Oxidation Numbers and Redox Reactions

What is it?

A redox reaction is one in which electrons are transferred from one substance to another. It always has two halves:

Oxidation: a loss of electrons. The oxidation number increases.
Reduction: a gain of electrons. The oxidation number decreases.

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

Zinc loses 2 electrons (oxidized); each $\ce{Cu^2+}$ ion gains 2 electrons (reduced). Oxidation and reduction always happen together: electrons lost by one substance are gained by another.

Key idea

OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). The substance that is oxidized is the reducing agent; the substance that is reduced is the oxidizing agent.

Why does it matter?

Energy. Batteries, fuel cells, combustion and respiration are all redox reactions.
Materials. Extracting metals from ores, rusting and corrosion protection depend on redox chemistry.
Analysis. Redox titrations (for example with potassium permanganate) measure iron, vitamin C and many other substances.

How does it work?

1. Rules for oxidation numbers

An oxidation number is the charge an atom would have if all its bonds were ionic. Apply these rules in order:

An uncombined element has oxidation number 0 (Zn, $\ce{O2}$ , $\ce{Cl2}$ ).
A monatomic ion has an oxidation number equal to its charge ( $\ce{Na+}$ : +1; $\ce{Cl-}$ : −1).
Fluorine is always −1. Group 1 metals are +1 and group 2 metals are +2 in compounds.
Hydrogen is usually +1 (but −1 in metal hydrides such as NaH).
Oxygen is usually −2 (but −1 in peroxides such as $\ce{H2O2}$ ).
The oxidation numbers add up to 0 in a neutral compound and to the charge in an ion.

2. Half-equations

Each half-reaction can be written separately, with electrons shown:

Oxidation: $\ce{Zn -> Zn^2+ + 2e-}$
Reduction: $\ce{Cu^2+ + 2e- -> Cu}$

Electrons appear on the product side of an oxidation and the reactant side of a reduction.

3. Balancing redox equations in acidic solution

Split the reaction into two half-equations.
Balance atoms other than O and H.
Balance O by adding $\ce{H2O}$ .
Balance H by adding $\ce{H+}$ .
Balance charge by adding electrons.
Multiply the half-equations so the electrons lost = electrons gained, then add them and cancel.
Check that atoms and charge both balance.

Think of it like this

A redox reaction is like a sale: one person can only sell (give) something if another buys (takes) it. The seller is the reducing agent, giving away electrons; the buyer is the oxidizing agent, taking them. The number of items sold must equal the number bought.

More precisely

Oxidation numbers are a bookkeeping tool, not real charges: in covalent molecules the electrons are shared, not transferred. Some reactions are disproportionations, in which the same element is both oxidized and reduced; for example, in $\ce{2H2O2 -> 2H2O + O2}$ , oxygen goes from −1 to −2 and to 0. In basic solution, balance as in acid and then add $\ce{OH-}$ to both sides to neutralize the $\ce{H+}$ .

Visualise it

Oxidation raises the oxidation number; reduction lowers it.

Worked example

Worked example: Assigning oxidation numbers

Question: Find the oxidation number of (a) S in $\ce{H2SO4}$ (b) Mn in $\ce{MnO4-}$ .

(a) H is +1 and O is −2; the compound is neutral: $2(+1) + x + 4(-2) = 0$ , so $x =$ +6.
(b) O is −2; the ion has charge −1: $x + 4(-2) = -1$ , so $x =$ +7.

Worked example: Identifying the agents

Question: In $\ce{2Al + 3Cu^2+ -> 2Al^3+ + 3Cu}$ , which species is oxidized and which is the oxidizing agent?

Al: 0 → +3. The oxidation number increases, so Al is oxidized (it is the reducing agent).
Cu: +2 → 0. It decreases, so $\ce{Cu^2+}$ is reduced: $\ce{Cu^2+}$ is the oxidizing agent.
Check: 2 Al lose 2 × 3 = 6 electrons; 3 $\ce{Cu^2+}$ gain 3 × 2 = 6 electrons. ✓

Worked example: Balancing in acidic solution

Question: Balance $\ce{MnO4- + Fe^2+ -> Mn^2+ + Fe^3+}$ in acid.

Oxidation: $\ce{Fe^2+ -> Fe^3+ + e-}$
Reduction: $\ce{MnO4- -> Mn^2+}$ . Add 4 $\ce{H2O}$ for oxygen, then 8 $\ce{H+}$ for hydrogen, then 5 electrons for charge:
$\begin{aligned} &\ce{MnO4- + 8H+ + 5e-} \\[2pt] &\quad \ce{-> Mn^2+ + 4H2O} \end{aligned}$
Multiply the oxidation by 5 so that 5 electrons are lost and 5 gained, then add:
$\begin{aligned} &\ce{MnO4- + 8H+ + 5Fe^2+} \\[2pt] &\quad \ce{-> Mn^2+ + 4H2O} \\[2pt] &\quad\quad + \ce{5Fe^3+} \end{aligned}$
Check charge: left −1 + 8 + 10 = +17; right +2 + 15 = +17. ✓

Common mistake

Common mistake: Mixing up oxidation and reduction

Reduction is a gain of electrons, even though the oxidation number goes down. Use OIL RIG, and remember that the oxidizing agent is itself reduced.

Common mistake: Balancing atoms but not charge

$\ce{Fe^3+ + Cu -> Fe^2+ + Cu^2+}$ has balanced atoms but not charge (+3 on the left, +4 on the right). The correct equation is $\ce{2Fe^3+ + Cu -> 2Fe^2+ + Cu^2+}$ .

Common mistake: Assuming oxygen is always −2

In peroxides such as $\ce{H2O2}$ , oxygen is −1; in $\ce{OF2}$ it is +2, because fluorine always takes −1.

Notation note

Oxidation numbers are written sign first (+2); ionic charges are written number first (2+).
Some books use Roman numerals for oxidation numbers in names: iron(III) chloride is $\ce{FeCl3}$ , with Fe at +3.
“e⁻” stands for an electron.

Remember this

OIL RIG: oxidation is loss of electrons (oxidation number up); reduction is gain (oxidation number down).
Reducing agent = oxidized; oxidizing agent = reduced.
Oxidation numbers: element 0; ion = charge; F −1; H +1; O −2; sum = overall charge.
Balance in acid: atoms, then O with H₂O, H with H⁺, charge with e⁻; equalize electrons; check charge.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/redox-reactions/