Representation of reactions at electrodes as half equations (HT only) — AQA GCSE Combined Science
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Representation of reactions at electrodes as half equations (HT only) explained
At the cathode during electrolysis, positively charged ions (cations) gain electrons from the electrode.
Read the full explanation
This electron gain is defined as reduction. For example, in molten lead(II) bromide, Pb²⁺ ions gain two electrons to form lead metal: Pb²⁺ + 2e⁻ → Pb. In aqueous solutions, hydrogen ions (H⁺) may gain electrons to form hydrogen gas: 2H⁺ + 2e⁻ → H₂. You represent these changes as half equations, balancing both atoms and charges. The cathode is the negative electrode, so it attracts cations. Reduction always involves electron gain, which can be remembered using the OIL RIG mnemonic (Oxidation Is Loss, Reduction Is Gain of electrons).
At the anode (positive electrode), negatively charged ions lose electrons and so the reactions are oxidations.
Electrolysis uses a direct current supply to drive non-spontaneous changes. The anode is the positive electrode, so negatively charged ions (anions) are attracted towards it. When an anion reaches the anode, it transfers one or more electrons to the electrode. Loss of electrons is oxidation, so every anode reaction is an oxidation. For example, in molten lead(II) bromide, bromide ions are oxidised: 2Br⁻ → Br₂ + 2e⁻. In aqueous solutions, hydroxide ions may be oxidised instead, producing oxygen: 4OH⁻ → O₂ + 2H₂O + 4e⁻. The electrons flow through the external circuit from anode to cathode. Note that representing these reactions as half equations is assessed at Higher Tier only.
Reactions at electrodes can be represented by half equations, for example:
A half equation shows what happens at one electrode only. It includes the reactant ion or molecule, the product, and the electrons gained or lost. At the cathode (negative electrode), positive ions gain electrons and are reduced, for example Cu²⁺ + 2e⁻ → Cu. At the anode (positive electrode), negative ions lose electrons and are oxidised, for example 2Cl⁻ → Cl₂ + 2e⁻. To write a correct half equation, balance atoms first, then balance charge by adding electrons to the more positive side. State symbols may be included. The two half equations can be combined to give the overall electrolysis equation. Representing electrode reactions as half equations is assessed at Higher Tier only.
2H+ + 2e⁻ → H₂
This half equation shows hydrogen ions, H⁺, being reduced to hydrogen gas, H₂, at the cathode during electrolysis. The equation must balance in both charge and atoms. Two H⁺ ions each gain one electron, giving two electrons in total, so the charge on the left is 2⁺ plus 2⁻, which is zero, matching the neutral H₂ on the right. The two hydrogen atoms on the left become the two hydrogen atoms in the H₂ molecule on the right. In aqueous solutions, H⁺ ions are attracted to the negative cathode, where they gain electrons. For example, in the electrolysis of dilute sulfuric acid, hydrogen is produced at the cathode. Representing these reactions as half equations is a Higher Tier only skill.
and
In Higher Tier electrolysis, students must be able to write half equations for reactions at the electrodes. The specification uses 'and' to join two key examples: the reduction of lead ions (Pb²⁺ + 2e⁻ → Pb) and the oxidation of bromide ions (2Br⁻ → Br₂ + 2e⁻). This highlights that students must be familiar with both types of processes—gain of electrons at the cathode and loss of electrons at the anode. Students may be asked to write these from scratch or complete and balance supplied half equations. Always verify that the total charge on the left equals the total charge on the right.
4OH⁻ → O₂ + 2H₂O + 4e⁻
This Higher Tier half equation describes oxidation at the anode during the electrolysis of aqueous solutions when hydroxide ions (OH⁻) are discharged. Four hydroxide ions lose one electron each, giving four electrons in total, and rearrange to form one oxygen molecule (O₂) plus two water molecules (2H₂O). The equation is balanced for atoms: four oxygen and four hydrogen atoms appear on both sides. It is also balanced for charge: the left side has four negative charges, while the right side has four electrons, each with a 1⁻ charge, so both sides total 4⁻. Because electrons are released, this is oxidation, occurring at the positive anode. Always count atoms and total charge separately to ensure the half equation is complete.
or
In Higher Tier electrolysis, students must be able to write half equations for electrode reactions. The specification uses 'or' to show two acceptable formats for writing oxidation half equations at the anode. For example, the oxidation of bromide ions can be written by adding electrons to the products (2Br⁻ → Br₂ + 2e⁻) OR by subtracting electrons from the reactants (2Br⁻ - 2e⁻ → Br₂). Both formats correctly represent the loss of electrons. Students should be comfortable interpreting and writing either format, ensuring that both atoms and electrical charges are fully balanced across the equation.
4OH⁻ – 4e⁻ → O₂ + 2H₂O
This half-equation describes oxidation at the anode during the electrolysis of aqueous solutions. Hydroxide ions, OH⁻, lose electrons. Four OH⁻ ions each lose one electron, represented by subtracting 4e⁻ on the left side (– 4e⁻). The remaining atoms rearrange to form one O₂ molecule and two H₂O molecules, balancing the oxygen and hydrogen atoms. Loss of electrons is oxidation, confirming this is an anodic process. To check the charge balance: if written as 4OH⁻ – 4e⁻ → O₂ + 2H₂O, the left side has a 4⁻ charge minus a 4⁻ charge, giving zero, matching the neutral products. If written as 4OH⁻ → O₂ + 2H₂O + 4e⁻, both sides have a net charge of 4⁻. This reaction occurs when OH⁻ is discharged instead of sulfate or nitrate ions.
Your focus
- Describe the process at the cathode in terms of positively charged ions gaining electrons.
- Write balanced half equations for reduction reactions at the cathode.
- Explain why reactions at the cathode are classified as reductions.
Show all 24 objectives
- Identify the anode as the positive electrode and describe the movement of anions towards it.
- Explain that anode reactions involve loss of electrons and are therefore oxidations.
- Write or recognise a balanced anode half equation showing electrons as products.
- Write balanced half equations for reactions at the cathode and anode.
- Classify cathode reactions as reduction and anode reactions as oxidation.
- Combine two half equations to produce the overall equation for electrolysis.
- Write the half equation for the reduction of hydrogen ions to hydrogen gas.
- Balance a half equation for both atoms and charge.
- Explain why hydrogen ions are reduced at the cathode during electrolysis.
- Write balanced half equations for electrode reactions from scratch.
- Complete and balance supplied half equations by adding electrons and balancing atoms.
- Verify that half equations are balanced for both mass and electrical charge.
- Write the balanced half equation for the oxidation of hydroxide ions at the anode.
- Explain how the half equation for hydroxide discharge demonstrates oxidation.
- Verify that the hydroxide half equation is balanced for both atoms and charge.
- Write oxidation half equations using either the addition or subtraction of electrons.
- Balance half equations for both mass and electrical charge.
- Interpret alternative formats for half equations in exam questions.
- Construct a balanced half-equation for the oxidation of hydroxide ions at an anode.
- Explain why the hydroxide half-equation involves loss of electrons and how charge balance is achieved.
- Apply the half-equation to predict the product at the anode during electrolysis of an aqueous solution.
Representation of reactions at electrodes as half equations (HT only) exam tips
Marking Points
- Identify the cathode as the negative electrode where positively charged ions (cations) are attracted.
- State that positively charged ions gain electrons at the cathode, and that this electron gain is reduction.
- Write balanced half equations for cathode reactions, ensuring both atoms and charges are balanced.
- Apply the concept of reduction as electron gain to explain products formed at the cathode in different electrolytes.
- The anode is the positive electrode in electrolysis.
- Negatively charged ions (anions) move towards the anode.
- At the anode, anions lose electrons to the electrode.
- Loss of electrons is oxidation, so anode reactions are oxidations.
- A half equation at the anode shows electrons as products, for example 2Br⁻ → Br₂ + 2e⁻.
- In aqueous solutions, hydroxide ions may be oxidised to oxygen rather than a simple anion.
- A half equation represents the reaction at one electrode only.
- Cathode half equations show reduction: positive ions gain electrons, for example Cu²⁺ + 2e⁻ → Cu.
- Anode half equations show oxidation: negative ions lose electrons, for example 2Cl⁻ → Cl₂ + 2e⁻.
- Atoms and charge must balance in a half equation.
- Electrons are shown as reactants in reduction and as products in oxidation.
- Combining two half equations gives the overall electrolysis equation.
- Identifies the reactant as hydrogen ions, H⁺, and the product as hydrogen molecules, H₂.
- States that reduction occurs at the cathode because electrons are gained.
- Balances atoms: two H⁺ ions produce one H₂ molecule, so the equation has 2H⁺ and H₂.
- Balances charge: two H⁺ ions carry a total charge of 2⁺, so two electrons, 2e⁻, are needed to give zero charge on both sides.
- Explains that the electrons come from the external circuit and are supplied at the negative electrode.
- Recognises that this half equation represents the cathode reaction in electrolysis of aqueous solutions containing H⁺ ions.
- Identifies that reduction occurs at the cathode, involving the gain of electrons (e.g., Pb²⁺ + 2e⁻ → Pb).
- Identifies that oxidation occurs at the anode, involving the loss of electrons (e.g., 2Br⁻ → Br₂ + 2e⁻).
- Completes a supplied half equation by adding the correct number of electrons to the appropriate side.
- Balances a supplied half equation by ensuring the number of atoms and total electrical charge are equal on both sides.
- Recognises that OH⁻ ions are oxidised at the anode, releasing electrons and forming O₂ and H₂O.
- Balances oxygen and hydrogen atoms so that 4OH⁻ gives O₂ and 2H₂O.
- Balances charge by including 4e⁻ on the product side, giving a total charge of 4⁻ on each side.
- States that loss of electrons is oxidation and that this half equation represents the anode reaction.
- Uses the equation to explain why oxygen gas is observed at the positive electrode in aqueous electrolysis.
- Writes an oxidation half equation by adding electrons to the product side (e.g., 2Br⁻ → Br₂ + 2e⁻).
- Writes an oxidation half equation by subtracting electrons from the reactant side (e.g., 2Br⁻ - 2e⁻ → Br₂).
- Balances the number of atoms of each element on both sides of the half equation.
- Verifies that the total electrical charge is identical on both the reactant and product sides of the half equation.
- Recognise that OH⁻ ions are oxidised because they lose electrons, and that loss of electrons is oxidation.
- Balance the oxygen atoms: four OH⁻ contain four oxygen atoms, which appear as one O₂ (two oxygen atoms) plus two H₂O (two oxygen atoms).
- Balance the hydrogen atoms: four OH⁻ contain four hydrogen atoms, which appear as four hydrogen atoms in two H₂O molecules.
- Balance charge by subtracting four electrons on the reactant side (– 4e⁻) to give a net zero charge on both sides, or adding them to the product side (+ 4e⁻) to give a net 4⁻ charge on both sides.
- Link the half-equation to the anode process in aqueous electrolysis, where hydroxide ions may be discharged instead of other anions.
Examiner Tips
- 💡Always check that half equations are balanced for both atoms and charge before finalising your answer.
- 💡Use the phrase 'gain of electrons' when defining reduction, and link it directly to the cathode.
- 💡State the electrode sign first, then name the ion and the electron change.
- 💡Check that charge and mass balance in any half equation you write.
- 💡Use the phrase 'loss of electrons is oxidation' to justify why an anode reaction is an oxidation.
- 💡Remember that writing half equations for electrode reactions is a Higher Tier only skill.
- 💡Balance atoms first, then charge, then check the final equation.
- 💡Label each half equation as oxidation or reduction to show understanding.
- 💡Use state symbols where they help clarify the product, such as Br₂(l) or O₂(g).
- 💡Be prepared to write half equations if you are taking the Higher Tier paper.
- 💡Check both atom balance and charge balance before writing the final half equation.
- 💡Remember the mnemonic 'reduction at the cathode' because both reduction and cathode begin with consonants, while oxidation and anode begin with vowels.
- 💡If asked to identify the electrode, link the gain of electrons to the negative electrode and state that hydrogen gas is formed.
- 💡Note that writing and balancing half equations is assessed at Higher Tier only.
- 💡When completing a half equation, first balance the atoms, then add electrons to balance the total charge.
- 💡Use the OIL RIG acronym (Oxidation Is Loss, Reduction Is Gain) to check if electrons should be reactants or products.
- 💡Check atom balance first, then charge balance; a Higher Tier half equation must satisfy both.
- 💡Link the side on which electrons appear to oxidation or reduction: electrons on the right means oxidation at the anode.
- 💡While both formats for oxidation are acceptable, adding electrons to the product side (e.g., 2Br⁻ → Br₂ + 2e⁻) is often less confusing and reduces sign errors.
- 💡Always check that your half equation produces the correct species, such as a stable diatomic molecule for halogens (e.g., Br₂) or a neutral atom for metals (e.g., Cu).
- 💡Check the charge on each side after balancing atoms: 4OH⁻ → O₂ + 2H₂O + 4e⁻ has a net charge of 4⁻ on both sides, ensuring charge is conserved.
- 💡If the question asks for the anode reaction in aqueous solution, consider hydroxide discharge as well as other possible anions, and choose the half-equation that matches the information given.
- 💡Write the equation exactly as required, using Unicode superscripts for charges and subscripts for formulae, and do not add state symbols unless they are requested.
Common Mistakes
- Confusing the cathode with the positive electrode. Correction: the cathode is the negative electrode; cations move towards it.
- Writing half equations that are not charge-balanced. Correction: add electrons to the left side for reduction so that total charge is equal on both sides.
- Stating that reduction is loss of electrons. Correction: reduction is gain of electrons; use OIL RIG to remember.
- Thinking the anode is negative because negative ions go there; correction: anions are attracted to the positive anode.
- Writing electrons as reactants in an anode half equation; correction: oxidation releases electrons, so they appear as products.
- Confusing oxidation with gain of oxygen only; correction: in electrode reactions, oxidation is loss of electrons.
- Putting electrons on the wrong side; correction: reduction gains electrons as reactants, oxidation releases electrons as products.
- Forgetting to balance atoms before charge; correction: balance atoms first, then add electrons to balance charge.
- Writing a half equation for both electrodes in one equation; correction: each half equation describes only one electrode.
- Writing H⁺ + e⁻ → H₂, which is incorrect because it does not balance hydrogen atoms or charge; the correct equation is 2H⁺ + 2e⁻ → H₂.
- Writing 2H⁺ + 2e⁻ → H₂²⁺ or H₂⁺, which incorrectly assigns a charge to the hydrogen molecule; H₂ is neutral.
- Placing this reaction at the anode instead of the cathode; oxidation occurs at the anode, while reduction of H⁺ occurs at the cathode.
- Adding electrons to the wrong side of a supplied equation; remember that reduction (cathode) gains electrons on the left, while oxidation (anode) loses electrons on the right.
- Failing to balance the charges correctly, such as writing Br⁻ → Br₂ + e⁻ instead of 2Br⁻ → Br₂ + 2e⁻; always check that total charge is equal on both sides.
- Forgetting to balance the atoms before balancing the charges, leading to an incorrect number of electrons being added to the half equation.
- Writing 2OH⁻ → O₂ + H₂O + 2e⁻: this is not atom-balanced because the oxygen and hydrogen counts do not match; correct it to 4OH⁻ → O₂ + 2H₂O + 4e⁻.
- Placing electrons on the left, as in 4OH⁻ + 4e⁻ → O₂ + 2H₂O: this would represent reduction; move the electrons to the product side because hydroxide ions lose electrons.
- Omitting the state or charge symbols, such as writing OH or e instead of OH⁻ and e⁻: include the negative charge on hydroxide ions and on electrons so the charge balance is clear.
- Confusing the signs when moving electrons across the arrow; writing 2Br⁻ + 2e⁻ → Br₂ instead of 2Br⁻ - 2e⁻ → Br₂ for oxidation.
- Forgetting to balance the diatomic molecules, such as writing Br⁻ → Br + e⁻ instead of 2Br⁻ → Br₂ + 2e⁻ when a halogen is produced.
- Confusing oxidation and reduction; remember that oxidation (loss of electrons) occurs at the anode and reduction (gain of electrons) at the cathode.
- Writing 4OH⁻ + 4e⁻ → O₂ + 2H₂O: the error is adding electrons to the reactant side, which implies reduction; correction is to subtract electrons on the left (– 4e⁻) or add them on the right (+ 4e⁻) to show oxidation.
- Using 2OH⁻ instead of 4OH⁻: the error is failing to balance oxygen and hydrogen; correction is to use four hydroxide ions so that one O₂ and two H₂O balance.
- Omitting the minus sign on hydroxide: the error is treating OH as a neutral group; correction is to write OH⁻ with a Unicode superscript minus sign.