Electrolysis is a reaction driven by electricity, and every electron can be counted. Predict which ions are discharged from E values and concentration, then turn current and time into moles, grams and cubic decimetres, and even into the Avogadro constant.
In electrolysis the cathode (negative) reduces and the anode (positive) oxidises. In aqueous solution the ions compete with water: at the cathode the species with the more positive E⊖ is reduced (Cu2+ and Ag+ before H2O; H2 forms instead of Na or K), and at the anode the species with the less positive E⊖ is oxidised, unless a high concentration of Cl− makes chlorine form instead of oxygen. The charge passed is Q = It, and one mole of electrons carries the Faraday constant, F = Le = 96 500 C mol−1. So 2.00 A for 965 s passes 1930 C = 0.0200 mol e−, depositing 0.0100 mol (0.635 g) of copper. Measuring the charge and the mass deposited gives L = F ÷ e.
Electrolyse molten and aqueous electrolytes in the simulator. For each one, record the cathode and anode products and compare them with the E⊖ values below.
| Electrolyte (inert electrodes) | Cathode product | Anode product | Reason |
|---|---|---|---|
| molten PbBr2 | Pb | Br2 | only Pb2+ and Br− present |
| CuSO4(aq) | Cu | O2 | Cu2+/Cu (+0.34 V) more positive than H2O/H2; SO42− not oxidised |
| Na2SO4(aq) | H2 | O2 | Na+/Na (−2.71 V) far too negative |
| dilute NaCl(aq) | H2 | mainly O2 | O2/H2O (+1.23 V) less positive than Cl2/Cl− (+1.36 V) |
| concentrated NaCl(aq) | H2 | Cl2 | high [Cl−] makes Cl− easier to oxidise than water |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
What is the quantity of charge passed when a current of 2.00 A flows for 965 s?
Which product forms at the cathode when aqueous sodium sulfate is electrolysed with inert electrodes?
What does the Faraday constant represent?
How many moles of electrons are needed to deposit one mole of copper from Cu2+(aq)?
Change the current and time in the electrolysis simulator and watch how much copper is deposited. Then check the numbers with the method below.
| Step | Copper from CuSO4(aq), 2.00 A for 965 s | Hydrogen at room conditions, same charge |
|---|---|---|
| Q = It | 2.00 × 965 = 1930 C | 1930 C |
| n(e−) = Q ÷ F | 1930 ÷ 96 500 = 0.0200 mol | 0.0200 mol |
| half-equation ratio | Cu2+ + 2e− → Cu: 0.0100 mol | 2H+ + 2e− → H2: 0.0100 mol |
| amount to mass or volume | 0.0100 × 63.5 = 0.635 g | 0.0100 × 24.0 = 0.240 dm3 |
Time must be in seconds (minutes times 60). Use the half-equation to convert moles of electrons into moles of product: 2 e− per Cu, 3 e− per Al, 1 e− per Ag, 4 e− per O2. In aqueous solution, reactive metals such as Na and K are never deposited: hydrogen forms instead. Dilute chloride solutions give mainly oxygen at the anode, not chlorine.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
A current of 0.500 A is passed through aqueous silver nitrate for 30.0 minutes. What mass of silver is deposited? (Ar Ag = 107.9; F = 96 500 C mol−1)
Concentrated aqueous sodium chloride is electrolysed with inert electrodes. What are the products at the cathode and the anode?
In an experiment, 0.500 A for 1200 s deposits 0.198 g of copper. Using e = 1.60 × 10−19 C, what value of the Avogadro constant does this give?
The same charge is passed through molten NaCl, molten MgCl2 and molten AlCl3 in series. What is the ratio of moles of Na : Mg : Al deposited?
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