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Electrolysis of Aqueous Compounds

Aqueous electrolysis: water vs solute ions, preferential discharge rules, and how concentration and electrode type change the products.

  • SEC G3 Pure Chemistry 2027

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In aqueous electrolysis, ions from the solute are not the only possible charge carriers: water also affects which products form. Predict each electrode product from the ions present, concentration and electrode material.

1. Definition

A. Aqueous Electrolyte

An aqueous electrolyte is a solution containing mobile ions. These can come from a dissolved ionic compound, acid or alkali, while water also contributes a very small concentration of ions.

Water ionises slightly: H₂O(l) ⇌ H + (aq) + OH-(aq)

B. Preferential (Selective) Discharge

Preferential (selective) discharge means that one species reacts more readily than another at an electrode. In an aqueous electrolyte, the competing species can include solute ions and water. An ion can carry charge through the solution without being discharged.

2. Key Ideas

  • The products depend on three factors: reactivity series, ion concentration, and electrode type.
  • At the cathode (reduction), either a metal is deposited or hydrogen is produced.
  • At an inert anode (oxidation), either oxygen is produced (from OH⁻/water) or a halogen is produced (from Cl⁻, Br⁻, I⁻).
  • In the required aqueous sulfate examples, SO₄²⁻ is not discharged at an inert anode; oxygen forms instead.
Recall: Electrode type matters

An inert electrode provides a conducting surface without being consumed in the stated reaction. A copper anode in copper(II) sulfate can instead dissolve and change the anode reaction. Graphite is treated as inert in these syllabus examples; it is not unreactive under every condition.

Types of Electrodes

3. Detailed Explanations

Select the reaction at each electrode
  • Start with the electrolyte, concentration and electrode material stated in the question.
  • At the cathode, compare metal deposition with hydrogen formation from water or hydrogen ions.
  • At an inert anode, compare the relevant anions with oxygen formation from water or hydroxide ions.
  • In the required sodium chloride comparison, dilute solution gives oxygen at the anode; concentrated brine gives chlorine.
  • A reactive copper anode needs a different prediction: the copper itself can be oxidised.

A. A Reliable Decision Method

  1. List the solute ions and include water as a possible reacting species. Water also supplies small amounts of H⁺ and OH⁻.
  2. Select the cathode reaction: metal-ion reduction or hydrogen formation.
  3. Check the anode material before using the anion and concentration rules.
  4. Connect each product to its half-equation and observation or gas test. Use Writing and Checking Electrode Half-Equations for the balancing method.

B. Cathode Rule (Reduction)

For the required qualitative predictions, compare the metal with hydrogen in the reactivity series. For example, copper ions can form copper, whereas sodium or magnesium ions remain in solution while hydrogen forms. This is a syllabus prediction guide, not a universal rule for every metal ion and set of operating conditions; use any additional information supplied for an unfamiliar system.

Cation presentWhat happens at the cathode (in aqueous solution, inert electrodes)
Ion of a metal less reactive than hydrogenThe metal ion is discharged (metal deposited), e.g., Cu²⁺ → Cu
Ion of a metal more reactive than hydrogenHydrogen is produced (from water / H⁺), e.g., with Na⁺, Mg²⁺
Recall: Reactivity series

Use the reactivity series to compare metals with hydrogen.

Reactivity Series

C. Anode Rule (Oxidation, Inert Electrodes)

Anions presentWhat happens at the anode (aqueous, inert electrodes)
Concentrated NaCl(aq) (brine)Chlorine is produced from chloride ions
Dilute NaCl(aq)Oxygen is produced from OH⁻/water
SO₄²⁻ present, with no competing halideSulfate is not discharged; oxygen forms instead

For another aqueous electrolyte, use the ions, concentration and any relevant information supplied in the question. Do not turn the sodium chloride comparison into a blanket rule for every halide solution.

This is the concentration trap

In dilute NaCl(aq), oxygen is produced at the anode. In concentrated NaCl(aq) (brine), chlorine is produced at the anode.

Electrolyse dilute sodium chloride, then switch to concentrated, and watch which ion is discharged at the anode and which gas collects.

t = 0 s

Dilute sodium chloride with carbon electrodes and a current of 1.0 A. Cations will move to the cathode and anions to the anode when it is switched on.

Charge, Q = It
0 C
n(e⁻) = Q/F
0 mol
At the cathode
—
At the anode
—
Set-up
A
More settings

Try this

0 of 4 done

  1. Switch on with a solid salt (heat off), then melt it and switch on again. (not done yet)

  2. Electrolyse dilute, then concentrated, sodium chloride. Compare the gas at the anode. (not done yet)

  3. In copper(II) sulfate, record the cathode's gain in mass at three different charges. (not done yet)

  4. Build a simple cell that gives the largest voltage you can. (not done yet)

Aqueous electrolysis: product decision guide

Use the question’s stated electrolyte, concentration and electrode material. These are qualitative syllabus rules for the stated examples, not numerical cut-offs or universal predictions for every electrolyte.

1 · Cathode (reduction)

Compare the cations

  • For a metal less reactive than hydrogen, such as copper in the required examples, its ions can be discharged to form the metal.
  • For ions of a very reactive metal, such as Na+, water is reduced and H2 forms in aqueous solution.

2 · Anode (oxidation)

Check electrode and anions

  • A reactive anode may itself be oxidised; a copper anode can form Cu2+.
  • With an inert anode, use the stated anions and concentration. In the required NaCl comparison, dilute solution gives O2 while concentrated brine gives Cl2.

3 · Verify the answer

Write and check

  • Write one balanced half-equation at each electrode, including states.
  • Check both atoms and total charge, then state the observation or gas test if asked.

D. What the Setup Looks Like (Acidified Water Example)

Labelled electrolytic cell and charge movementA direct-current power supply is connected to two inert electrodes in an electrolyte. The positive anode attracts anions and is where oxidation occurs. The negative cathode attracts cations and is where reduction occurs. Electrons travel in the external wires while ions carry charge through the liquid.DC power supply+−electrons move in the external circuitelectrolyte: mobile ions carry chargeANODE (+)CATHODE (−)oxidation: loses e−reduction: gains e−−anion+cationions move through the liquid, not through the wiresLabelled electrolytic cell and charge movementMobile layout of a direct-current supply connected to a positive anode and negative cathode in an electrolyte. Anions move to the anode, cations move to the cathode, and electrons move through the wires.DC power supply+−electrons move through the wireselectrolyte: mobile ionsANODE (+)CATHODE (−)−anion+cationoxidationloses e−reductiongains e−
For acidified water with inert electrodes, hydrogen forms at the cathode and oxygen at the anode. Ion colours and sizes are symbolic, and the apparatus is schematic, not to scale. Gas-collection apparatus is omitted so the charge pathways remain clear.

4. Common Mistakes

  • Writing Na⁺ → Na in aqueous NaCl(aq) (sodium is too reactive; hydrogen forms instead).
  • Ignoring the stated concentration: concentrated sodium chloride produces chlorine at an inert anode in the required comparison.
  • Using 2H + (aq) + 2e⁻ → H₂(g) in alkaline solution without explaining where H⁺ comes from (use water reduction instead).
  • Forgetting gas tests (pop / relights / bleaches).
  • Forgetting to mention the electrode type when it matters (copper electrodes vs graphite).

5. Exam Tips

Explain why an ion is selected

Use the phrases preferentially discharged and discharged. Then justify with (i) reactivity series, (ii) concentration, or (iii) electrode type.

  • For hydrogen at the cathode in neutral or alkaline solution, write: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH-(aq)
  • For oxygen at an inert anode in alkaline solution, write: 4OH-(aq) → O₂(g) + 2H₂O(l) + 4e⁻

In acidic solution, oxygen formation can instead be written 2H₂O(l) → O₂(g) + 4H + (aq) + 4e⁻. These equations describe the same oxygen product in different conditions. Water continuously supplies reacting particles; the reaction is not limited to the tiny amount of ions initially present in pure water.

6. Worked Examples

Modelled example 1

Dilute Sulfuric Acid (Acidified Water), Inert Electrodes

Core

Problem

Dilute sulfuric acid is electrolysed using graphite electrodes. State both gases, write both half-equations and give the gas tests.
Study the worked solution

  1. Identify aqueous ions

    Method

    Include H⁺, SO₄²⁻ and water-derived OH⁻.

    Reason

    Aqueous electrolysis includes ions from both solute and water.

    Working

    Discharge candidates include H⁺ and water/OH⁻; sulfate remains.

  2. Reduce hydrogen ions

    Method

    Form hydrogen at the cathode.

    Reason

    Hydrogen ions gain electrons by reduction.

    Working

    2H + (aq) + 2e⁻ → H₂(g); a lighted splint gives a pop.

  3. Oxidise water at the anode

    Method

    Form oxygen rather than discharging sulfate.

    Reason

    Sulfate is not discharged at the inert anode under these conditions.

    Working

    2H₂O(l) → O₂(g) + 4H + (aq) + 4e⁻; a glowing splint relights.

Guided practice 2

Aqueous Sodium Hydroxide, Inert Electrodes

About 7 min

Problem

Aqueous NaOH is electrolysed using graphite electrodes. State both gases and write the half-equations.

Choose both products, then write the half-equations on paper

Cathode gas
Anode gas

Hints

Hint 1: cathode competition
Sodium is too reactive; water supplies hydrogen instead.
Hint 2: anode species
Oxidise hydroxide ions to oxygen.
View solution step by step

  1. Reduce water

    Method

    Produce hydrogen at the cathode.

    Reason

    Water is reduced preferentially instead of Na⁺.

    Working

    2H₂O(l) + 2e⁻ → H₂(g) + 2OH-(aq).

  2. Oxidise hydroxide

    Method

    Produce oxygen at the anode.

    Reason

    Hydroxide ions lose electrons at the inert positive electrode.

    Working

    4OH-(aq) → O₂(g) + 2H₂O(l) + 4e⁻.

Common misconception 3

Aqueous Sodium Chloride: Dilute vs Concentrated (Anode Products)

Find and correct the mistake

Learner claim

A student says chlorine always forms at the anode whenever aqueous sodium chloride is electrolysed. Correct the claim for dilute solution and concentrated brine, including gas tests.

Use chloride concentration

Dilute anode gas
Concentrated anode gas

View solution step by step

  1. Keep the cathode result fixed

    Method

    Use hydrogen at the cathode for both concentrations.

    Reason

    Water is reduced instead of sodium ions.

    Working

    2H₂O + 2e⁻ → H₂ + 2OH⁻; lighted splint gives a pop.

  2. Correct the dilute case

    Method

    Use oxygen at the dilute anode.

    Reason

    Hydroxide discharge is favoured at low chloride concentration.

    Working

    4OH⁻ → O₂ + 2H₂O + 4e⁻; glowing splint relights.

  3. Correct the concentrated case

    Method

    Use chlorine for concentrated brine.

    Reason

    High chloride concentration favours chloride discharge.

    Working

    2Cl⁻ → Cl₂ + 2e⁻; damp blue litmus turns red then bleaches white.

Guided practice 4

Aqueous Copper(II) Sulfate, Inert Electrodes

About 7 min

Ion-selection transfer

Copper(II) sulfate solution is electrolysed using graphite electrodes. State both products, write the half-equations and give one observation.

Choose the products, then write both equations on paper

Cathode product
Anode product

Hints

Hint 1: cathode hierarchy
Copper is less reactive than hydrogen and is deposited.
Hint 2: no halide
With no halide present, oxygen forms at the inert anode.
View solution step by step

  1. Reduce copper ions

    Method

    Deposit copper at the cathode.

    Reason

    Cu²⁺ is preferentially discharged.

    Working

    Cu²⁺(aq) + 2e⁻ → Cu(s).

  2. Oxidise hydroxide ions

    Method

    Form oxygen at the inert anode.

    Reason

    Sulfate is not discharged under these conditions.

    Working

    4OH-(aq) → O₂(g) + 2H₂O(l) + 4e⁻.

  3. Connect products to observations

    Method

    Report a reddish-brown cathode coating and paler blue solution.

    Reason

    Copper metal forms while Cu²⁺ ions are removed without replacement.

    Working

    Copper coats the cathode; the blue colour fades.

7. Mind Stretchers

Mind stretcher 1: No Halide PresentExtension

Predict the products when aqueous Na₂SO₄(aq) is electrolysed using inert electrodes.

Show Answer

At the cathode: hydrogen (sodium is too reactive to be discharged).

At the anode: oxygen (sulfate is not discharged).

Mind stretcher 2: Reactive Electrodes Change the AnodeExtension

Copper(II) sulfate solution is electrolysed using copper electrodes. State what happens at the anode, and why oxygen is not the main anode product.

Show Answer

The copper anode dissolves (is oxidised) to form Cu²⁺: Cu(s) → Cu²⁺(aq) + 2e⁻

Oxygen is not the main anode product because the anode is reactive and is oxidised more readily than OH⁻ being discharged.

Try independently: A learner predicts sodium at the cathode whenever sodium chloride is electrolysed. Compare molten sodium chloride with its aqueous solution, using inert electrodes. Explain which extra reacting species changes the aqueous prediction.

Show answer and reasoning

Molten sodium chloride has no water: Na⁺ gains electrons and sodium forms. In aqueous sodium chloride, water is preferentially reduced and hydrogen forms; sodium ions remain in solution. The formula of the dissolved salt alone cannot decide the product.

Practise and check

Practise and check

Practise selecting products from the stated ions, concentration and electrode material.

Open the Redox Chemistry topic check
Syllabus and review details

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