Oxidation and reduction

When magnesium burns in carbon dioxide there is no oxygen anywhere in the reactants, and the magnesium is oxidised anyway, which is a hint that the useful definition is not about oxygen.

The previous lessons moved protons between molecules. This one and the next move electrons, which is harder to see, because an electron shifted along a bond leaves the formula unchanged. The first job is a device that makes the shift visible.

What actually changes hands

The oldest definition is Lavoisier's: oxidation is combination with oxygen. It works for rusting and burning, and it fails for $2Mg + C \to 2MgO + C$ , where magnesium takes oxygen away from carbon, and it says nothing at all about $Mg + C \to MgC$ , which is chemically the same event.

The modern definition is about electrons. Oxidation is loss of electrons; reduction is gain. The two always occur together, since electrons do not accumulate anywhere, so the whole class is called redox. The species that takes electrons is the oxidising agent, and it is itself reduced; the species that supplies them is the reducing agent, and it is oxidised. Both halves of that sentence catch people out, and the reliable way to keep them straight is to track the electrons rather than the words.

For an ionic reaction the transfer is literal: magnesium atoms become $M$ and chlorine molecules become $C$ , two electrons each. For a covalent one, such as methane burning, nothing is transferred completely. Electron density shifts along polar bonds, which is a matter of degree. That is what the next section exists to handle.

Oxidation numbers: a fiction that works

The oxidation number of an atom is the charge it would have if every bond it took part in were fully ionic, with each shared pair assigned entirely to the more electronegative atom. It is an accounting convention, not a measurement. The carbon in methane has an oxidation number of $- 4$ and a real partial charge of a few tenths of an electron. Nothing is being claimed about physical charge.

The rules follow from that definition, and are worth deriving rather than memorising. An element on its own has zero, since the sharing is symmetric. A monatomic ion has its charge. Fluorine, the most electronegative element, always takes $- 1$ . Oxygen takes $- 2$ except when bonded to fluorine or to itself. Hydrogen takes $+ 1$ except when bonded to a metal, where it is the more electronegative partner and takes $- 1$ . And the numbers must sum to the charge on the species, which is the constraint that determines everything else.

Example. Assign oxidation numbers to sulfur in $S$ and in , to iron in $F$ , and to oxygen in and in $O$ .

In sulfuric acid, two hydrogens at $+ 1$ and four oxygens at $- 2$ leave $x + 2 - 8 = 0$ , so sulfur is $+ 6$ . In thiosulfate, $2 x - 6 = - 2$ gives $x = + 2$ . In magnetite, $3 x - 8 = 0$ gives $x = + 8 / 3$ , which is not a possible charge on any atom. In hydrogen peroxide the oxygens are bonded to each other, so that bond is shared equally and each oxygen is $- 1$ . In oxygen difluoride, fluorine outranks oxygen, so the fluorines are $- 1$ and oxygen is $+ 2$ .

Now you. Assign oxidation numbers to chromium in $C$ , to manganese in $Mn$ , to nitrogen in $N$ , and to carbon in glucose, .

Answer

Dichromate: $2 x - 14 = - 2$ , so chromium is $+ 6$ . Permanganate: $x - 8 = - 1$ , so manganese is $+ 7$ . Ammonium: $x + 4 = + 1$ , so nitrogen is $- 3$ . Glucose: $6 x + 12 - 12 = 0$ , so carbon averages $0$ , though the individual carbons in the real molecule range from $- 1$ to $+ 1$ .

Two of those answers show the limits honestly. The $+ 8 / 3$ in magnetite is an average over two iron(III) and one iron(II) in the real lattice. The $+ 2$ in thiosulfate is an average over one sulfur at about $+ 5$ and one at about $- 1$ , which is why isotopic labelling shows the two sulfurs behaving quite differently in reactions. Oxidation numbers are a device for counting electrons transferred, and they do that job perfectly; they are not a description of a molecule.

With them in hand, redox becomes visible: any reaction in which an oxidation number changes is a redox reaction. Carbon goes from $- 4$ in methane to $+ 4$ in carbon dioxide, so combustion is redox. In an acid and base neutralisation nothing changes at all, which is why proton transfer is a separate family.

Half reactions in acid

Balancing a redox equation by inspection is painful and unnecessary. Splitting it into two half reactions, balancing each, and combining them so the electrons cancel is mechanical.

The procedure in acidic solution is: balance the atom being oxidised or reduced, then balance oxygen by adding water, then hydrogen by adding , then charge by adding electrons. The electrons should come out on the left for a reduction and on the right for an oxidation, which is a check on the work.

Example. Balance the reaction of permanganate with oxalate in acid, $Mn + \to M + C$ .

Reduction half: manganese goes from $+ 7$ to $+ 2$ . Balancing oxygen with water and hydrogen with protons gives $Mn + 8 \to M + 4 O$ , whose charges are $+ 7$ on the left and $+ 2$ on the right, so five electrons are added to the left.

Oxidation half: $\to 2 C$ is already balanced in atoms, with charges $- 2$ and $0$ , so two electrons go on the right.

To cancel electrons, multiply the first by two and the second by five, giving ten each:

2 Mn + 5 + 16 \to 2 M + 10 C + 8 O

Check the charge: $- 2 - 10 + 16 = + 4$ on the left, $+ 4$ on the right. Check the atoms: manganese two each side, carbon ten, oxygen $8 + 20 = 28$ on the left and $20 + 8 = 28$ on the right.

Now you. Balance the oxidation of iron(II) by permanganate in acid, $Mn + F \to M + F$ .

Answer

The manganese half is as before, needing five electrons. The iron half is $F \to F + e^{-}$ , one electron, so it is multiplied by five:

Mn + 5 F + 8 \to M + 5 F + 4 O

Charges: $- 1 + 10 + 8 = + 17$ on the left, $+ 2 + 15 = + 17$ on the right.

Half reactions in base

In alkaline solution is not available at any useful concentration, so the balanced equation must not contain it. The quickest route is to balance in acid as above, then add enough $O$ to both sides to convert every into water, and cancel whatever water appears on both sides.

Take the oxidation of iodide by permanganate in base, which stops at manganese dioxide rather than going to $M$ . The half reactions are $Mn + 2 O + 3 e^{-} \to Mn + 4 O$ and $+ 6 O \to I + 3 O + 6 e^{-}$ . Doubling the first to match six electrons and combining gives

2 Mn + + O \to 2 Mn + I + 2 O

with charge $- 3$ on each side and nine oxygens on each side.

That the product is $Mn$ in base and $M$ in acid is not a detail of the method: it is chemistry, and it says that the oxidising power of permanganate depends on pH. The next lesson makes that dependence quantitative.

Ranking the agents

Some species take electrons more eagerly than others, and the ordering can be established by experiment before any theory. Put a strip of zinc into copper sulfate solution and it darkens with a deposit of copper while the blue fades, so zinc gives electrons to copper ions. Put copper into zinc sulfate and nothing happens. Repeating this over many pairs gives the activity series, in which each metal displaces every metal below it.

The series explains a great deal at a glance: why potassium and sodium have to be kept out of water, why iron rusts and gold does not, why aluminium is protected by an oxide film rather than by being unreactive, and why the metals known to antiquity are gold, silver and copper, the three that occur native because they are the hardest to oxidise. It also explains extraction: metals near the top are won by electrolysis, ones in the middle by reduction with carbon, and the ones at the bottom are simply dug up.

Among non-metals the halogens rank the same way. Chlorine displaces bromide from solution and bromine displaces iodide, so oxidising power falls down the group, which is the ordering used in the next lesson to build a numerical scale.

Redox titration

Because the electron count is exact, a redox reaction of known stoichiometry measures concentration as precisely as an acid and base titration does, and often more conveniently.

Permanganate is the classic titrant, for a reason worth noticing: it is intensely purple and its product $M$ is almost colourless, so the first drop in excess turns the flask permanently pink and no indicator is needed. It is not a primary standard, since solid potassium permanganate always carries some manganese dioxide, so it is standardised against pure sodium oxalate using the equation balanced above.

Example. $25.00$ mL of a solution of iron(II) in dilute sulfuric acid requires $22.45$ mL of $0.02000$ mol dm⁻³ potassium permanganate to reach the first permanent pink. Find the concentration of the iron(II).

The permanganate supplies $0.02245 \times 0.02000 = 4.490 \times 10^{- 4}$ mol. The balanced equation has five iron per permanganate, so the iron is $5 \times 4.490 \times 10^{- 4} = 2.245 \times 10^{- 3}$ mol, in $25.00$ mL. That gives $2.245 \times 10^{- 3} / 0.02500 = 0.08980$ mol dm⁻³.

Now you. A $0.5000$ g sample of iron ore is dissolved and all the iron reduced to iron(II). Titration needs $18.30$ mL of $0.02000$ mol dm⁻³ permanganate. Find the percentage of iron by mass, taking the molar mass of iron as $55.845$ g mol⁻¹.

Answer

Permanganate: $0.01830 \times 0.02000 = 3.660 \times 10^{- 4}$ mol, so iron is $1.830 \times 10^{- 3}$ mol, weighing $1.830 \times 10^{- 3} \times 55.845 = 0.1022$ g. As a percentage of the sample, $0.1022 / 0.5000 = 20.44$ per cent.

Dichromate is the other common titrant, less powerful but a genuine primary standard, stable in solution and usable in hydrochloric acid, which permanganate is not because it oxidises chloride. It needs a separate indicator, since chromium(III) is green rather than colourless. Iodine titrations form a third family, in which iodine liberated by an oxidising agent is titrated with thiosulfate to a starch end point, and this is how dissolved oxygen and residual chlorine in water are routinely measured.

When a substance reacts with itself

An element in an intermediate oxidation state can sometimes be oxidised and reduced at once, which is disproportionation. Copper(I) does it in water, $2 C \to Cu + C$ , which is why copper(I) salts are stable only as insoluble solids or complexes. Chlorine does it in cold alkali, $C + 2 O \to C + Cl + O$ , going to $- 1$ and $+ 1$ from $0$ , which is the industrial route to bleach. Warm the same mixture and hypochlorite disproportionates further to chloride and chlorate.

The reverse, comproportionation, brings two different states together to a middle one, as when iodate and iodide give iodine in acid.

Whether either happens is a thermodynamic question with a definite answer, and so far this lesson has provided none: nothing here says why copper(I) disproportionates and iron(II) does not, or why permanganate is a stronger oxidising agent in acid. Those need a number attached to each half reaction rather than a rank order.

Separating the two halves into different beakers, so the electrons have to travel through a wire, supplies exactly that number, and turns it into a voltage a meter can read. That is the last lesson.