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In a redox reaction, one species is oxidised by losing electrons while another is reduced by gaining electrons. Free AQA notes and 15 questions answered.
Half-Equations and Redox is part of Oxidation, reduction and redox equations in A Level AQA Chemistry.
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In a redox reaction, one species is oxidised by losing electrons while another is reduced by gaining electrons. Free AQA notes and 15 questions answered.
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Use this lesson to stay inside Oxidation, reduction and redox equations while connecting the detail back to the wider Physical chemistry topic.
Read the core explanations from Half-Equations and Redox before testing yourself from memory.
Picture a reaction between iron and copper sulphate solution. Iron dissolves and copper metal deposits on the surface. Something is being transferred between the two species — and that something is electrons.
Redox reactions are, at their heart, electron-transfer events. One species loses electrons; another gains them. These two processes always happen together — you cannot have one without the other.
The memory tool that holds this together is OIL RIG: Oxidation Is Loss, Reduction Is Gain — of electrons. Burn that into your memory now, because every half-equation you write depends on it.
Now that you know what oxidation and reduction mean, the definitions of oxidising and reducing agents follow directly. An oxidising agent is an electron acceptor — it accepts electrons from another species, and in doing so, it causes that species to be oxidised.
A reducing agent is an electron donor — it donates electrons to another species, causing that species to be reduced. The reducing agent is itself oxidised in the process.
The relationship is reciprocal and worth stating clearly: the oxidising agent gets reduced; the reducing agent gets oxidised. This is where many students trip up in exam questions, so let's address it head-on.
With the definitions of oxidising and reducing agents secure, you are ready to write the equations that show electron transfer explicitly. These are called half-equations.
A half-equation shows either the oxidation process or the reduction process in isolation, with electrons written explicitly as reactants or products. This makes the electron transfer visible in a way that an overall equation does not.
The method for writing a half-equation follows a clear sequence. Work through each stage deliberately — rushing leads to unbalanced charges, which costs marks.
Worked example: Writing a Half-Equation: Oxidation of Iron(II) to Iron(III)
Write the half-equation for the oxidation of Fe²⁺(aq) to Fe³⁺(aq).
Answer: $\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-$
Now apply the same logic to a reduction half-equation. The only difference is that electrons appear on the left-hand side — they are consumed by the species being reduced.
Worked example: Writing a Half-Equation: Reduction of Manganate(VII) in Acid
Write the half-equation for the reduction of MnO₄⁻(aq) to Mn²⁺(aq) in acidic solution.
Answer: $\text{MnO}_4^{-}(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$
Once you have both half-equations, combining them is a matter of ensuring the electrons cancel completely. The number of electrons lost in the oxidation half-equation must equal the number gained in the reduction half-equation.
If the electron counts do not match, multiply one or both half-equations by appropriate integers to equalise them. Then add the two equations together and cancel any species that appear on both sides.
Worked example: Combining Half-Equations: Fe²⁺ and MnO₄⁻ in Acid
Combine the oxidation of Fe²⁺ to Fe³⁺ with the reduction of MnO₄⁻ to Mn²⁺ in acidic solution to give the overall redox equation.
Answer: $5\text{Fe}^{2+}(aq) + \text{MnO}_4^{-}(aq) + 8\text{H}^+(aq) \rightarrow 5\text{Fe}^{3+}(aq) + \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$
Notice that electrons do not appear in the final overall equation. If you can still see electrons after combining, something has gone wrong — either the electron counts were not equalised, or the cancellation was incomplete.
The manganate(VII) example is the most common in AQA papers, but dichromate(VI) ions, $\text{Cr}_2\text{O}_7^{2-}$, appear regularly too. The method is identical — balance atoms, then charges, then combine.
Worked example: Half-Equation: Reduction of Dichromate(VI) in Acid
Write the half-equation for the reduction of Cr₂O₇²⁻(aq) to Cr³⁺(aq) in acidic solution.
Answer: $\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l)$
This half-equation involves 6 electrons. If you were combining it with the Fe²⁺ oxidation half-equation (which produces 1 electron per iron), you would multiply the iron half-equation by 6 to equalise the electrons before adding.
Both atom balance and charge balance are required for a fully correct half-equation; neither condition on its own is sufficient.
When combining half-equations, the most common error is forgetting to multiply all coefficients in a half-equation when scaling it up. If you multiply the iron half-equation by 5, every species in that equation — including the Fe²⁺ and Fe³⁺ — must be multiplied by 5.
| Stage | What to check | Common error |
|---|---|---|
| Writing a half-equation | Electrons on the correct side: products for oxidation, reactants for reduction | Placing electrons on the wrong side — loses the mark for electron placement |
| Balancing atoms | Use H₂O for oxygen, H⁺ for hydrogen (in acid) | Forgetting to add H⁺ after adding H₂O, leaving hydrogen unbalanced |
| Balancing charge | Count total charge on each side; add electrons to equalise | Miscounting charge when multiple ions are present |
| Combining half-equations | Multiply to equalise electron counts before adding | Multiplying only some species in a half-equation, not all |
| Final overall equation | No electrons remaining; charges and atoms both balance | Leaving electrons in the final equation — automatic error |
AQA questions on this topic often ask you to 'identify the oxidising agent' or 'state which species is reduced'. These are one-mark answers, but they require precise language. Name the specific species (e.g. 'MnO₄⁻') rather than a vague description.
Half-equations and the ability to combine them sit at the heart of AQA redox chemistry. Every topic that follows — electrode potentials, electrolysis, and transition metal chemistry — builds directly on the electron-transfer language you have practised here.
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Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
A redox reaction is one in which electron transfer occurs simultaneously: one species is oxidised by losing electrons and another is reduced by gaining electrons. The two processes cannot occur independently, as summed up by OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons.
An oxidising agent is an electron acceptor: it accepts electrons from another species, causing that species to be oxidised, and is itself reduced. A reducing agent is an electron donor that causes another species to be reduced and is itself oxidised.
A reducing agent is an electron donor: it donates electrons to another species, causing that species to be reduced. The reducing agent is itself oxidised in the process, just as an oxidising agent, an electron acceptor, is itself reduced.
To write a half-equation, write the species that is changing, balance the atoms, balance oxygen with H₂O, balance hydrogen with H⁺ in acidic solution, then balance the charge by adding electrons. Electrons are products in an oxidation half-equation and reactants in a reduction half-equation.
OIL RIG stands for Oxidation Is Loss, Reduction Is Gain, of electrons. It is the memory tool behind every half-equation: a species that loses electrons is oxidised, and a species that gains electrons is reduced.
To combine half-equations, multiply one or both so the electrons lost in oxidation equal the electrons gained in reduction, add them together, then cancel species that appear on both sides. For Fe²⁺ and MnO₄⁻, multiply the iron half-equation by 5 so both involve 5 electrons; no electrons should remain in the overall equation.
The half-equation for dichromate(VI) in acid is Cr₂O₇²⁻(aq) + 14 H⁺(aq) + 6 e⁻ → 2 Cr³⁺(aq) + 7 H₂O(l). Chromium is balanced first, 7 H₂O balances oxygen, 14 H⁺ balances hydrogen, and 6 electrons bring the left charge from +12 down to +6.
The half-equation for zinc is Zn(s) → Zn²⁺(aq) + 2 e⁻. Zinc loses 2 electrons, so it is oxidised and the electrons are written as products. The matching reduction for copper(II) ions is Cu²⁺(aq) + 2 e⁻ → Cu(s).
The half-equation for the reduction of manganate(VII) in acid is MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O. Manganese goes from +7 to +2, a gain of 5 electrons; oxygen is balanced with 4 H₂O, hydrogen with 8 H⁺, and the charge with 5 electrons.
The oxidising agent is reduced, because it accepts electrons from another species. The reducing agent is oxidised, because it donates electrons. This reciprocal relationship trips up many students: the oxidising agent gets reduced and the reducing agent gets oxidised.
The half-equation for Fe²⁺ to Fe³⁺ is Fe²⁺(aq) → Fe³⁺(aq) + e⁻. Iron goes from +2 to +3, so it is oxidised and the electron appears as a product. The charges balance: +2 on the left and +3 + (−1) = +2 on the right.
The overall equation for MnO₄⁻ and Fe²⁺ in acid is 5 Fe²⁺ + MnO₄⁻ + 8 H⁺ → 5 Fe³⁺ + Mn²⁺ + 4 H₂O. The iron half-equation is multiplied by 5 so both involve 5 electrons, which then cancel; the charges balance at +17 on each side.
Electrons go on the right-hand side, as products, in an oxidation half-equation, and on the left-hand side, as reactants, in a reduction half-equation. Placing electrons on the wrong side loses the mark for electron placement.
Oxygen in a half-equation is balanced by adding H₂O, and hydrogen is then balanced by adding H⁺, which is valid in acidic solution. For MnO₄⁻ → Mn²⁺, 4 H₂O balance the oxygen and 8 H⁺ balance the hydrogen, before electrons are added to balance the charge.
Common errors when combining half-equations are multiplying only some species in a half-equation instead of every one, forgetting to equalise the electrons before adding, and leaving electrons in the final equation. A correct overall equation has no electrons left and balances in both atoms and charge.