Short on time? This page is the recap. For explanations, diagrams and worked examples, use the full study guide.
Recap
- In any electrochemical cell, the amount of substance oxidised or reduced is fixed by the number of electrons that pass.
- You measure electrons indirectly: current (amperes) and time (seconds) give the charge (coulombs).
- The Faraday constant, F = 96,485 C mol⁻¹, is the charge on one mole of electrons. It turns coulombs into moles of electrons.
- The half-reaction gives the electrons needed per particle. For a metal ion, this is the size of its charge.
- The same counting works for metal deposited at a cathode, metal removed from an anode, gas formed, and the anode used up in a galvanic cell.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| I = q / t | I in A, q in C, t in s | find charge (q = I × t), time (t = q / I) or current |
| n(e⁻) = q / F | F = 96,485 C mol⁻¹ | turn charge into moles of electrons (or q = n(e⁻) × F) |
| n(substance) = n(e⁻) ÷ electrons per particle | from the half-reaction | link electrons to metal or gas |
| m = n × M | M in g mol⁻¹ | turn moles of product into mass |
| electrons per ion = n(e⁻) ÷ n(metal) | from measured mass and charge | find the charge on an unknown metal ion |
Assumptions behind the numbers
- The current is steady for the whole time (or you use the average current).
- All the charge goes into the reaction you are calculating: no side reactions such as hydrogen forming at the cathode. The answer is therefore a maximum.
- The solution does not run out of the ion being reduced.
- Cells joined in series share one path, so they pass the same charge.
Mistakes to avoid
- Time in minutes or hours. Convert to seconds before q = I × t.
- Treating coulombs as moles. Divide by F.
- Ignoring the ion’s charge. Cu²⁺ needs 2 mol e⁻ per mol Cu; Au³⁺ needs 3.
- Gas electron counts. 2 e⁻ per Cl₂ or H₂, but 4 e⁻ per O₂ from water.
- Equal masses at both electrodes. Equal moles of electrons, not equal masses.
- Thinking concentration or electrode size changes the yield. For a fixed charge, they do not.
Quick self-check
- How much charge passes when 1.00 A flows for 1.00 h, and how many moles of electrons is that? (3600 C; 0.0373 mol)
- How many moles of electrons deposit 1 mol of aluminium from Al³⁺? (3 mol)
- What mass of copper does 0.100 mol of electrons deposit from Cu²⁺? (0.0500 mol Cu, so 3.18 g)
Next: practice questions.