Describe an electrolytic determination of Avogadro's constant
Combine measured charge and electrode mass change to estimate charge per mole of electrons and then the Avogadro constant.
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
What two measurements give charge in an electrolysis experiment?
What you will learn
- Describe the measurements in an electrolytic method
- Use electrode stoichiometry to find moles of electrons
- Use F = Le to estimate Avogadro's constant
Measure charge and amount of deposited substance
A suitable electrolytic method deposits a metal from its ions, such as silver from Ag⁺ using an inert cathode. Measure current and time to find Q = It, and determine the cathode mass increase to find the amount of metal deposited.
For Ag⁺ + e⁻ → Ag, each deposited Ag atom corresponds to one electron. Thus the amount of deposited silver equals the amount of electrons transferred, and the charge per mole of electrons is F = Q/n(e⁻).
Convert Faraday constant to a particle count
Once F is estimated, use F = Le, so L = F/e. This gives the number of electrons in one mole from a known elementary charge.
Reliable measurements require a clean, dry electrode, a stable measured current and careful mass readings. Any uncertainty in current, time or mass affects the result; this is an explanation of the syllabus method, not a guarantee of a precise experimental value.
Pause and explain
After estimating F, how is L found if e is known?
Worked example
An electrolysis deposits 1.08 g Ag from Ag⁺ in 1930 s at 0.50 A. Use Aᵣ(Ag)=107.9 and e=1.60×10⁻¹⁹ C to estimate L.
Show the worked solution
- Q = It = 0.50×1930 = 965 C.
- n(Ag)=1.08/107.9≈0.0100 mol; the half-equation gives the same n(e⁻).
- F=Q/n(e⁻)≈96 500 C mol⁻¹, then L=F/e.
Answer L ≈ 96 500/(1.60×10⁻¹⁹) ≈ 6.03×10²³ mol⁻¹.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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.
Describe how mass change at a silver cathode can be linked to electron charge in an electrolysis experiment.
Give me a hint
Use Ag⁺ + e⁻ → Ag and Q=It.
Compare my reasoning
- Measure current and electrolysis time to find Q.
- Measure deposited silver mass and convert to moles.
- The 1:1 half-equation gives moles of electrons; calculate F and then L=F/e.
Find Q=It and n(Ag)=mass/107.9. Since Ag⁺ + e⁻ → Ag, n(e⁻)=n(Ag); calculate F=Q/n(e⁻), then L=F/e.
Look for these in your work
- Electrical measurements stated
- Mass converted to moles
- Stoichiometry linked to L
A measured experiment gives Q=1930 C and deposits 2.16 g Ag. Estimate F and L using Aᵣ(Ag)=107.9 and e=1.60×10⁻¹⁹ C.
Give me a hint
Convert silver mass to moles, then divide charge by electron moles.
Compare my reasoning
- n(Ag)=2.16/107.9≈0.0200 mol.
- n(e⁻)=0.0200 mol and F=1930/0.0200.
- Divide F by e.
F≈96 500 C mol⁻¹; L≈96 500/(1.60×10⁻¹⁹)=6.03×10²³ mol⁻¹.
Look for these in your work
- Mass conversion correct
- Faraday constant calculated
- Avogadro constant estimated
Why should the cathode be cleaned and dried before it is weighed after deposition?
Give me a hint
Think about what the balance is intended to measure and what else can add mass.
Compare my reasoning
- The measured change should be deposited metal only.
- Water or salts left on the electrode add extra mass.
- Contamination biases the calculated amount and Avogadro estimate.
Cleaning and drying reduces water or salt contamination, so the mass increase better represents deposited metal; otherwise the calculated mole amount and L are biased.
Look for these in your work
- Target quantity identified
- Contamination source noted
- Effect on calculation explained
Common mistakes
- Using the electrode mass as a mole count without dividing by molar mass
- Forgetting the electron-to-metal ratio
- Using F directly as Avogadro's constant
- Ignoring current, time or balance uncertainty
What can you explain now?
What two measurements give charge in an electrolysis experiment?
Compare with the explanation
Current and time
Multiply current by elapsed time using Q = It.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
How do a measured metal deposit and passed charge provide F, then L?
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.