Electrolysis

Specification Reference Physical Chemistry, Electrochemistry 24.1

Quick Notes

Electrolysis uses electricity to drive non-spontaneous redox reactions.
In molten electrolytes, only the cations and anions of the ionic compound are present:
- Cations are reduced at the cathode (gain electrons).
- Anions are oxidised at the anode (lose electrons).
In aqueous electrolytes, H⁺(aq) and OH⁻(aq) can also be reduced or oxidised:
- At the cathode:
  - If the metal is more reactive than hydrogen then H₂ gas forms.
  - If the metal is less reactive than hydrogen then metal forms.
- At the anode:
  - OH⁻ ions are usually oxidised to form O₂ gas.
  - If halide ions (Cl⁻, Br⁻, I⁻) are present in high enough concentration, they are oxidised instead (e.g., Cl₂ gas).
  - Concentration matters: In dilute halide solutions, OH⁻ ions may be oxidised instead of halides.
Predicting products at the electrodes: We can use standard electrode potentials (E°) and the reactivity series.
Key formulas:
- F = L × e: Links Faraday constant, Avogadro constant, and charge on an electron.
- Q = It: Charge = current × time.
Determining Avogadro’s constant: Electrolysis can be used experimentally to determine Avagadro’s constant.

Full Notes

Electrolysis is the process of using electricity to break down a compound and drive a non-spontaneous reaction.

It works by passing an electric current through a liquid or molten ionic substance called an electrolyte, which contains free ions that can move and carry charge.

Electrodes are solid conductors that allow electricity to enter and leave the electrolyte:

CIE A-Level Chemistry labelled electrolysis apparatus showing power supply, electrodes, and electrolyte.

The cathode is the negative electrode, where reduction (gain of electrons) happens.
The anode is the positive electrode, where oxidation (loss of electrons) happens.

During electrolysis, positive ions move towards the cathode to gain electrons, while negative ions move towards the anode to lose electrons.

Matt’s Exam Tip

You should always remember the cathode is where reduction takes place and anode where oxidation takes place to avoid any confusion.

This is because in voltaic cells the cathode is positively charged and the anode is negatively charged (the opposite way round to electrolysis) however still the cathode is where reduction takes place and the anode where oxidation takes place. If you remember cathode = reduction and anode = oxidation, you will always be correct, regardless of whether the question is about electrolysis or voltaic cells.

Predicting Products of Electrolysis

We can predict the products formed at each electrode based on the type of electrolyte being used.

Molten Electrolytes

The ionic compound is melted to a liquid state (molten), only the cations and anions of the compound are present:

Cation is reduced at the cathode.
Anion is oxidised at the anode.

Example Electrolysis of molten NaCl

CIE A-Level Chemistry molten sodium chloride electrolysis showing Na formed at cathode and Cl2 at anode.

Cathode: Na⁺ + e⁻ → Na
Anode: 2Cl⁻ → Cl₂ + 2e⁻

Aqueous Electrolytes

The ionic compound is dissolved in water and is aqueous (aq). H⁺(aq) and OH⁻(aq) ions from water are also present.

As a result, water may compete with the ions from the compound at the electrodes.

We can use standard electrode potentials (E° values) or reactivity trends to predict which species is discharged.

At the cathode:
- H⁺(aq) ions are reduced to H₂(g) if the metal is more reactive than hydrogen (lower E° value).
- The metal ion is reduced if it is less reactive than hydrogen (higher E° value).
At the anode:
- Generally, OH⁻(aq) is oxidised to form O₂ gas.
- If halide ions (e.g. Cl⁻, Br⁻) are present, they are oxidised instead.
- Note: Ions like SO₄²⁻ and NO₃⁻ do not get oxidised.

Example Electrolysis of aqueous NaCl

CIE A-Level Chemistry electrolysis of aqueous sodium chloride showing hydrogen at cathode and chlorine at anode.

Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻
Anode: 2Cl⁻ → Cl₂ + 2e⁻

Concentration matters:

If halide concentration is very low (e.g., dilute NaCl), OH⁻ from water may be oxidised instead.
If the metal ion concentration is very low, hydrogen gas may form instead of metal at the cathode.

Relationship Between F, L, and e

The Faraday constant (F) is the amount of electric charge carried by 1 mole of electrons.

It connects the charge on one electron to the charge on a mole of electrons:

CIE A-Level Chemistry relation F = L × e linking Faraday constant with Avogadro constant and elementary charge.

F = L × e

where:

F = Faraday constant = 96500 C mol⁻¹
L = Avogadro constant (6.022 × 10²³ mol⁻¹)
e = elementary charge (1.60 × 10⁻¹⁹ C)

Electrolysis Calculations

Charge Passed:
The total electrical charge that flows (charged passed) during electrolysis can be calculated using:

CIE A-Level Chemistry formula Q = I t for charge passed in electrolysis.

Q = It

where:

Q = charge (Coulombs)
I = current (Amperes)
t = time (seconds)

Mass or Volume of Substance Formed

By using both F = L × e and Q = It, we can determine how much a given substance has formed at an electrode during electrolysis.

Steps:

Calculate total charge: Q = It
Calculate moles of electrons: n(e⁻) = Q ÷ F
Use half-equations to relate electrons to product.
Convert moles of product to mass or volume.

Worked Example 1: Mass of metal deposited

An aqueous solution of CuSO₄ was electrolysed with a current of 2.0 A for 30 minutes. How much copper formed at the cathode?

Q = 2.0 × 1800 = 3600 C
n(e⁻) = 3600 ÷ 96500 ≈ 0.0373 mol
For Cu²⁺ + 2e⁻ → Cu, moles Cu = 0.0373 ÷ 2 = 0.0186 mol
Mass Cu = 0.0186 × 63.5 = 1.18 g

Answer: 1.18 g of copper is deposited.

Worked Example 2: Volume of gas produced

An aqueous solution of NaCl was electrolysed with a current of 0.50 A for 30 minutes. What volume of H₂(g) was produced? (r.t.p = 24 dm³ mol⁻¹).

Q = 0.50 × 30 × 60 = 900 C
n(e⁻) = 900 ÷ 96500 ≈ 0.00933 mol
For 2H⁺ + 2e⁻ → H₂, moles H₂ = 0.00933 ÷ 2 ≈ 0.004665 mol
Volume H₂ = 0.004665 × 24 = 0.112 dm³

Answer: 0.112 dm³ of hydrogen gas is produced.

Determining Avogadro’s Constant Experimentally

We can determine L by electrolysing a known solution and measuring the substance formed:

Measure mass (or volume) of the product.
Calculate moles of electrons involved.
Measure total charge (Q = It).
Calculate e = Q ÷ number of electrons.
Rearranging F = L × e, find L = F ÷ e.

This links experimental mass/volume measurements to fundamental constants like L and e.

Summary

Electrolysis drives non-spontaneous redox reactions using electrical energy.
Molten electrolytes: cations reduce at cathode and anions oxidise at anode.
Aqueous electrolytes: H⁺ and OH^- ions can compete with ions from the compound present. We can use E°/reactivity and concentration to predict products.
Key relations: F = L × e and Q = It
Electrolysis can be used to determine Avogadro’s constant experimentally.