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K324 / 2027
Redox Chemistry

Lesson 4 of 6

Aqueous electrolysis: more than one possible ion

Water introduces competing products.

O-Level 6092 (2026) / SEC G3 K324 (2027)

Pure only
  • I can apply selective discharge.Syllabus

    Syllabus K324, 7.2(e). Apply selective discharge

  • I can predict aqueous products.Syllabus

    Syllabus K324, 7.2(f). Predict aqueous products

  • I can construct electrode equations.Syllabus

    Syllabus K324, 7.2(g). Construct electrode equations

Aqueous electrolysis: more than one possible ion

Water introduces competing products.

Make a guess. You are not marked.

Concentrated aqueous sodium chloride is electrolysed with inert electrodes. What forms at the cathode?

  1. Sodium metal, from Na+
  2. Hydrogen gas, from water
  3. Chlorine gas, from Cl-
Show the answer

Hydrogen gas, from water

Sodium is very reactive, so sodium ions are hard to reduce. Hydrogen ions from water are reduced instead, giving hydrogen gas.

In aqueous electrolysis, consider the solute ions and the water. With inert electrodes, a metal less reactive than hydrogen, such as copper, is usually deposited at the cathode. For a very reactive metal ion such as Na+, hydrogen forms instead. At the anode, sulfate is generally not discharged; oxygen forms from water/hydroxide. Halide concentration matters: concentrated chloride favours chlorine, while dilute chloride solution is treated here as giving oxygen.

Apply selective discharge to a new solution
  1. List species and electrode material

    For aqueous potassium iodide with inert electrodes, consider K+, I- and water (H+ and OH-). Molten KI has no water competing.

  2. Choose the cathode product

    Potassium is above hydrogen in the reactivity series. In the school aqueous model, hydrogen forms rather than potassium metal. For a metal below hydrogen, such as copper or silver, the metal is normally deposited.

  3. Choose the anode product

    Halide discharge forms the corresponding halogen: bromide gives bromine; iodide gives iodine. For this iodide solution iodine forms; 2I-(aq) -> I2(aq) + 2e-. Chloride needs the dilute/concentrated comparison shown below. Sulfate is not discharged in these examples: oxygen forms from water/hydroxide.

  4. Check atoms, charge and observations

    The iodide half-equation has two iodine atoms and total charge -2 on each side. The anode region becomes brown as iodine dissolves; hydrogen bubbles at the cathode. Do not describe every halogen product as a gas.

Products with inert electrodes
ElectrolyteCathodeAnode
Aqueous copper(II) sulfateCopperOxygen
Dilute aqueous sodium chlorideHydrogenOxygen
Concentrated aqueous sodium chlorideHydrogenChlorine

Label an aqueous electrolysis cell

Two graphite electrodes dip into blue aqueous copper(II) sulfate. Sulfate and hydroxide ions move to the anode, where oxygen bubbles form. Copper(II) and hydrogen ions move to the cathode, which becomes coated with pink-brown copper. The blue colour fades as copper(II) ions are removed.

Label the d.c. supply, the anode (+) and cathode (-), what each electrode is made of, the electrolyte and its state, and the product at each electrode. With inert electrodes, copper deposits at the cathode and oxygen bubbles off at the anode, so the blue colour fades and the solution becomes acidic.
What happens to the electrolyte (inert electrodes)
ElectrolyteColourpHReason
Copper(II) sulfate(aq)Blue fadesFalls: the solution becomes acidicCu2+ ions are removed as copper. OH- ions are removed as oxygen, so H+ and SO42- ions are left: sulfuric acid forms.
Concentrated sodium chloride(aq)Stays colourlessRises: the solution becomes alkalineHydrogen forms from water at the cathode and chlorine forms at the anode, so Na+ and OH- ions are left: sodium hydroxide forms.
Dilute sodium chloride(aq)Stays colourlessStays about 7 overallHydrogen and oxygen both come from water, so water is used up and the solution slowly becomes more concentrated.
Dilute sulfuric acidStays colourlessFalls slowlyHydrogen and oxygen both come from water. The same amount of acid is left in less water, so it becomes more concentrated.

Collect the two gases to compare their volumes

Two inverted tubes full of dilute sulfuric acid sit over the two electrodes. The tube over the cathode holds twice as much gas, hydrogen, as the tube over the anode, which holds oxygen.

When water is electrolysed, the volume of hydrogen at the cathode is twice the volume of oxygen at the anode.

Why 2 : 1? Count the electrons. The same electrons that leave the anode arrive at the cathode, so both electrodes handle the same number of moles of electrons. Write both half-equations with the same number of electrons. Cathode: 4H+(aq) + 4e- → 2H2(g). Anode: 4OH-(aq) → O2(g) + 2H2O(l) + 4e-. Four moles of electrons make 2 mol of hydrogen but only 1 mol of oxygen. Equal moles of gas take up equal volumes at the same temperature and pressure, so the hydrogen volume is twice the oxygen volume. In a real experiment the oxygen volume is a little less than half, because oxygen is slightly more soluble in water than hydrogen.

Worked exampleLink the mass at one electrode to the gas at the other

Aqueous copper(II) sulfate is electrolysed with graphite electrodes. The cathode gains 0.640 g of copper. Calculate the volume of oxygen formed at the anode, at room temperature and pressure. (Ar: Cu = 64; one mole of gas occupies 24 dm3 at r.t.p.)

Try it first. Then show the worked stepsHide the worked steps
  1. Moles of copper deposited = 0.640 ÷ 64 = 0.0100 mol.
  2. Cathode: Cu2+ + 2e- → Cu, so moles of electrons = 2 × 0.0100 = 0.0200 mol.
  3. Anode: 4OH- → O2 + 2H2O + 4e-, so moles of oxygen = 0.0200 ÷ 4 = 0.00500 mol.
  4. Volume of oxygen = 0.00500 × 24 = 0.120 dm3.
Answer

0.120 dm3 (120 cm3) of oxygen. The electrons are the link: the same 0.0200 mol of electrons passes through both electrodes.

Useful balanced half-equations
ProcessHalf-equation
Copper depositionCu2+(aq) + 2e- -> Cu(s)
Hydrogen from water2H2O(l) + 2e- -> H2(g) + 2OH-(aq)
Oxygen from hydroxide4OH-(aq) -> O2(g) + 2H2O(l) + 4e-
Chlorine from chloride2Cl-(aq) -> Cl2(g) + 2e-

Worked exampleConstruct a half-equation from an unfamiliar ion charge

An aqueous solution contains Ag+ ions and silver is discharged at an inert cathode. A separate molten lead(II) bromide cell produces bromine at the anode. Construct the relevant half-equations.

Try it first. Then show the worked stepsHide the worked steps
  1. At the cathode an Ag+ ion gains one electron to make a neutral silver atom: Ag+(aq) + e- -> Ag(s).
  2. At the anode bromide ions lose electrons. Bromine is Br2, so begin 2Br- -> Br2, then add 2e- on the right to balance charge.
  3. For the molten-cell anode use Br-(l) and Br2(g); the molten electrolyte is hot enough for bromine to be a gas. Check both atoms and net charge.
Answer

Cathode in the silver solution: Ag+(aq) + e- -> Ag(s). Anode in molten lead(II) bromide: 2Br-(l) -> Br2(g) + 2e-. Each equation must balance atoms and charge.

Worked exampleCompare dilute and concentrated brine

Explain why changing NaCl concentration can change one electrode product.

Try it first. Then show the worked stepsHide the worked steps
  1. Na+ is not deposited from either aqueous solution; hydrogen forms at the cathode.
  2. In dilute solution, oxygen forms at the inert anode.
  3. At high chloride concentration, chloride is preferentially discharged and chlorine forms.
Answer

Cathode product remains hydrogen; anode product changes. Always state aqueous/molten, concentration and electrode material.

Check your understanding

What happens to the blue colour during electrolysis of copper(II) sulfate solution with inert electrodes?

  1. It fades, because Cu2+ ions are removed at the cathode and nothing replaces them.
  2. It stays the same, because the anode releases Cu2+ as fast as the cathode removes it.
  3. It deepens, because water is used up and the copper(II) sulfate becomes more concentrated.
Show the answer

It fades, because Cu2+ ions are removed at the cathode and nothing replaces them.

Cu2+ is discharged as copper at the cathode. An inert anode makes oxygen instead of dissolving, so the Cu2+ that gives the blue colour is not replenished.

Check your understanding

Dilute sulfuric acid is electrolysed with inert electrodes. 36 cm3 of hydrogen collects at the cathode. What volume of oxygen collects at the anode in the same time?

  1. 18 cm3, because 4 mol of electrons make 2 mol H2 but 1 mol O2
  2. 36 cm3, because exactly the same amount of charge passes through both of the electrodes
  3. 72 cm3, because each oxygen molecule is made from two oxygen atoms joined together
Show the answer

18 cm3, because 4 mol of electrons make 2 mol H2 but 1 mol O2

Cathode: 2H+ + 2e- -> H2. Anode: 4OH- -> O2 + 2H2O + 4e-. The same 4 mol of electrons give 2 mol of H2 and 1 mol of O2, so H2 : O2 is 2 : 1 by volume. 36 / 2 = 18 cm3.

Check your understanding

A few drops of Universal Indicator are added to concentrated aqueous sodium chloride, which is then electrolysed with inert electrodes. What happens to the colour near the cathode?

  1. It turns red, because acid forms around the cathode
  2. It turns violet, because hydroxide ions are left there
  3. It is bleached, because chlorine forms near it
  4. It stays green, because sodium chloride is neutral
Show the answer

It turns violet, because hydroxide ions are left there

At the cathode, hydrogen forms from water: 2H2O + 2e- -> H2 + 2OH-. Hydroxide ions build up with the Na+ ions, so the solution there becomes alkaline.