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Electrolysis Predictions and Faraday’s Law

Learn and apply Electrolysis Predictions and Faraday’s Law in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Electrolysis Predictions and Faraday’s Law: Orientation

Electrolysis questions are prediction + calculation: decide what is discharged (molten vs aqueous), write half-equations, then use Q = It and n(e⁻) = Q/F to find product amounts.

Before the harder applications, review Writing Redox Equations from Half-Equations and keep the Electrochemistry hub as your route map.

Definitions (Must Know)

A. Electrolysis

Electrolysis is the conduction of electricity through an ionic compound (electrolyte), leading to chemical changes at the electrodes.

Key labels:

  • cathode (negative): reduction occurs
  • anode (positive): oxidation occurs

B. Faraday constant, F

The Faraday constant is the charge per mole of electrons: F ≈ 9.65 × 10⁴ C mol⁻¹

It links the Avogadro constant, L, and the elementary charge, e: F = Le

Detailed Explanations

A. Why electrode signs matter in electrolysis

Because the cathode is negative in electrolysis, therefore it attracts cations and supplies electrons (reduction). Because the anode is positive, therefore it attracts anions and removes electrons (oxidation).

B. Predicting products: molten vs aqueous

1) Molten ionic compounds (general rule)

  • at the cathode: the cation is reduced
  • at the anode: the anion is oxidised

Example (molten NaCl): Na + (l) + e⁻ → Na(l) 2Cl-(l) → Cl₂(g) + 2e⁻

2) Aqueous solutions (common syllabus patterns)

At the cathode (reduction), common competing reductions include:

  • H⁺ / water → H₂
  • metal ions → metal (for less reactive metals)

At the anode (oxidation), common competing oxidations include:

  • halide ions → halogen (often if concentrated)
  • water / OH⁻ → O₂

What to write in exams:

  • identify which species is preferentially discharged and justify it (reactivity / electrode potential ideas + concentration if given)

Aqueous electrolysis: product decision guide

Use the question’s stated electrolyte, concentration and electrode material. These are qualitative syllabus rules, not numerical cut-offs.

1 · Cathode (reduction)

Compare the cations

  • A less reactive metal ion, such as Cu2+, is discharged to form the metal.
  • For a more reactive metal ion, such as Na+, water is reduced and H2 forms.

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.

C. Faraday’s Law calculations (workflow)

  1. Find Q from current and time: Q = It
  2. Find moles of electrons: n(e⁻) = Q/F
  3. Use the electrode half-equation to convert n(e⁻) to moles of product
  4. Convert to mass or gas volume if needed:
    • mass: m = nM
    • gas: use pV = nRT with the temperature, pressure and units stated in the question

Mini example: If Q = 9650 C, then n(e⁻) = Q/F = 9650/(9.65 × 10⁴) = 0.100 mol.

D. Industrial electrolysis

Anodising aluminium: the aluminium object is made the anode. Oxidation forms a thicker, adherent aluminium oxide layer on its surface. The layer protects the underlying metal against further corrosion; technical operating details are not required.

Electrolytic purification of copper: impure copper is the anode and pure copper is the cathode in a solution containing Cu²⁺. At the anode, Cu(s) → Cu²⁺(aq) + 2e⁻; at the cathode, Cu²⁺(aq) + 2e⁻ → Cu(s). Copper therefore transfers from the impure anode to the pure cathode. Less reactive impurities may form an anode sludge while more reactive impurities may remain as ions; the assessed explanation rests on the electrode reactions.

Electrolytic purification of copperImpure copper is the positive anode and pure copper is the negative cathode in copper(II) sulfate solution. Copper atoms leave the anode as copper(II) ions and copper(II) ions gain electrons at the cathode. Insoluble impurities collect below the anode.Electrolytic purification of copperDC power supply+−ANODE (+)CATHODE (−)impure copperpure copper sheetCuCuCu²⁺Cu²⁺Cu²⁺Cu²⁺metal ions move to the cathodeAnode: Cu → Cu²⁺ + 2e⁻Cathode: Cu²⁺ + 2e⁻ → Cuinsoluble anode sludge
Track copper rather than memorising electrode signs: copper atoms leave the impure anode as ions, and copper ions gain electrons at the pure cathode.

Worked Examples

Modelled example 1

Write Molten-Electrolyte Half-Equations

Core

Problem

Molten PbBr₂ is electrolysed using inert electrodes. Write both half-equations.
Study the worked solution

  1. Cathode reduction

    Method

    Add electrons to lead(II) ions.

    Reason

    Positive ions gain electrons at the cathode.

    Working

    Pb²⁺(l) + 2e⁻ → Pb(l)

  2. Anode oxidation

    Method

    Remove electrons from bromide ions and pair bromine atoms.

    Reason

    Bromide is the only anion in the molten compound.

    Working

    2Br-(l) → Br₂(g) + 2e⁻

Guided practice 2

Calculate Moles of Electrons

About 6 min

Problem

A current of 2.00 A passes for 30.0 minutes. Calculate moles of electrons transferred.

Try this before viewing the solution

Hints

Hint 1: seconds
Current uses coulombs per second.
Hint 2: faraday
After Q = It, use n(e⁻) = Q/F.
View solution step by step

  1. Find charge

    Method

    Convert 30.0 min to 1800 s, then multiply by current.

    Reason

    An ampere is one coulomb per second.

    Working

    Q = (2.00)(1800) = 3600 C

  2. Convert charge to amount

    Method

    Divide by Faraday’s constant.

    Reason

    F is charge per mole of electrons.

    Working

    n(e⁻) = 3600/(9.65 × 10⁴) = 3.73 × 10⁻² mol

Common misconception 3

Correct the Time-Unit Error

Find and correct the mistake

Learner attempt

For 1.50 A flowing for 20.0 min, a learner writes Q = (1.50)(20.0) = 30.0 C. Correct the first error and the charge.

Try this before viewing the solution

Time unit for Q = It

View solution step by step

  1. Convert time

    Method

    Multiply minutes by 60.

    Reason

    The current unit contains seconds.

    Working

    t = (20.0)(60) = 1200 s

  2. Recalculate

    Method

    Multiply 1.50 A by 1200 s.

    Reason

    Current × time gives charge.

    Working

    Q = 1800 C

Examiner practice 4

Convert Electron Amount to Aluminium

2 marks

Problem

For Al³⁺ + 3e⁻ → Al, calculate aluminium produced by 0.150 mol electrons. [2 marks]

Try this before viewing the solution

View solution step by step

  1. Use the ratio

    1 mark

    Method

    Relate 3 mol electrons to 1 mol aluminium.

    Reason

    The half-equation coefficients give the mole ratio.

    Working

    n(Al) = n(e⁻)/3

  2. Calculate

    1 mark

    Method

    Divide 0.150 by 3.

    Reason

    Three electron moles are required per aluminium mole.

    Working

    n(Al) = 0.0500 mol

Examiner practice 5

Compare copper purification with aluminium anodising

6 marks

Problem

In electrolytic purification, impure copper is made the anode and pure copper is deposited at the cathode. In anodising, an aluminium object is made the anode. Explain the purpose and electrode process in each use. [6 marks]

Try this before viewing the solution

View solution step by step

  1. Oxidise the impure copper anode

    1 mark

    Method

    Write the loss of electrons from copper atoms.

    Reason

    Oxidation always occurs at the anode.

    Working

    Cu(s) → Cu²⁺(aq) + 2e⁻.

  2. Deposit copper at the pure cathode

    2 marks

    Method

    Reduce copper(II) ions onto the pure copper sheet.

    Reason

    The cathode supplies electrons, so copper transfers from the impure electrode to the pure one.

    Working

    Cu²⁺(aq) + 2e⁻ → Cu(s); the cathode gains pure copper.

  3. Account for impurities

    1 mark

    Method

    State that less reactive impurities can collect below the anode while more reactive ones may remain as ions.

    Reason

    They do not plate onto the cathode under the chosen operating conditions in the same way as copper.

    Working

    The process separates copper from its impurities rather than merely moving the whole anode.

  4. Explain anodising

    2 marks

    Method

    Make the aluminium object the positive anode so oxidation at its surface builds a thicker adherent oxide layer.

    Reason

    The oxide layer protects the underlying aluminium from further corrosion.

    Working

    Purification removes impurities from copper; anodising deliberately changes and protects the aluminium surface.

Challenge 6

Predict Brine Products

Minimal support

Problem

Concentrated aqueous NaCl is electrolysed with inert electrodes. Predict both electrode products and write half-equations.

Try this before viewing the solution

Cathode product
Anode product

Hints

Hint 1: aqueous
Water introduces competing discharge possibilities absent from a melt.
Hint 2: concentrated-halide
Use the concentrated-halide rule at the anode.
View solution step by step

  1. Cathode

    Method

    Reduce water to hydrogen.

    Reason

    Sodium ions are not discharged from aqueous solution under these conditions.

    Working

    2H₂O(l) + 2e⁻ → H₂(g) + 2OH-(aq)

  2. Anode

    Method

    Oxidise chloride to chlorine.

    Reason

    Concentrated halide is preferentially discharged at the inert anode.

    Working

    2Cl-(aq) → Cl₂(g) + 2e⁻

Mind Stretchers

Mind stretcher 1Extension

Why can the products of aqueous electrolysis differ from molten electrolysis for the same compound?

Show Hint

List every species actually present in the molten and aqueous cases; water adds competing discharge possibilities.

Show Answer

Mark scheme:

  • In aqueous solutions, water provides H⁺/OH⁻ which can be discharged instead of the ions from the compound.
  • The preferred discharge depends on relative ease of oxidation/reduction and concentration, so products can change.

Mind stretcher 2: Using charge conservation across cells in seriesExtension

Question. Two electrolytic cells are connected in series. One deposits copper by Cu²⁺ + 2e⁻ → Cu while the other produces oxygen by 2H₂O → O₂ + 4H⁺ + 4e⁻. If 0.0200 mol of copper is deposited, determine the amount of oxygen formed and explain why current and time need not be given.

Show Hint

Use the copper half-equation to infer electron amount. Cells in series receive the same charge.

Show Answer

Copper deposition uses 2(0.0200) = 0.0400 mol of electrons. The same charge, and therefore the same amount of electrons, passes through both series cells. Four moles of electrons produce one mole of oxygen, so n(O₂) = 0.0400/4 = 0.0100 mol. Current and time are unnecessary because the copper deposit already measures the charge passed.