equations, are written in terms of H⁺ ions and at others in terms of OH^– ions; for example
2H⁺ + 2e^– ® H₂ (acidic form)
2H₂O + 2e^–  ® H₂ + 2OH^– (alkaline form)
We generally write the form that is most applicable to the solutions being discussed – whether they are acidic or alkaline.

To convert an alkaline form to an acidic form add H⁺s to both sides and proceed similarly.

Electrolysis of water
Such conversions are often required in writing equations for so-called electrolysis of water experiments. 

For the electrolysis of dilute sulfuric acid (hydrolysis of water) we would use acidic forms:
At the cathode: 2H⁺ + 2e^– ® H₂
At the anode: 2H₂O ® O₂ + 4H⁺ + 4e^–
Overall (double the first and add to the second): 4H⁺ + 2H₂O ® O₂ + 4H⁺ + 2H₂
Cancelling out 4H⁺ gives: 2H₂O ® 2H₂ + O₂

For electrolysis of sodium hydroxide solution (also hydrolysis of water) we would use alkaline forms:
At the cathode: 2H₂O + 2e^– ® H₂ + 2OH^–
At the anode: 4OH^– ® O₂ + 2H₂O + 4e^–
Overall: 2H₂O ® 2H₂ + O₂ (double the 1st, add to 2nd, cancel out 2H₂O + 4OH^–)

For electrolysis of near neutral solutions of salts such as sodium nitrate, magnesium sulfate or potassium fluoride which are also electrolysis of water we could use either the acidic pair or the alkaline pair (but don't use one of each!). 

Oxidation by O₂ in corrosion contexts
In corrosion contexts oxidation is often by oxygen gas either dissolved in water or from the air. Again we can write the reduction half equation in either acidic or alkaline forms:
O₂ + 4H⁺ + 4e^– ® 2H₂O
O₂ + 2H₂O + 4e^– ® 4OH^–

4. Different E^os for different forms of similar half reactions

From a table of standard electrode potentials we find that the two forms of what is essentially the one process have different E^os: for example

The reason lies in the 'standard' part of the name. The standard electrode potential is the value when all solutes in the reaction (equation) are present at concentrations of 1.00 mol/L (and gases present at partial pressures of 1.00 atmosphere). So the value for equation (i) is for [H⁺] = 1.00 mol/L (and so with [OH^–]  = 1.0 x 10^–14 mol/L).  The E^o value for equation (ii) is for [OH^–] = 1.0 mol/L and so [H⁺] = 1.0 x 10^–14 mol/L).

We can calculate a (non-standard) electrode potential for both of these half reactions for [H⁺] = [OH^–] = 1.0 x 10^–7 mol/L (that is pH = 7.0; neutral solution). It is 0.82 V for both of them.

Getting the same value for both should be no surprise: after all both equations simply represent the reduction of oxygen gas in water. Or looked at from the other direction, they both represent the oxidation of water to oxygen gas.

Similarly for

the first value refers to [H⁺] = 1.00 mol/L while for the second [OH^–] = 1.0 mol/L. Again we can calculate a value for pH = 7.0

Again the same value should be no surprise because both half equations just represent the reduction of water to hydrogen gas.

Remember, it does not matter whether we write the equations as above or with one electron only: H⁺ + e^– ® ½H₂ . The E^o value is the same (see CCHSC page 59).

5. Anodisation of aluminium

On page 406 it was stated that aluminium was a passivating metal, meaning that it readily forms a protective oxide layer that prevents further oxidation (corrosion). This protective oxide layer can be made thicker and hence more effective in protecting the underlying metal by a process called anodising.

The aluminium object to be protected is made the anode in an electrolysis cell: the electrolyte is a solution of sulfuric acid and the cathode a piece of unreactive metal (for example, stainless steel or platinum). Under these conditions the process that occurs at the anode is
2Al + 3H₂O ® Al₂O₃ + 6H⁺ + 6e^–
This thickens the oxide layer and so make the aluminium more resistant to corrosion.

If certain dyes (coloured materials) are also present in the electrolyte, they can become incorporated in the oxide layer and so give it a definite colour. In this way aluminium objects can be given distinctive and durable colours. Gold and bronze coloured anodised aluminium objects, such as car and door keys, are fairly common.  Less fashionable now than they were 20 years ago, brightly coloured saucepan lids and cake containers can probably still be seen in your grandmother's kitchen.