Oxidation, reduction and redox equations

In a nutshell

A redox reaction is one in which electrons are transferred. One species loses electrons (is oxidised) and, at the same time, another gains them (is reduced).

This subtopic is about three skills: spotting what has been oxidised and what has been reduced using oxidation states, writing half-equations for each process, and combining them into the overall redox equation.

Assumed knowledge: Atomic structure, Amount of substance, Bonding.

Core content

Oxidation, reduction and the two agents

Every redox reaction is a transfer of electrons from one species to another.

• Oxidation is the loss of electrons. It is an increase in oxidation state.
• Reduction is the gain of electrons. It is a decrease in oxidation state.

The memory hook is OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).

The two species that drive the reaction are named for what they do to each other, not for what happens to themselves:

• An oxidising agent (oxidant) is an electron acceptor. It oxidises the other species, and in doing so it is itself reduced.
• A reducing agent (reductant) is an electron donor. It reduces the other species, and in doing so it is itself oxidised.

Take the displacement of copper by zinc:

$Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$

• Zn goes from 0 to +2: it loses 2 electrons, so it is oxidised. Zinc is the reducing agent.
• Cu²⁺ goes from +2 to 0: it gains 2 electrons, so it is reduced. The copper(II) ion is the oxidising agent.

Still don't get it? · why an oxidising agent is itself reduced

Picture a party. One guest keeps handing their sweets away; another keeps taking them.

The guest handing sweets out ends up with none, they have given theirs away. In electron terms this guest has lost electrons, so it has been oxidised. Because it causes the other guest to end up with more, we call it the reducing agent.

The guest taking the sweets ends up with a pile, they have gained. In electron terms this one has been reduced, and because it forces the other to give theirs up, we call it the oxidising agent.

The confusing part is only the naming. To make something else lose electrons (to oxidise it), a species has to take those electrons itself, and taking electrons is being reduced. So an oxidising agent is always itself reduced, and a reducing agent is always itself oxidised. In the exam: the oxidising agent is the electron acceptor and is reduced; the reducing agent is the electron donor and is oxidised.

The rules for assigning oxidation states

The oxidation state (oxidation number) of an atom is the number of electrons it has lost (positive) or gained (negative) compared with the free element. Work through these rules in order; where two rules clash, the one for the more electronegative element wins.

Several elements take fixed oxidation states in almost all of their compounds, which is what lets you work out the unknown one:

• Group 1 metals = +1; Group 2 metals = +2; Al = +3
• Fluorine = −1 (always, it is the most electronegative element)
• Hydrogen = +1, except in metal hydrides where it is −1 (e.g. NaH)
• Oxygen = −2, except in peroxides where it is −1 (e.g. H₂O₂), and in compounds with fluorine where it is positive
• Cl, Br, I = −1, except in compounds with oxygen or fluorine (where they are positive)

Oxidation state is always quoted per atom. In CaCl₂ each Cl is −1, not −2; the two chlorines together balance the +2 on calcium.

Working out an oxidation state from a formula

Set the known oxidation states against the rule that the whole species must add up to zero (a compound) or to the ion's charge (an ion), then solve for the unknown.

For the manganate(VII) ion, MnO₄⁻:

$Mn + 4(-2) = -1$ $Mn = -1 + 8 = +7$

So manganese is +7 in MnO₄⁻. The Roman numeral in the name, manganate(VII), is this oxidation state, which is how AQA names ions of elements with variable oxidation states (iron(III), copper(II), iodate(V)).

Writing half-equations

A half-equation shows only the oxidation or only the reduction, with the electrons written in.

• In an oxidation half-equation the electrons are lost, so they go on the right (the product side).
• In a reduction half-equation the electrons are gained, so they go on the left (the reactant side).

For simple electron transfer, balance the atom then add electrons equal to the change in oxidation state:

• Oxidation: $Fe \rightarrow Fe^{2+} + 2e^-$
• Reduction: $Cl_2 + 2e^- \rightarrow 2Cl^-$

Always check that the total charge is the same on both sides, the electrons are what make this balance.

When the species gains or loses oxygen (for example MnO₄⁻ ⟶ Mn²⁺), balance it in acidic conditions using H₂O and H⁺:

1. Balance the element being oxidised or reduced.
2. Add electrons equal to the total change in oxidation state.
3. Add H₂O to balance the oxygen atoms.
4. Add H⁺ to balance the hydrogen atoms.
5. Check the charges balance on both sides.

Still don't get it? · balancing a half-equation with H+ and H2O

You have the main ingredient right but the side bits are off, so you top up with water and tiny tokens until both sides match. The water (H2O) fixes the oxygen; the tokens (H+) fix the hydrogen.

Take MnO4(-) turning into Mn(2+).

Step 1, the manganese is already one each side.

Step 2, manganese falls from +7 to +2, a drop of 5, which is a gain of 5 electrons (reduction), so put 5e- on the left: MnO4(-) + 5e- -> Mn2+.

Step 3, there are 4 oxygens on the left and none on the right, so add 4 H2O on the right: MnO4(-) + 5e- -> Mn2+ + 4H2O.

Step 4, that adds 8 hydrogens on the right, so add 8 H+ on the left: MnO4(-) + 8H+ + 5e- -> Mn2+ + 4H2O.

Step 5, check the charge. Left: (-1) + (+8) + (-5) = +2. Right: +2. They match, so it is balanced.

In the exam: in acidic conditions, balance oxygen with water, balance hydrogen with H+, add electrons for the oxidation-state change, then prove the total charges are equal on each side.

Combining half-equations into an overall redox equation

To build the full equation, add an oxidation half-equation to a reduction half-equation so that the electrons cancel. They only cancel if the numbers are equal, so first multiply each half-equation up to the lowest common number of electrons, then add and cancel.

The electrons must disappear completely from the final equation: a redox equation with electrons still in it is not finished.

Worked examples

Example 1, oxidation states from a formula (three types).

A neutral compound, nitrogen in HNO₃. H is +1 and each O is −2, and the compound is neutral:

$(+1) + N + 3(-2) = 0$ $N = +6 - 1 = +5$

Nitrogen is +5.

A polyatomic ion, chromium in Cr₂O₇²⁻. Each O is −2 and the states add up to the ion's charge, −2:

$2Cr + 7(-2) = -2$ $2Cr = -2 + 14 = +12$ $Cr = +6$

Each chromium is +6.

An exception, oxygen in the peroxide H₂O₂. H is +1 and the compound is neutral, so the two oxygens cannot be the usual −2:

$2(+1) + 2O = 0$ $O = -1$

Oxygen is −1, because H₂O₂ is a peroxide. Spot the exceptions (peroxides, metal hydrides) before you assume the default value.

Example 2, writing a reduction half-equation with H⁺ and H₂O.

Build the half-equation for MnO₄⁻ ⟶ Mn²⁺.

Mn changes from +7 to +2, a gain of 5 electrons:

$MnO_4^- + 5e^- \rightarrow Mn^{2+}$

Balance the 4 oxygens with 4H₂O, then the 8 hydrogens with 8H⁺:

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

Check the charge: left (−1) + (+8) + (−5) = +2; right +2. Balanced for atoms and charge.

Example 3, combining half-equations.

Acidified manganate(VII) ions oxidise iron(II) ions. The two half-equations are:

• Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
• Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$

The reduction uses 5 electrons and the oxidation releases 1, so multiply the oxidation by 5 to match:

$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$

Add the two together and cancel the 5 electrons on each side:

$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$

Final check. Atoms: 1 Mn, 4 O, 8 H, 5 Fe on each side. Charge: left (−1) + (+8) + (+10) = +17; right (+2) + 0 + (+15) = +17. Balanced, and no electrons remain.

Common exam mistakes

• Giving the nitrogen in a nitrate ion as +7; it is +5. (+7 is the single most common wrong answer here.)
• Defining an oxidising agent as an "electron pair acceptor" or as just "gain of electrons". The mark needs electron acceptor (a species that accepts/gains electrons); "electron pair" and "gain of electrons" on its own are rejected.
• Defining a reducing agent as an "electron pair donor" or as just "loss of electrons". The mark needs electron donor; "electron pair donor" and "loss of electrons" alone are rejected.
• Mixing up the agents. The oxidising agent is itself reduced and the reducing agent is itself oxidised, the opposite of what the name feels like.
• Treating oxidation state as a total, not per atom: writing Cl in CaCl₂ as −2. Each Cl is −1.
• Multiplying an element's oxidation state by the number of atoms when you quote it. Work out the value for one atom, then multiply only when summing to zero or to the ion's charge.
• Charges not balancing in a half-equation or overall equation. Constructing redox equations is one of the lowest-scoring skills; always finish by checking the total charge is equal on both sides.
• Putting the electrons on the wrong side: reduction has electrons on the left (gained), oxidation has them on the right (lost).
• Leaving electrons in the final overall equation, or adding half-equations without first making the electron numbers equal so they cancel.
• Forgetting an uncombined element is 0, including diatomic molecules (O₂, Cl₂) and metals.
• Naming an agent as just the element: "iodine" instead of "the iodine in iodate ions", or the element rather than the full species. Name the whole substance.

Key definitions

• Oxidation - the loss of electrons; an increase in oxidation state.
• Reduction - the gain of electrons; a decrease in oxidation state.
• Oxidising agent (oxidant) - an electron acceptor; it is itself reduced in the reaction.
• Reducing agent (reductant) - an electron donor; it is itself oxidised in the reaction.
• Oxidation state (oxidation number) - the number of electrons an atom has lost (positive) or gained (negative) relative to the free element; the charge the atom would have if the bonding electrons were given to the more electronegative atom.
• Half-equation - an equation showing only the oxidation process or only the reduction process, including the electrons transferred.
• Redox reaction - a reaction in which electrons are transferred from the reducing agent to the oxidising agent, so that oxidation and reduction happen together.

Specification

• I can state that oxidation is the loss of electrons and that oxidising agents are electron acceptors.
• I can state that reduction is the gain of electrons and that reducing agents are electron donors.
• I can identify the species oxidised and reduced from the change in oxidation state, and name the oxidising and reducing agents.
• I can recall and apply the rules for assigning oxidation states.
• I can work out the oxidation state of an element in a compound or ion from its formula.
• I can write half-equations identifying the oxidation and reduction processes in a redox reaction.
• I can combine half-equations to give an overall redox equation.

Related notes

• Atomic structure
• Amount of substance
• Group 2, the alkaline earth metals
• Group 7, the halogens