Electron transfer reactions

📚The chemistry

Four ways to describe oxidation

All four descriptions are equivalent; which is most convenient depends on the reaction. For organic chemistry the oxygen/hydrogen version is usually quickest; for everything else, electrons and oxidation states. OIL RIG — oxidation is loss, reduction is gain — if you need a mnemonic.

Two more terms, and they run in the opposite direction to intuition:

• The oxidising agent is the species that causes oxidation, so it gains the electrons and is itself reduced.
• The reducing agent causes reduction, so it loses electrons and is itself oxidised.

The rules for assigning oxidation states are in S3.1. Remember that the strength of the method is that it works for covalent species too: carbon is +4 in \( \mathrm{CO_2} \) and −4 in \( \mathrm{CH_4} \), which is why combustion counts as oxidation.

Half-equations

Half-equations separate the two processes and show the electrons explicitly. To construct one in acidic solution:

1. balance the atoms other than O and H;
2. balance oxygen by adding \( \mathrm{H_2O} \);
3. balance hydrogen by adding \( \mathrm{H^{+}} \);
4. balance the charge by adding electrons to the more positive side.

For example, the reduction of the manganate(VII) ion:

To combine two half-equations into an overall equation, multiply each so that the electrons cancel, then add and cancel anything appearing on both sides. Electrons must never appear in a final overall equation.

Reactivity: metals and halogens

Metals react by being oxidised, and reactivity increases down a group and to the left across the table, because ionization energy falls. A more reactive metal displaces a less reactive one from a solution of its ions:

The blue solution fades and a brown deposit forms on the zinc. The guide says explicitly that the reactivity series does not have to be learned — appropriate data will be supplied — but you must be able to interpret a table of which metal displaces which.

Halogens react by being reduced, and reactivity decreases down group 17, so a more reactive halogen displaces a less reactive one from its halide (S3.1).

Reactive metals with dilute acids give a salt and hydrogen:

This is redox: the metal is oxidised, and \( \mathrm{H^{+}} \) is reduced to \( \mathrm{H_2} \). Metals below hydrogen in the reactivity series — copper, silver, gold — do not react, because they cannot reduce \( \mathrm{H^{+}} \).

Electrochemical cells

The one rule that never changes:

The signs, however, are different in the two kinds of cell, and that is what confuses people:

In a voltaic cell, two half-cells each containing a metal in a solution of its ions are joined by an external circuit and a salt bridge. Electrons flow through the external circuit from anode to cathode, always from the more reactive metal towards the less reactive one. The salt bridge completes the circuit and maintains electrical neutrality by allowing ions to move — anions towards the anode compartment, cations towards the cathode.

A secondary (rechargeable) cell uses redox reactions that can be reversed by applying an external voltage. To write the charging reactions, simply reverse the discharge equations and swap oxidation for reduction. Advantages and disadvantages are examinable: primary cells are cheap and simple but disposable; secondary cells are reusable but heavier and lose capacity over many cycles; fuel cells run indefinitely on an external supply but require a fuel infrastructure (R1.3).

Electrolysis of a molten salt is the simplest case, because only the ions of the salt are present. Molten lead(II) bromide:

Oxidation and reduction of organic compounds

Alcohols are oxidised by an acidified oxidising agent — the names and formulas of specific oxidising agents will not be assessed, so \( [\mathrm{O}] \) may be used in equations.

• Primary alcohol → aldehyde → carboxylic acid — a two-step oxidation. To stop at the aldehyde, use distillation, so the aldehyde (which has the lower boiling point, having no hydrogen bonding) evaporates and is condensed away as soon as it forms. To go all the way to the acid, use reflux, where a vertical condenser returns the vapour to the flask so that it stays in contact with the oxidising agent.
• Secondary alcohol → ketone, and no further, because the carbonyl carbon carries no hydrogen to remove.
• Tertiary alcohols are not oxidised under these conditions at all, for the same reason: there is no hydrogen on the carbon bearing the –OH.

Reduction runs the other way, and is brought about by a reducing agent supplying hydride ions, \( \mathrm{H^{-}} \), which attack the \( \delta+ \) carbonyl carbon:

• carboxylic acid → aldehyde → primary alcohol;
• ketone → secondary alcohol;
• alkene + H₂ → alkane, and alkyne + 2H₂ → alkane — reduction by addition of hydrogen, which lowers the degree of unsaturation. Usually run with a nickel or platinum catalyst.

A useful ladder for the oxidation state of carbon, which the guide itself offers as a linking question: \( \mathrm{CH_4} \) (−4), \( \mathrm{CH_3OH} \) (−2), \( \mathrm{HCHO} \) (0), \( \mathrm{HCOOH} \) (+2), \( \mathrm{CO_2} \) (+4) — increasing oxidation at every step.

AHL Standard electrode potentials

The half-cell \( \mathrm{H^{+}}(aq) + \mathrm{e^{-}} \rightleftharpoons \tfrac{1}{2}\mathrm{H_2}(g) \) is assigned \( E^{\ominus} = 0 \) by convention, and everything else is measured against it. Standard conditions: 298 K, 100 kPa, 1 mol dm⁻³ solutions.

All values in the data booklet are reduction potentials. Reading them:

• a more positive \( E^{\ominus} \) means the species is more easily reduced — it is a stronger oxidising agent;
• a more negative \( E^{\ominus} \) means the reverse (oxidation) is more favourable — a stronger reducing agent.

A positive \( E^{\ominus}_{\text{cell}} \) means the reaction is spontaneous in the direction written. If it comes out negative, the reverse reaction is the spontaneous one. And note that \( E^{\ominus} \) values are not multiplied when you scale a half-equation — potential is an intensive property, unlike enthalpy.

where \( n \) is the number of electrons transferred and \( F \) is the Faraday constant, in the data booklet. A positive \( E^{\ominus}_{\text{cell}} \) therefore gives a negative \( \Delta G^{\ominus} \) — the same conclusion as R1.4, reached electrochemically.

AHL Electrolysis of aqueous solutions

Water complicates matters: it can itself be oxidised or reduced, so at each electrode there is competition. Use the electrode potentials to decide which species reacts.

Two effects modify this, and the guide limits them to two named solutions, NaCl(aq) and CuSO₄(aq):

• Concentration. In concentrated sodium chloride solution chlorine is released at the anode rather than oxygen, despite the electrode potentials, because the chloride concentration is so high. In dilute solution, oxygen is released.
• Nature of the electrode. With inert (graphite or platinum) electrodes in \( \mathrm{CuSO_4}(aq) \), oxygen is released at the anode. With a copper anode, the copper itself dissolves: \( \mathrm{Cu}(s) \rightarrow \mathrm{Cu^{2+}}(aq) + 2\mathrm{e^{-}} \). That is the basis of electroplating and of the electrolytic refining of copper.

Electroplating coats an object with a thin metal layer. The object to be plated is made the cathode, the plating metal is the anode, and the electrolyte contains ions of that metal. Silver-plating a spoon: