Electrolysis of aqueous solutions — AQA GCSE Chemistry
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Electrolysis of aqueous solutions explained
Electrolysis of aqueous solutions covers 6 specification points, including The ions discharged when an aqueous solution… and At the negative electrode (cathode), hydrogen….
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Open a point below for its explanation and revision guidance.
The ions discharged when an aqueous solution is electrolysed using inert electrodes depend on the relative reactivity of the elements involved.
In aqueous solution, water molecules dissociate slightly to give H⁺ and OH⁻ ions alongside the ions from the dissolved compound. At the cathode, the more easily reduced positive ion is discharged. If the metal is more reactive than hydrogen, hydrogen is produced instead of the metal; if the metal is less reactive than hydrogen, the metal is produced. At the anode, oxygen is usually produced unless a halide ion is present, in which case the halide is discharged. Reactivity therefore decides which ions are discharged. For example, in sodium chloride solution, hydrogen forms at the cathode and chlorine at the anode; in copper chloride solution, copper forms at the cathode and chlorine at the anode.
At the negative electrode (cathode), hydrogen is produced if the metal is more reactive than hydrogen.
When an aqueous solution is electrolysed, both metal ions and H⁺ ions from water are attracted to the negative electrode. The ion that gains electrons more readily is discharged. If the metal is more reactive than hydrogen, its ions are less easily reduced, so H⁺ ions are reduced instead and hydrogen gas is produced: 2H⁺ + 2e⁻ → H₂. For example, in sodium chloride solution, hydrogen is produced at the cathode rather than sodium. If the metal is less reactive than hydrogen, such as copper, the metal ion is discharged and the metal is deposited: Cu²⁺ + 2e⁻ → Cu. The gas can be identified by a lighted splint giving a squeaky pop.
At the positive electrode (anode), oxygen is produced unless the solution contains halide ions when the halogen is produced.
In aqueous electrolysis, the anode attracts negative ions. Hydroxide ions, OH⁻, are present from water and are discharged, producing oxygen gas. However, if the solution contains halide ions such as Cl⁻, Br⁻ or I⁻, these are discharged in preference, so the halogen forms instead. For example, electrolysing sodium chloride solution gives chlorine at the anode, while electrolysing sodium sulfate solution gives oxygen. The rule is therefore: halide present means halogen at the anode; otherwise oxygen. This links ion discharge to the relative ease of oxidation of the ions present. Higher Tier students must also write half-equations for these processes, e.g., 4OH⁻ → O₂ + 2H₂O + 4e⁻ or 2Cl⁻ → Cl₂ + 2e⁻.
This happens because in the aqueous solution water molecules break down producing hydrogen ions and hydroxide ions that are discharged.
Water is a weak electrolyte because only a small fraction of its molecules dissociate into ions: H₂O ⇌ H⁺ + OH⁻. In an aqueous solution, these ions join those from the dissolved ionic compound, so both positive and negative ions are available at the electrodes. At the cathode, hydrogen ions gain electrons to form hydrogen gas: 2H⁺ + 2e⁻ → H₂. At the anode, hydroxide ions lose electrons to form oxygen gas: 4OH⁻ → O₂ + 2H₂O + 4e⁻. This explains why hydrogen and oxygen can be produced even when the dissolved compound contains neither hydrogen nor oxygen as simple ions. The ions discharged depend on which species is easiest to oxidise or reduce under the conditions used.
Students should be able to predict the products of the electrolysis of aqueous solutions containing a single ionic compound.
To predict products, first identify the ions present: the compound supplies a positive ion and a negative ion, while water supplies H⁺ and OH⁻. At the cathode, the less reactive positive ion is usually discharged; if the metal is more reactive than hydrogen, hydrogen is produced instead. At the anode, halide ions are discharged in preference to hydroxide ions, giving a halogen; otherwise oxygen is produced. For example, electrolysing copper chloride solution gives copper at the cathode and chlorine at the anode, while electrolysing sodium sulfate solution gives hydrogen and oxygen. The prediction must name the product at each electrode and can be supported by half-equations.
Required practical 3: investigate what happens when aqueous solutions are electrolysed using inert electrodes. This should be an investigation involving developing a hypothesis.
In this practical, you electrolyse different aqueous solutions using inert electrodes such as graphite and observe the products at each electrode. You develop a hypothesis linking the ions present to the products, for example that a solution containing halide ions will produce a halogen at the anode. You then test the hypothesis by electrolysing solutions such as sodium chloride, copper chloride and sodium sulfate, recording observations at the cathode and anode. Observations may include gas bubbles, colour changes, litmus or indicator tests, and identification of gases. You evaluate whether results support the hypothesis and explain any anomalies using ion discharge rules.
Your focus
- Predict the products of electrolysis of aqueous solutions using inert electrodes.
- Explain how relative reactivity determines which ions are discharged.
- Describe simple tests to identify the gases formed at each electrode.
Show all 18 objectives
- Predict whether hydrogen or a metal is produced at the cathode in a given aqueous solution.
- Write the half-equation for the production of hydrogen at the cathode.
- Explain the link between metal reactivity and discharge at the negative electrode.
- Identify the anode product for a named aqueous solution.
- Explain why a halogen forms in preference to oxygen when halide ions are present.
- (HT only) Write a balanced anode half-equation for oxygen or a halogen.
- Describe how water molecules provide hydrogen and hydroxide ions in aqueous solution.
- Write balanced half-equations for hydrogen and oxygen production.
- Explain how water-derived ions compete with ions from a dissolved compound.
- Predict the cathode and anode products for a named aqueous solution.
- Justify predictions using ion discharge rules.
- Write half-equations for the predicted products.
- Develop and test a hypothesis about aqueous electrolysis products.
- Record and interpret observations at inert electrodes.
- Evaluate results against the hypothesis using ion discharge rules.
Electrolysis of aqueous solutions exam tips
Marking Points
- Aqueous solutions contain H⁺ and OH⁻ ions from water in addition to the ions from the dissolved substance.
- At the cathode, the ion that is discharged depends on the relative reactivity of the metal compared with hydrogen.
- If the metal is more reactive than hydrogen, hydrogen gas is produced at the cathode rather than the metal.
- If the metal is less reactive than hydrogen, the metal is produced at the cathode.
- At the anode, oxygen is usually produced from OH⁻ ions unless a halide ion is present, when the halide is discharged.
- The products can be identified by testing gases, for example a lighted splint for hydrogen and damp litmus for chlorine.
- At the cathode, both metal ions and H⁺ ions from water are attracted.
- If the metal is more reactive than hydrogen, H⁺ ions are discharged instead of the metal ions.
- The cathode half-equation for hydrogen production is 2H⁺ + 2e⁻ → H₂.
- If the metal is less reactive than hydrogen, the metal ion is discharged and the metal is deposited at the cathode.
- Hydrogen can be identified by a lighted splint producing a squeaky pop.
- State that the anode is the positive electrode and attracts negative ions.
- Identify hydroxide ions from water as the species discharged when no halide is present, giving oxygen.
- Explain that halide ions are discharged in preference to hydroxide ions, giving the halogen.
- (HT only) Write or recognise the anode half-equation for oxygen production: 4OH⁻ → O₂ + 2H₂O + 4e⁻.
- (HT only) Write or recognise a halogen half-equation, for example 2Cl⁻ → Cl₂ + 2e⁻.
- Apply the rule to a named solution, such as predicting chlorine from sodium chloride solution and oxygen from sodium sulfate solution.
- State that water molecules break down into hydrogen ions and hydroxide ions.
- Recognise that these ions are discharged at the electrodes, giving hydrogen at the cathode and oxygen at the anode.
- Write the cathode half-equation: 2H⁺ + 2e⁻ → H₂.
- Write the anode half-equation: 4OH⁻ → O₂ + 2H₂O + 4e⁻.
- Explain that the ions from water compete with ions from the dissolved compound.
- Link the discharge of water-derived ions to the products observed in a named example.
- List all ions present, including H⁺ and OH⁻ from water.
- Apply the cathode rule: metal below hydrogen in reactivity gives the metal; otherwise hydrogen is produced.
- Apply the anode rule: halide present gives the halogen; otherwise oxygen is produced.
- Name the products at both electrodes for a given single ionic compound.
- Support predictions with correct half-equations where required.
- Explain the prediction using discharge of the appropriate ions.
- Develop a testable hypothesis linking the ions in solution to the products at each electrode.
- Select appropriate aqueous solutions and inert electrodes for the investigation.
- Record observations at both electrodes, such as bubbles, colour or gas test results.
- Identify products using suitable tests, for example a glowing splint for oxygen or damp litmus for chlorine.
- Compare results with the hypothesis and explain any differences using ion discharge rules.
- Describe how to control variables such as solution concentration, voltage and electrode material.
Examiner Tips
- 💡State the rule first, then apply it to the specific solution given.
- 💡Name the products at each electrode and give a simple test for each gas.
- 💡Use the reactivity series to justify why a metal or hydrogen is discharged.
- 💡Compare the metal's position in the reactivity series with hydrogen before predicting the cathode product.
- 💡Give the half-equation for hydrogen production when hydrogen is the product.
- 💡State a simple test for hydrogen to confirm the product.
- 💡Name the solution and identify all ions present before predicting the anode product.
- 💡Use the phrase 'halide present, halogen produced; otherwise oxygen' to structure your answer.
- 💡(HT only) When asked to write half-equations, ensure both atoms and charges are balanced.
- 💡Quote the water dissociation equilibrium H₂O ⇌ H⁺ + OH⁻ when explaining the source of the ions.
- 💡Match each half-equation to the correct electrode before writing it.
- 💡Use the words 'discharged' and 'electrons gained or lost' to show understanding of redox at electrodes.
- 💡Write the ion list before making any prediction.
- 💡State the product at each electrode separately and name the electrode.
- 💡Use reactivity series position for the cathode and halide presence for the anode as your two decision steps.
- 💡State your hypothesis clearly before describing the method.
- 💡Link each observation to a named product and the ion responsible.
- 💡Describe a specific confirmatory test for each gas rather than saying 'a gas was produced'.
Common Mistakes
- Assuming the metal is always produced at the cathode; correct by comparing the metal's reactivity with hydrogen.
- Writing that hydrogen is produced at the anode; correct by placing hydrogen production at the negative electrode.
- Ignoring water as a source of ions; correct by including H⁺ and OH⁻ from water in the explanation.
- Writing that a reactive metal such as sodium is deposited at the cathode; correct by stating that hydrogen is produced instead.
- Placing hydrogen production at the positive electrode; correct by placing it at the negative electrode (cathode).
- Omitting electrons from the half-equation; correct by writing 2H⁺ + 2e⁻ → H₂.
- Thinking oxygen is always produced at the anode regardless of the ions present; correct this by checking for halide ions first and naming the halogen if they are present.
- Confusing the anode with the cathode; correct this by remembering that the anode is positive and attracts negative ions.
- (HT only) Writing the oxygen half-equation without balancing electrons or water; correct this by balancing atoms and charge so the equation reads 4OH⁻ → O₂ + 2H₂O + 4e⁻.
- Believing that water is a strong electrolyte and fully ionises; correct this by describing water as only slightly ionised, so few ions are present.
- Writing hydrogen ions as H⁺ without showing electron gain at the cathode; correct this by including 2e⁻ in the half-equation.
- Assuming hydroxide ions are positive; correct this by using the formula OH⁻ with a superscript minus sign.
- Ignoring water-derived ions and predicting only from the dissolved compound; correct this by always including H⁺ and OH⁻ in the ion list.
- Predicting a reactive metal such as sodium at the cathode; correct this by stating that hydrogen is produced when the metal is more reactive than hydrogen.
- Predicting oxygen at the anode when a halide is present; correct this by applying the halide preference rule.
- Using reactive metal electrodes that take part in the reaction; correct this by choosing inert graphite electrodes.
- Recording only one electrode's observations; correct this by observing and recording both the cathode and the anode.
- Treating a single trial as conclusive; correct this by repeating the investigation and checking reproducibility.