15.3 - Electrolysis - Aqueous Solutions
What forms in an aqueous electrolysis
- A dissolved ionic compound has ions that are free to move, so its solution can be electrolysed without melting it.
- Water breaks up slightly into hydrogen and hydroxide ions:
- H2O(l) ⇌ H+(aq) + OH-(aq)
- This means four kinds of ion are present: the positive and negative ions of the dissolved compound, plus H+ and OH- from the water.
- At each electrode, only one of the two competing ions is discharged to form an element, while the other stays in solution.
- Inert electrodes (such as platinum or graphite) are used so they don't react.
Three example solutions
- Dilute sulfuric acid contains H+, SO42- and OH-. Hydrogen forms at the cathode and oxygen at the anode. OH- is discharged in preference to SO42-, so water is decomposed.
- Copper(II) chloride contains Cu2+, Cl-, H+ and OH-. Copper is deposited at the cathode and chlorine forms at the anode.
- Concentrated sodium chloride contains Na+, Cl-, H+ and OH-. Hydrogen forms at the cathode and chlorine at the anode. Na+ and OH- stay in solution, leaving sodium hydroxide.
| Solution | Ions present | At the cathode | At the anode | Left in the solution |
|---|---|---|---|---|
| Dilute sulfuric acid | H+, SO42-, OH- | Hydrogen | Oxygen | Sulfate ions |
| Copper(II) chloride | Cu2+, Cl-, H+, OH- | Copper | Chlorine | Water, as the compound is used up |
| Concentrated sodium chloride | Na+, Cl-, H+, OH- | Hydrogen | Chlorine | Sodium hydroxide |
Industrial use of sodium chloride electrolysis
Electrolysing concentrated sodium chloride produces three useful products:
- Sodium hydroxide - Used to make soap.
- Chlorine - Used for bleach and plastics.
- Hydrogen - Used as a fuel.
Half equations at aqueous electrodes
- A half equation shows the reaction at a single electrode, including the electrons transferred.
- Cathode - Positive ions gain electrons (reduction).
- Anode - Negative ions lose electrons (oxidation).
Half equations with inert electrodes
- At the cathode:
- 2H+ + 2e- → H2
- Cu2+ + 2e- → Cu
- At the anode:
- 4OH- → O2 + 2H2O + 4e-
- 2Cl- → Cl2 + 2e-
Non-inert copper anode
When the anode is made of copper, copper atoms lose electrons and dissolve into solution as ions:
Cu → Cu2+ + 2e-
Worked example - Half equations for dilute sulfuric acid
Dilute sulfuric acid is electrolysed. Write the half equation at each electrode and state which is oxidation and which is reduction.
Step 1 - Name the ion discharged at each electrode
- The solution contains H+, SO42- and OH-.
- H+ is discharged at the cathode and OH- at the anode.
Step 2 - Build the cathode equation
Hydrogen gas is H2, requiring two H+ ions and two electrons:
2H+ + 2e- → H2
Step 3 - Build the anode equation
Four OH- ions form one O2 molecule and two H2O molecules, releasing four electrons:
4OH- → O2 + 2H2O + 4e-
Step 4 - Identify oxidation and reduction
- The hydrogen ions gain electrons, so the cathode reaction is reduction.
- The hydroxide ions lose electrons, so the anode reaction is oxidation.
TermDefinition
Inert electrodeAn unreactive electrode, such as platinum or graphite, that does not take part in the electrolysis reaction.
Half equationAn equation that shows the electrons lost or gained by one substance in a reaction, such as the reaction at one electrode. It is also called an ion-electron equation.