Calculate amounts and masses of electrolysis products
Use electrode half-equations and Faraday's constant to convert charge into amounts, masses or gas volumes.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 24. Treat all tabulated electrode potentials as reduction potentials; state conditions, phases, signs and units. Written practice is self-checked
Before you begin
- Use AS redox half-equations, oxidation states, mole ratios, current and charge.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many moles of electrons form one mole of Cu from Cu²⁺?
What you will learn
- Relate electrons to product amount using a half-equation
- Calculate mass of a deposited or liberated substance
- Calculate a gas volume from electrolysis charge and stated conditions
Use the balanced electrode half-equation
First calculate n(e⁻) = Q/F. Then use the half-equation to determine how many moles of electrons produce one mole of product. For Mᶻ⁺ + z e⁻ → M, n(M) = n(e⁻)/z.
For a deposited metal, mass m = n(M) × Mᵣ. For a gas, convert product amount to volume using the molar volume stated for the question's conditions.
Keep the electron ratio attached to the product
For hydrogen formation, 2H⁺ + 2e⁻ → H₂, so two moles of electrons produce one mole H₂. For chlorine formation, 2Cl⁻ → Cl₂ + 2e⁻, two moles of electrons are released per mole Cl₂.
Check that the electrode reaction and product match the electrolysis conditions before calculating. Charge alone does not identify a product; the balanced half-equation supplies the ratio.
Pause and explain
If 2.00 mol of electrons reduce Cu²⁺ to Cu, how much Cu forms?
Worked example
A charge of 1.93 × 10⁵ C passes through Cu²⁺ solution. Calculate the mass of copper deposited. Use F = 9.65 × 10⁴ C mol⁻¹ and Aᵣ(Cu) = 63.5.
Show the worked solution
- n(e⁻) = Q/F = 2.00 mol.
- Cu²⁺ + 2e⁻ → Cu, so n(Cu) = 1.00 mol.
- Multiply by the molar mass.
Answer m(Cu) = 1.00 × 63.5 = 63.5 g.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate the mass of silver deposited by 1930 C using Ag⁺ + e⁻ → Ag, F = 9.65 × 10⁴ C mol⁻¹ and Aᵣ(Ag) = 107.9.
Give me a hint
Find moles of electrons, use the 1:1 ratio, then calculate mass.
Compare my reasoning
- n(e⁻) = 1930/(9.65 × 10⁴) = 0.0200 mol.
- The half-equation gives n(Ag)=0.0200 mol.
- m = nMᵣ.
m(Ag) = 0.0200 × 107.9 = 2.16 g.
Look for these in your work
- Charge converted
- Electron ratio applied
- Mass and unit correct
A charge of 19 300 C produces hydrogen by 2H⁺ + 2e⁻ → H₂. Find the gas volume at room conditions using 24.0 dm³ mol⁻¹.
Give me a hint
Convert charge to electron moles, divide by two for H₂, then use molar volume.
Compare my reasoning
- n(e⁻) = 19 300/(9.65 × 10⁴) = 0.200 mol.
- n(H₂) = 0.100 mol.
- V = 0.100 × 24.0.
V(H₂) = 2.40 dm³ at room conditions.
Look for these in your work
- Electron amount correct
- Half-equation ratio applied
- Volume conditions and units stated
A student deposits 0.50 mol of a metal from M²⁺ but uses n(e⁻)=0.50 mol. Correct the charge requirement.
Give me a hint
Use the two-electron reduction ratio for a 2+ ion.
Compare my reasoning
- Each M²⁺ gains two electrons.
- 0.50 mol M requires 1.00 mol electrons.
- Multiply by F to obtain charge if needed.
0.50 mol M requires 1.00 mol e⁻ because M²⁺ + 2e⁻ → M. The charge is 0.50 × 2 × F = 9.65 × 10⁴ C.
Look for these in your work
- Charge ratio identified
- Electron amount corrected
- Optional charge calculation consistent
Common mistakes
- Using one electron per Mᶻ⁺ ion when z is larger than one
- Confusing moles of electrons with moles of product
- Using the wrong gas molar volume
- Applying a cathode equation to an anode product
What can you explain now?
How many moles of electrons form one mole of Cu from Cu²⁺?
Compare with the explanation
Two
Cu²⁺ ions gain two electrons to become neutral copper atoms.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why must product calculations use both Q/F and the electron coefficient in the half-equation?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.