Electrolysis of aqueous solutions — AQA GCSE Combined Science
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Electrolysis of aqueous solutions explained
In aqueous solution, water molecules dissociate slightly to give hydrogen ions, H⁺, and hydroxide ions, OH⁻, as well as the ions from the dissolved compound.
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At the cathode, positive ions compete: the less reactive element is discharged. For example, in copper chloride solution, Cu²⁺ ions are discharged rather than H⁺ because copper is less reactive than hydrogen, so copper metal forms. At the anode, negative ions compete: if a halide ion such as Cl⁻ is present, it is usually discharged; otherwise OH⁻ is discharged and oxygen forms. For example, in sodium chloride solution, Cl⁻ is discharged to give chlorine, while in sodium sulfate solution, OH⁻ is discharged to give oxygen. The relative reactivity of the elements involved therefore determines which ions are discharged.
At the negative electrode (cathode), hydrogen is produced if the metal is more reactive than hydrogen.
In aqueous electrolysis, the negative electrode (cathode) attracts positive ions. Water provides H⁺ ions alongside metal ions such as Cu²⁺, Na⁺ or Zn²⁺. The ion discharged depends on relative reactivity: if the metal is more reactive than hydrogen, its ions are too stable to be reduced easily, so H⁺ ions gain electrons instead and hydrogen gas forms. For example, in sodium chloride solution, sodium is more reactive than hydrogen, so the cathode reaction is 2H⁺ + 2e⁻ → H₂. In copper chloride solution, copper is less reactive than hydrogen, so Cu²⁺ + 2e⁻ → Cu deposits copper metal. Students should compare the metal with hydrogen in the reactivity series, then write the correct half-equation and identify the product by its observations.
At the positive electrode (anode), oxygen is produced unless the solution contains halide ions when the halogen is produced.
At the positive electrode (anode), negative ions lose electrons (oxidation). Aqueous solutions contain OH⁻ ions from water and possibly halide ions such as Cl⁻, Br⁻ or I⁻. If no halide ions are present, OH⁻ ions are discharged and oxygen gas is produced: 4OH⁻ → O₂ + 2H₂O + 4e⁻. If halide ions are present, the halogen is produced instead, for example 2Cl⁻ → Cl₂ + 2e⁻. So sodium chloride solution gives chlorine at the anode, while sodium sulfate solution gives oxygen. Students should identify the ions present, check for halide ions, then name the product and write the correct half-equation. Observations help: chlorine bleaches damp litmus paper, while oxygen relights a glowing splint.
This happens because in the aqueous solution water molecules break down producing hydrogen ions and hydroxide ions that are discharged.
In an aqueous solution, water ionises slightly: H₂O ⇌ H⁺ + OH⁻. These ions compete with those from the dissolved compound. At the cathode, hydrogen is produced if the metal is more reactive than hydrogen. At the anode, oxygen is produced unless halide ions are present, in which case the halogen is produced. For example, in aqueous sodium chloride, H⁺ is discharged at the cathode to form hydrogen gas (2H⁺ + 2e⁻ → H₂), and Cl⁻ is discharged at the anode to form chlorine gas (2Cl⁻ → Cl₂ + 2e⁻). The remaining Na⁺ and OH⁻ ions stay in solution, forming sodium hydroxide. This explains why the products of aqueous electrolysis often differ from those of the molten compound.
Students should be able to predict the products of the electrolysis of aqueous solutions containing a single ionic compound.
To predict products, first write the formula of the dissolved compound and list its ions, then add H⁺ and OH⁻ from water. At the cathode, the positive ions compete: usually hydrogen is produced unless the metal is less reactive than hydrogen, in which case the metal is deposited. At the anode, the negative ions compete: usually oxygen is produced from OH⁻ unless a halide ion is present, in which case the halogen is formed. For example, in copper(II) chloride solution, Cu²⁺ and H⁺ compete at the cathode and Cu²⁺ is discharged, giving copper metal; Cl⁻ and OH⁻ compete at the anode and Cl⁻ is discharged, giving chlorine gas. In sodium chloride solution, hydrogen and chlorine are formed, leaving sodium hydroxide in solution. The method is the same for any single ionic compound: identify ions, compare ease of discharge, then name the electrode products.
Required practical activity 9: investigate what happens when aqueous solutions are electrolysed using inert electrodes. This should be an investigation involving developing a hypothesis.
You plan and carry out an investigation into electrolysis of aqueous solutions using inert electrodes, developing a hypothesis. You set up a circuit with a DC supply, inert electrodes (such as graphite), and an aqueous electrolyte. You observe and record products at each electrode, then explain them using ion discharge rules. For example, electrolysing copper(II) chloride solution may deposit copper at the cathode and produce chlorine at the anode, while electrolysing sodium chloride solution may produce hydrogen at the cathode and chlorine at the anode. You test your hypothesis by comparing observations with predictions based on reactivity and concentration. You assess risks and control variables to ensure valid results.
Your focus
- List the ions present in an aqueous solution, including H⁺ and OH⁻ from water.
- Use relative reactivity to predict which ions are discharged at the cathode and anode.
- Write half-equations for the discharge of chosen ions and name the products formed.
Show all 18 objectives
- Identify the cathode as the negative electrode and describe reduction of positive ions there.
- Use the reactivity series to predict whether hydrogen or a metal forms at the cathode.
- Write a balanced cathode half-equation for the discharge of H⁺ or a metal ion.
- Identify the anode as the positive electrode and describe oxidation of negative ions there.
- Decide whether oxygen or a halogen forms by checking for halide ions in the solution.
- Write a balanced anode half-equation and suggest a test for the product.
- State that water molecules ionise to produce H⁺ and OH⁻ in aqueous solution.
- Predict the products of electrolysis of aqueous solutions based on reactivity and the presence of halide ions.
- Explain how sodium hydroxide is formed during the electrolysis of aqueous sodium chloride.
- List all ions present in an aqueous solution of a single ionic compound.
- Apply the rules for discharge at the cathode and anode to name the products.
- Justify predictions using reactivity, ease of discharge and half-equations.
- Plan an investigation to electrolyse aqueous solutions using inert electrodes and formulate a testable hypothesis.
- Carry out the electrolysis safely and record observations at the cathode and anode accurately.
- Explain the products formed at each electrode using ion discharge rules and the reactivity series, and evaluate the hypothesis against evidence.
Electrolysis of aqueous solutions exam tips
Marking Points
- Aqueous solutions contain ions from the compound plus H⁺ and OH⁻ from water.
- At the cathode, the less reactive positive ion or element is discharged; for example, Cu²⁺ is discharged instead of H⁺.
- At the anode, halide ions such as Cl⁻ are usually discharged if present; otherwise OH⁻ is discharged and oxygen is produced.
- The relative reactivity of the elements involved determines which ions are discharged.
- Examples: copper chloride solution gives copper at the cathode and chlorine at the anode; sodium sulfate solution gives hydrogen at the cathode and oxygen at the anode.
- Identify the negative electrode as the cathode, where positively charged ions gain electrons (reduction).
- Compare the reactivity of the metal in the solution with hydrogen using the reactivity series.
- If the metal is more reactive than hydrogen, state that hydrogen gas, not the metal, is produced at the cathode.
- Write the cathode half-equation 2H⁺ + 2e⁻ → H₂ when hydrogen is discharged.
- If the metal is less reactive than hydrogen, state that the metal is deposited at the cathode.
- Link the product to an observation, such as bubbles of gas for hydrogen or a coloured solid deposit for a metal.
- Identify the positive electrode as the anode, where negative ions lose electrons (oxidation).
- List the negative ions present, including OH⁻ from water and any halide ions from the dissolved compound.
- State that oxygen is produced when no halide ions are present, from discharge of OH⁻ ions.
- State that the halogen is produced when halide ions are present, for example chlorine from Cl⁻.
- Write a correct anode half-equation, such as 4OH⁻ → O₂ + 2H₂O + 4e⁻ or 2Cl⁻ → Cl₂ + 2e⁻.
- Describe a suitable test or observation to identify the product, such as relighting a glowing splint for oxygen or bleaching damp litmus for chlorine.
- State that water molecules ionise to a small extent (H₂O ⇌ H⁺ + OH⁻), adding H⁺ and OH⁻ to the electrolyte.
- Explain that at the cathode, hydrogen is produced if the metal is more reactive than hydrogen.
- Explain that at the anode, oxygen is produced unless halide ions are present, in which case the halogen is produced.
- Describe how in aqueous sodium chloride, H⁺ and Cl⁻ are discharged, leaving Na⁺ and OH⁻ in solution to form sodium hydroxide.
- Write half-equations for the discharge of ions, such as 2H⁺ + 2e⁻ → H₂ and 2Cl⁻ → Cl₂ + 2e⁻.
- Write the formula of the dissolved compound and list the ions it provides, for example CuCl₂ gives Cu²⁺ and Cl⁻.
- Add H⁺ and OH⁻ from water to the list of ions present in the aqueous solution.
- At the cathode, compare the metal ion with H⁺; hydrogen is usually formed unless the metal is less reactive than hydrogen, when the metal is deposited.
- At the anode, compare the non-metal ion with OH⁻; oxygen is usually formed unless a halide ion is present, when the halogen is formed.
- Name the products and, where required, support each with a half-equation such as Cu²⁺ + 2e⁻ → Cu or 2Cl⁻ → Cl₂ + 2e⁻.
- Recognise that ions left in solution can form a new compound, for example sodium hydroxide remains after electrolysis of aqueous sodium chloride.
- Develops a hypothesis linking the ions present in the aqueous solution to the products expected at each inert electrode.
- Sets up a safe electrolysis circuit using a DC power supply, inert electrodes, and an aqueous electrolyte, ensuring electrodes do not touch.
- Records observations at both electrodes, such as gas bubbles, colour changes, or metal deposits, and identifies the products where possible.
- Explains cathode products by linking to the reactivity series and the discharge of positive ions, including hydrogen from water when appropriate.
- Explains anode products by considering negative ion discharge and, where relevant, the effect of concentration on which halogen or oxygen is produced.
- Evaluates the hypothesis against evidence, suggesting improvements or further tests to increase confidence in conclusions.
Examiner Tips
- 💡State the ions present, including H⁺ and OH⁻ from water, before predicting the products.
- 💡Use the reactivity series to decide which positive ion is discharged at the cathode.
- 💡For the anode, check for a halide ion first; if none is present, discharge OH⁻ to form oxygen.
- 💡State the rule first, then apply it to the named solution, for example 'sodium is more reactive than hydrogen, so hydrogen is produced'.
- 💡Give the half-equation as well as the product name to show the electron transfer clearly.
- 💡Use the reactivity series explicitly in your answer rather than saying 'it depends on the metal'.
- 💡Name the ions present before predicting the product, so your reasoning is clear and checkable.
- 💡Quote the rule directly: oxygen unless halide ions are present, then the halogen.
- 💡Include the state symbol or observation where possible, for example 'chlorine gas bleaches damp litmus paper'.
- 💡Always check if the electrolyte is molten or aqueous; if aqueous, remember to consider H⁺ and OH⁻ ions.
- 💡For aqueous sodium chloride, remember the products are hydrogen, chlorine, and sodium hydroxide.
- 💡Use a two-column ion list for cathode and anode before writing any product.
- 💡State the rule you are applying, such as 'copper is less reactive than hydrogen, so copper is deposited'.
- 💡Include state symbols or gas names where the product is a gas, and balance half-equations if they are requested.
- 💡State your hypothesis clearly and link it to the ions present in the solution, not just to the name of the solution.
- 💡Describe observations precisely, for example 'bubbles of gas at the cathode' or 'pink-brown deposit at the cathode', rather than vague statements.
- 💡When explaining products, refer to the reactivity series and the relative ease of discharge of ions, including hydrogen and hydroxide ions from water.
- 💡Include a risk assessment and identify the independent, dependent, and control variables in your plan.
- 💡Use the terms 'inert electrodes', 'electrolyte', 'cathode', and 'anode' correctly throughout your answer.
Common Mistakes
- Assuming the metal ion is always discharged at the cathode; correction: a reactive metal ion such as Na⁺ remains in solution while H⁺ is discharged to form hydrogen.
- Thinking oxygen is always produced at the anode; correction: a halide ion such as Cl⁻ is usually discharged in preference to OH⁻.
- Forgetting that water contributes H⁺ and OH⁻ ions; correction: include these ions when explaining the competition at each electrode.
- Thinking the more reactive metal is always discharged at the cathode; correct this by explaining that more reactive metals form stable ions that are harder to reduce, so hydrogen is released instead.
- Confusing the cathode with the anode; correct this by recalling that the cathode is the negative electrode and attracts positive ions.
- Writing the hydrogen half-equation as H⁺ + e⁻ → H₂ without balancing; correct this to 2H⁺ + 2e⁻ → H₂ so atoms and charge balance.
- Assuming oxygen is always produced at the anode; correct this by checking for halide ions first, because halide ions are discharged in preference to OH⁻.
- Confusing the anode with the cathode; correct this by recalling that the anode is the positive electrode and attracts negative ions.
- Writing chlorine as Cl rather than Cl₂; correct this by showing that two chloride ions lose electrons to form one chlorine molecule.
- Assuming the metal is always produced at the cathode. Correction: hydrogen is produced if the metal is more reactive than hydrogen.
- Stating that oxygen is always produced at the anode. Correction: if halide ions (like chloride) are present, the halogen (like chlorine) is produced.
- Thinking sodium chloride solution leaves sodium chloride behind. Correction: H⁺ and Cl⁻ are discharged, leaving Na⁺ and OH⁻ to form sodium hydroxide.
- Predicting the metal at the cathode for every aqueous solution; correct this by checking whether the metal is less reactive than hydrogen before deciding.
- Predicting oxygen at the anode when a halide is present; correct this by applying the rule that a halide ion is discharged in preference to OH⁻.
- Forgetting water as a source of ions; correct this by always adding H⁺ and OH⁻ to the ion list before comparing discharge.
- Error: assuming the positive electrode is the cathode. Correction: the cathode is the negative electrode where reduction occurs; the anode is the positive electrode where oxidation occurs.
- Error: predicting that reactive metals such as sodium are produced at the cathode from aqueous solutions. Correction: hydrogen is usually produced instead because water is reduced more readily than sodium ions.
- Error: ignoring concentration effects at the anode. Correction: in concentrated halide solutions, the halogen may form; in dilute solutions, oxygen may form instead.
- Error: using reactive metal electrodes rather than inert ones. Correction: inert electrodes such as graphite or platinum must be used so they do not react and affect the products.