Electrochemistry

Oxidation numbers: what they mean and the rules for finding them

The oxidation number of an element in a compound or ion, also called its oxidation state, is the charge the atom would carry if the bonding electrons were all assigned to the more electronegative atom in each bond. Because it describes charge, it is written with the sign in front of the number: +3 and -2, never 3+ and 2-. An oxidation number is a bookkeeping device, and its whole purpose is to let you follow electrons through a reaction, even in a covalent substance where no ions are ever formed.

Two rules start every calculation. First, the sum of the oxidation numbers of all the atoms equals the overall charge: 0 for a neutral compound, and the charge of the ion for a polyatomic ion. Second, most elements have a fixed oxidation number in most of their compounds, so learn the common ones and you can find the unknown one by subtraction.

Work through them in this order: put in the values you know, multiply each by the number of that atom, add the unknown once, and set the total equal to the overall charge. The table gives the values that do most of the work in the exam.

Calculating the oxidation number of an element in a compound or an ion

In sulphuric acid, H2SO4, hydrogen is +1 and oxygen is -2. Let the oxidation number of sulphur be x. Then 2(+1) + x + 4(-2) = 0, so x - 6 = 0 and x = +6. In the sulphate ion, SO4^2-, the same atoms must now sum to -2: x + 4(-2) = -2, so x = +6 again. The element keeps its oxidation number when the compound becomes an ion, because the extra electrons belong to the whole ion rather than to any one atom.

Now a metal. In the manganate(VII) ion, MnO4^-, let the oxidation number of manganese be x: x + 4(-2) = -1, so x - 8 = -1 and x = +7. In the dichromate(VI) ion, Cr2O7^2-, two chromium atoms carry the unknown: 2x + 7(-2) = -2, so 2x = 12 and x = +6. Notice that the charge on the ion is easy to forget: it is -1 for MnO4^- and -2 for Cr2O7^2-, not zero, and treating it as zero is the most common mistake in this calculation.

A negative oxidation number works the same way. In the ammonium ion, NH4+, nitrogen is the more electronegative atom, so x + 4(+1) = +1 and x = -3. In the nitrate ion, NO3-, x + 3(-2) = -1 and x = +5. The two nitrogen atoms in ammonium nitrate, NH4NO3, therefore have different oxidation numbers, -3 in the cation and +5 in the anion, which is why the compound is such a good oxidising agent once it is heated.

Watch the peroxides and the odd cases. In hydrogen peroxide, H2O2, the sum must be zero, so 2(+1) + 2x = 0 and oxygen is -1, not -2. In sodium thiosulfate, Na2S2O3, 2(+1) + 2x + 3(-2) = 0, so 2x = 4 and each sulphur is +2. In sodium hydride, NaH, the sum is 0 with sodium at +1, so hydrogen is -1.

Oxidation and reduction in terms of electron transfer

Oxidation is the loss of electrons, and reduction is the gain of electrons. Remember it as OIL RIG, or with the older memory aid that a reducing agent reduces something else by giving its own electrons away. A reaction in which one species loses electrons and another gains them is a redox reaction: the two halves cannot happen separately, because the electrons lost by one atom must be taken by another.

A half equation shows the electrons that one species loses or gains. When iron(II) ions are oxidised in acidified solution, Fe2+ -> Fe3+ + e-, an electron is lost, so iron is oxidised and its oxidation number rises from +2 to +3. When chlorine is reduced, Cl2 + 2e- -> 2Cl-, two electrons are gained, so each chlorine atom falls from 0 to -1. The number of electrons in a half equation always equals the change in oxidation number multiplied by the number of atoms involved.

A half equation that involves oxygen or hydrogen in aqueous solution must also be balanced with water molecules and hydrogen ions. For manganate(VII) in acid: MnO4^- + 8H+ + 5e- -> Mn2+ + 4H2O. Manganese falls from +7 to +2, a change of 5, so five electrons are gained; eight hydrogen ions supply the four water molecules on the right.

To combine two half equations into the overall equation, multiply each one so that the electrons lost equal the electrons gained, then add them and cancel the electrons. Five Fe2+ ions provide the five electrons that one MnO4^- ion needs: MnO4^- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+. Check the charges on both sides: -1 + 8 + 10 = +17 on the left and +2 + 15 = +17 on the right.

• Oxidation: loss of electrons, oxidation number increases, the species is the reducing agent.
• Reduction: gain of electrons, oxidation number decreases, the species is the oxidising agent.
• The number of electrons lost by the reducing agent must equal the number gained by the oxidising agent.
• Balance the charges as well as the atoms when you finish a half equation or an overall equation.

Oxidising and reducing agents, and redox seen as a change in oxidation number

An oxidising agent is a species that accepts electrons from something else, and in doing so is itself reduced. Its oxidation number falls. A reducing agent is a species that donates electrons to something else, and is itself oxidised: its oxidation number rises. This is the part candidates get backwards, so fix it with a sentence rather than the letters: the agent causes the change in the other substance, and is changed in the opposite direction itself.

The oxidation number definition of redox works even where electrons are not obviously transferred, which is why it is the more useful of the two. Oxidation is an increase in oxidation number and reduction is a decrease in oxidation number. When ethanol is oxidised to ethanoic acid by acidified potassium dichromate(VI), no simple ion gives up an electron, but carbon rises from -2 in ethanol to +3 in ethanoic acid while chromium falls from +6 in dichromate(VI) to +3 in the green chromium(III) ion. The orange colour of dichromate disappears and the solution turns green, and that colour change is how the reaction is followed.

Learn a small stock of real agents. Oxidising agents: acidified potassium manganate(VII), acidified potassium dichromate(VI), chlorine, hydrogen peroxide, oxygen, concentrated sulphuric acid and hot concentrated nitric acid. Reducing agents: hydrogen, carbon, carbon monoxide, the reactive metals such as zinc and iron, hydrogen sulphide, potassium iodide, sodium sulfite and tin(II) chloride.

Read the direction off the oxidation numbers. In Fe2O3 + 3CO -> 2Fe + 3CO2 the iron falls from +3 to 0, so iron(III) oxide is reduced and carbon monoxide is the reducing agent; the carbon rises from +2 to +4, so carbon monoxide is oxidised and iron(III) oxide is the oxidising agent. Each species is named from what it does to the other, not from what happens to itself.

Using changes in oxidation number to balance an equation

Oxidation number changes give you a quick route to the correct ratio of the two reacting species, and examiners ask for that ratio because it shows you understand why the numbers are what they are. The method is: find the atom that is oxidised and the atom that is reduced, work out the change in oxidation number for each, and make the total rise equal the total fall.

Take the reaction of iron(II) ions with acidified manganate(VII). Manganese goes from +7 to +2, a fall of 5, so one MnO4^- accepts five electrons; iron goes from +2 to +3, a rise of 1, so each Fe2+ loses one electron. Five Fe2+ are therefore needed for every MnO4^-, and the balanced equation is MnO4^- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+. The remaining atoms, the hydrogen ions and the water molecules, are balanced by inspection once the electron ratio is fixed.

The same method gives the dichromate(VI) equation. Each chromium falls from +6 to +3, a change of 3, and there are two chromium atoms in Cr2O7^2-, so the ion as a whole falls by 6 and accepts six electrons. Six Fe2+ are oxidised, and the equation is Cr2O7^2- + 14H+ + 6Fe2+ -> 2Cr3+ + 7H2O + 6Fe3+. Check the charge on both sides if you have time: -2 + 14 + 12 = +24 on the left, and 2(+3) + 6(+3) = +24 on the right.

Where more than one atom of the same element appears on one side, multiply the change by the number of atoms before you compare it with the other element. In the oxidation of hydrogen sulphide by iron(III) ions, the sulphur in H2S is -2 and in the sulphur produced it is 0, a rise of 2 per atom; the iron falls from +3 to +2, a fall of 1; two Fe3+ are needed for each H2S. That gives 2Fe3+ + H2S -> 2Fe2+ + S + 2H+, which balances in charge as well as in atoms.

Disproportionation: one element oxidised and reduced at once

Disproportionation is the oxidation and reduction of the same element in the same reaction, so a single species is both the oxidising agent and the reducing agent. The element must start at one oxidation number and finish at two different ones, one higher and one lower, and recognising that pattern is worth a mark on its own.

Chlorine with cold dilute sodium hydroxide is the standard example: Cl2 + 2NaOH -> NaCl + NaClO + H2O. Chlorine starts at 0 in Cl2 and finishes at -1 in sodium chloride and at +1 in sodium chlorate(I). The same element has moved in both directions, so this is disproportionation. The reaction matters industrially, because sodium chlorate(I) is the bleach and disinfectant made in bulk this way.

With hot concentrated sodium hydroxide the oxidation goes further: 3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O. Chlorine still goes to -1 in the chloride, but the oxidised product is now the chlorate(V) ion, in which chlorine is +5. Copper(I) does the same trick in water, 2Cu+ -> Cu2+ + Cu, with copper going from +1 to +2 and from +1 to 0.

Hydrogen peroxide disproportionates, 2H2O2 -> 2H2O + O2, and this is the reaction you see when hydrogen peroxide fizzes in a cut or when a little manganese(IV) oxide is dropped into it. Oxygen starts at -1 in the peroxide, falls to -2 in water and rises to 0 in the oxygen gas. Manganese(IV) oxide is a catalyst here, not a reactant, so do not quote its oxidation number as part of the disproportionation.

• One element, two products, two different oxidation numbers: the definition of disproportionation.
• Cold dilute sodium hydroxide with chlorine gives chloride and chlorate(I); hot concentrated sodium hydroxide gives chloride and chlorate(V).
• Copper(I) ions disproportionate in water to copper(II) ions and copper metal.
• Hydrogen peroxide disproportionates to water and oxygen, and manganese(IV) oxide catalyses it.

Roman numerals: showing the oxidation number of an element in a name

Some elements form more than one compound with the same partner, so the name on its own is not enough to identify the substance. A Roman numeral in brackets after the element name shows the magnitude of its oxidation number, with no sign: iron(II) means iron with oxidation number +2 and iron(III) means iron with oxidation number +3. The numeral is part of the name, not part of the formula, and it stays in brackets.

The elements you meet most often in this list are the transition metals, which is why their compounds are named this way. Iron(II) sulphate is FeSO4 and iron(III) chloride is FeCl3; copper(I) oxide is Cu2O and copper(II) oxide is CuO; manganese(IV) oxide is MnO2 and potassium manganate(VII) is KMnO4; chromium(III) oxide is Cr2O3 and potassium dichromate(VI) is K2Cr2O7. The Roman numerals run from I upwards, and for this course you rarely need to go beyond VII.

Work the numeral out from the formula rather than trying to remember it. In PbO2, oxygen supplies 2(-2) = -4, so lead is +4 and the compound is lead(IV) oxide. In tin(II) chloride, tin is +2, so the formula must have two chloride ions for charge balance and is SnCl2, while tin(IV) chloride is SnCl4.

Do not confuse the Roman numeral with the charge written on an ion. In Fe2+ the 2+ is the charge on that one particle, and iron(II) in a name also means +2, but a name such as iron(III) oxide tells you nothing about the charge on a free ion: the numeral is the oxidation number of the iron in the compound, worked out from the formula. Elements with only one common oxidation number take no numeral at all, so it is sodium chloride and aluminium oxide, never sodium(I) chloride.

Redox in the laboratory and in industry

Redox is not an abstract bookkeeping exercise: it is how a great deal of practical chemistry and heavy industry works. In the laboratory, a solution of acidified potassium manganate(VII) is used in a titration to find the concentration of iron(II) ions. The MnO4^- ion is reduced from +7 to the almost colourless Mn2+ while the Fe2+ is oxidised from +2 to +3, and the purple colour of the manganate disappears as it is added, so the first permanent pink is the end point and no indicator is needed.

Oxidation by acidified potassium dichromate(VI) turns a primary alcohol such as ethanol into ethanoic acid and a secondary alcohol such as propan-2-ol into propanone. Chromium falls from +6 to +3 and the solution changes from orange to green, while the carbon of the alcohol rises in oxidation number as the alcohol is oxidised first to the aldehyde and then to the acid. Distilling the product out as it forms stops the oxidation at the aldehyde.

The largest scale redox processes are industrial. In the extraction of iron in the blast furnace, iron(III) oxide is reduced by carbon monoxide: Fe2O3 + 3CO -> 2Fe + 3CO2, with iron falling from +3 to 0 and carbon rising from +2 to +4. In the Contact process for making sulphuric acid, sulphur dioxide is oxidised to sulphur trioxide over a vanadium(V) oxide catalyst, 2SO2 + O2 <-> 2SO3, and the sulphur rises from +4 to +6 while the vanadium moves between +5 and +4 and back again as the catalyst. In the manufacture of nitric acid, ammonia is oxidised to nitrogen monoxide over a hot platinum-rhodium gauze, 4NH3 + 5O2 -> 4NO + 6H2O, and the nitrogen rises from -3 to +2.

Everyday examples are redox as well. Iron rusts because it is oxidised by oxygen from 0 to +3 while the oxygen is reduced from 0 to -2, and the same chemistry is put to work deliberately in the lead-acid battery of a car, where lead is oxidised and lead(IV) oxide is reduced as the cell discharges. Hydrogen peroxide is used as an oxidising agent to disinfect a wound, and chlorine is added to water supplies both as an oxidising agent and through the disproportionation reaction that makes the disinfectant.

• Titration: acidified potassium manganate(VII) oxidises iron(II), Mn going +7 to +2 and Fe going +2 to +3, with no indicator needed.
• Alcohol oxidation: acidified potassium dichromate(VI) oxidises ethanol to ethanoic acid, Cr going +6 to +3, with an orange to green colour change.
• Blast furnace: Fe2O3 + 3CO -> 2Fe + 3CO2, iron reduced from +3 to 0 and carbon oxidised from +2 to +4.
• Contact process: 2SO2 + O2 <-> 2SO3 over vanadium(V) oxide, sulphur oxidised from +4 to +6.
• Nitrile and nitric acid chemistry: ammonia oxidised to nitrogen monoxide over a platinum-rhodium gauze, nitrogen going from -3 to +2.