Redox and Oxidation Numbers

Rusting, respiration, batteries, bleaching and the extraction of metals all depend on electrons moving from one substance to another. Reactions in which this happens are called redox reactions, and oxidation numbers are the bookkeeping system that tracks the electrons. This note defines oxidation and reduction in terms of electron transfer and oxidation number, gives the rules for working out oxidation numbers in any compound or ion, explains oxidising and reducing agents and disproportionation, and shows how Roman numerals in names give oxidation numbers. These ideas run through the inorganic chemistry of Periods 3, Group 2, Group 17 and nitrogen, and through redox titrations in Paper 3.

Oxidation and reduction

Oxidation was first used to mean "gaining oxygen" ($\ce{2Mg + O2 -> 2MgO}$) and reduction "losing oxygen". But many reactions with no oxygen at all behave the same way. Burning magnesium in chlorine is very like burning it in oxygen: in both, magnesium atoms lose two electrons to become $\ce{Mg^2+}$. The modern definitions are therefore in terms of electrons.

A useful memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).

Oxidation and reduction always happen together: electrons lost by one species must be gained by another. For example, when zinc is added to copper(II) sulfate solution:

$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$

• Zinc loses two electrons: $\ce{Zn -> Zn^2+ + 2e-}$. Zinc is oxidised.
• Copper(II) ions gain two electrons: $\ce{Cu^2+ + 2e- -> Cu}$. Copper(II) ions are reduced.

These two equations, each showing only one of the processes with electrons included, are called half-equations. You will balance more complicated ones in the next note.

Oxidation numbers

For simple ions, it is obvious who has gained or lost electrons. In covalent molecules, no electrons are transferred completely, but they are shared unequally. Oxidation numbers (also called oxidation states) treat every compound as if it were fully ionic, with the more electronegative atom taking all the shared electrons. The oxidation number is then the charge each atom would have.

The rules

Apply these in order. A rule higher in the list takes priority over a lower one.

Always write the sign in front of an oxidation number: $+2$, not $2$. (Ionic charges are written with the number first, $\ce{Fe^2+}$; oxidation numbers with the sign first, $+2$.)

For example, chromium in the dichromate(VI) ion, $\ce{Cr2O7^2-}$:

$$2x + 7(-2) = -2 \quad\Longrightarrow\quad 2x = 12 \quad\Longrightarrow\quad x = +6$$

species	element	working	oxidation number
$\ce{MnO4-}$	Mn	$x + 4(-2) = -1$	$+7$
$\ce{SO4^2-}$	S	$x + 4(-2) = -2$	$+6$
$\ce{SO2}$	S	$x + 2(-2) = 0$	$+4$
$\ce{H2S}$	S	$2(+1) + x = 0$	$-2$
$\ce{NH4+}$	N	$x + 4(+1) = +1$	$-3$
$\ce{NO3-}$	N	$x + 3(-2) = -1$	$+5$
$\ce{NO2}$	N	$x + 2(-2) = 0$	$+4$
$\ce{ClO3-}$	Cl	$x + 3(-2) = -1$	$+5$
$\ce{ClO-}$	Cl	$x + (-2) = -1$	$+1$
$\ce{H2O2}$	O	$2(+1) + 2x = 0$	$-1$
$\ce{S2O3^2-}$	S	$2x + 3(-2) = -2$	$+2$ (average)

Nitrogen ranges from $-3$ (in $\ce{NH3}$) to $+5$ (in $\ce{HNO3}$), and sulfur from $-2$ to $+6$. The maximum positive oxidation number of a main-group element usually equals its number of outer electrons (its group number): $+5$ for N, $+6$ for S, $+7$ for Cl.

Roman numerals in names

When an element can have more than one oxidation number, the name of the compound gives it as a Roman numeral in brackets, with no space.

name	formula	oxidation number shown
iron(II) sulfate	$\ce{FeSO4}$	Fe $+2$
iron(III) chloride	$\ce{FeCl3}$	Fe $+3$
copper(I) oxide	$\ce{Cu2O}$	Cu $+1$
sulfur(IV) oxide	$\ce{SO2}$	S $+4$
sulfur(VI) oxide	$\ce{SO3}$	S $+6$
potassium manganate(VII)	$\ce{KMnO4}$	Mn $+7$
potassium dichromate(VI)	$\ce{K2Cr2O7}$	Cr $+6$
sodium chlorate(I)	$\ce{NaClO}$	Cl $+1$
sodium chlorate(V)	$\ce{NaClO3}$	Cl $+5$
nitrogen(II) oxide	$\ce{NO}$	N $+2$

For oxyanions, the ending -ate shows that oxygen is present, and the numeral gives the oxidation number of the other element.

Oxidising and reducing agents

This is the part students most often get backwards. In $\ce{Zn + Cu^2+ -> Zn^2+ + Cu}$:

• $\ce{Cu^2+}$ is the oxidising agent: it takes electrons from zinc (it oxidises zinc) and is reduced.
• $\ce{Zn}$ is the reducing agent: it gives electrons to $\ce{Cu^2+}$ (it reduces copper ions) and is oxidised.

Common oxidising agents: $\ce{O2}$, $\ce{Cl2}$ and the other halogens, acidified $\ce{MnO4-}$ (purple to colourless $\ce{Mn^2+}$), acidified $\ce{Cr2O7^2-}$ (orange to green $\ce{Cr^3+}$), concentrated $\ce{H2SO4}$, $\ce{HNO3}$, $\ce{H2O2}$.

Common reducing agents: reactive metals, $\ce{H2}$, $\ce{C}$, $\ce{CO}$, $\ce{Fe^2+}$, $\ce{I-}$, $\ce{SO2}$.

Not every reaction is redox. Neutralisation ($\ce{HCl + NaOH -> NaCl + H2O}$), precipitation ($\ce{Ag+ + Cl- -> AgCl}$) and thermal decomposition of carbonates ($\ce{CaCO3 -> CaO + CO2}$) involve no change in oxidation number.

Disproportionation

The classic example is chlorine with cold, dilute aqueous sodium hydroxide:

$$\ce{Cl2(aq) + 2NaOH(aq) -> NaCl(aq) + NaClO(aq) + H2O(l)}$$

Chlorine starts at $0$. In $\ce{NaCl}$ it is $-1$ (reduced); in $\ce{NaClO}$ it is $+1$ (oxidised). This mixture is household bleach.

With hot, concentrated sodium hydroxide, chlorine disproportionates further:

$$\ce{3Cl2(aq) + 6NaOH(aq) -> 5NaCl(aq) + NaClO3(aq) + 3H2O(l)}$$

Chlorine goes from $0$ to $-1$ (in five $\ce{NaCl}$) and to $+5$ (in one $\ce{NaClO3}$). The total decrease ($5 \times 1 = 5$) equals the total increase ($1 \times 5 = 5$), as it must.

Other examples:

• Hydrogen peroxide: $\ce{2H2O2 -> 2H2O + O2}$; oxygen goes from $-1$ to $-2$ (in water) and to $0$ (in $\ce{O2}$).
• Copper(I) ions in aqueous solution: $\ce{2Cu+ -> Cu + Cu^2+}$; copper goes from $+1$ to $0$ and to $+2$.
• Nitrogen dioxide in water: $\ce{2NO2 + H2O -> HNO3 + HNO2}$; nitrogen goes from $+4$ to $+5$ and to $+3$.

Worked examples

Practice

Answers

1. Oxidation: loss of electrons, increase in oxidation number. Reduction: gain of electrons, decrease in oxidation number.
2. (a) $2x - 2 = 0$, $+1$. (b) $x - 8 = -2$, $+6$. (c) $x - 2 - 2 = 0$, $+4$. (d) $2x - 14 = 0$, $+7$. (e) Peroxide: Ba $+2$, so $2y = -2$, $-1$.
3. (a) $\ce{SnCl2}$. (b) $\ce{K2CrO4}$. (c) $\ce{MnO2}$. (d) $\ce{NaClO2}$.
4. Mg: $0 \to +2$ (oxidised): Mg is the reducing agent. C: $+4$ in $\ce{CO2}$ $\to 0$ (reduced): $\ce{CO2}$ is the oxidising agent.
5. (a) Not redox: Ca $+2$, O $-2$, H $+1$, Cl $-1$ throughout. (b) Redox: Na $0 \to +1$ (oxidised); H $+1 \to 0$ in $\ce{H2}$ (reduced). (c) Not redox: Cr is $+6$ in both $\ce{CrO4^2-}$ and $\ce{Cr2O7^2-}$; O and H unchanged.
6. In $\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O}$, chlorine (oxidation number 0) is converted into $\ce{NaCl}$, where it is $-1$ (reduced), and into $\ce{NaClO}$, where it is $+1$ (oxidised). The same element is simultaneously oxidised and reduced.
7. Iodine: $-1 \to 0$, an increase of 1 for each of 8 atoms: total $+8$. Sulfur: $+6 \to -2$, a decrease of 8 for one atom: total $-8$. Eight electrons are transferred from iodide to sulfur, so the changes balance.
8. N: $-3$ in $\ce{NH3}$ to $0$ in $\ce{N2}$: increase of 3 per N atom, $2 \times 3 = 6$ per $\ce{N2}$. Cu: $+2 \to 0$: decrease of 2 per Cu, $3 \times 2 = 6$. Six electrons are transferred per mole of $\ce{N2}$.
9. $3(+1) + x + 3(-2) = 0$, so $x = +3$. Equation: $\ce{4H3PO3 -> 3H3PO4 + PH3}$. Phosphorus goes from $+3$ to $+5$ in $\ce{H3PO4}$ (oxidised; three atoms, total $+6$) and to $-3$ in $\ce{PH3}$ (reduced; one atom, total $-6$). The same element in the same species is oxidised and reduced, so it is disproportionation.
10. V: $\dfrac{56.0}{50.9} = 1.100$; O: $\dfrac{44.0}{16.0} = 2.750$. Ratio $1 : 2.5 = 2 : 5$: $\ce{V2O5}$. V: $2x - 10 = 0$, $x = +5$. Reduced to $+4$, which is $\ce{VO2}$ (or $\ce{V2O4}$): $\ce{V2O5 + SO2 -> 2VO2 + SO3}$. Check: V $2 \times (-1) = -2$; S $+4 \to +6$, $+2$. Balanced. (This is the catalytic step in the Contact process.)