Loav · VCE Chemistry · Unit 3 / Area of Study 2
U3A2-04 · 2026 curriculum · Expanded original teaching guide · Study guide
1. Use electricity to drive chemical change
Electrolysis uses an external direct-current supply to drive a reaction that does not proceed spontaneously in the desired direction under the operating conditions. The anode is where oxidation occurs; the cathode is where reduction occurs. These definitions never change. In an electrolytic cell the anode is connected to the positive terminal and the cathode to the negative terminal. The supply withdraws electrons from the anode and supplies electrons to the cathode.
Electrons move through the external conductors, while ions carry charge through the electrolyte. Do not draw electrons passing through a solution or membrane. Cations generally migrate toward the cathode and anions toward the anode, but actual cell design and membrane selectivity control which ions can cross. Electrode reactions and electrolyte movement together maintain the current and avoid large charge accumulation.
2. Inventory the species before predicting products
A molten ionic compound contains mobile ions without water. For molten sodium chloride with suitable inert electrodes, Na⁺(l) + e⁻ → Na(l) occurs at the cathode and 2Cl⁻(l) → Cl₂(g) + 2e⁻ at the anode. Adding gives 2NaCl(l) → 2Na(l) + Cl₂(g). The state of sodium is liquid at the high temperature required to melt sodium chloride. A cold solid salt cannot conduct by freely moving ions in the same way.
An aqueous solution introduces water as a competing reactant. At a cathode, water can be reduced: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq). Thus aqueous sodium chloride does not simply produce sodium metal like the molten salt. At an anode, water may be oxidised, written in acidic form as 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻. In alkaline conditions, use 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻. These are medium-dependent forms; do not combine them carelessly with inconsistent conditions.
The electrochemical series lists standard reduction potentials. Among feasible cathode reductions, the more positive potential indicates a stronger thermodynamic tendency for reduction under standard conditions. At an anode consider oxidation, so reverse candidate reduction half-equations. A reactive electrode can itself oxidise: a copper anode may undergo Cu(s) → Cu²⁺(aq) + 2e⁻ rather than acting only as an inert conductor.
Predictions have limits. Standard potentials refer to standard conditions, not every concentration, pH, temperature or electrode surface. Competing reactions have different kinetic barriers and overpotentials, and applied voltage and electrode material matter. Concentrated brine can yield chlorine at an appropriate anode even where a simple standard-potential comparison points to oxygen. Use supplied experimental information and conditions rather than claiming the table guarantees a product. The table also does not directly establish the practical operating voltage, which must overcome additional losses.
3. Connect design to the desired product
A commercial cell needs a conductive electrolyte, suitable electrodes, a controlled power supply and a way to collect products. Molten electrolytes can enable metals that water would prevent depositing, but heating consumes energy. Dissolved additives can improve conductivity, alter melting behaviour or help control a desired process, depending on the stated chemistry. An additive is not automatically an inert spectator; check whether it can react.
Separators or membranes can prevent products mixing and reacting back, while allowing appropriate ionic charge transport. Removing products can improve collection and reduce unwanted secondary reactions. Electrode materials must conduct, withstand the conditions and either remain sufficiently inert or deliberately supply a reactant. Balance product quality, energy use, material durability and waste rather than calling any electrically driven process automatically sustainable.
4. Count charge, then electrons, then product
Charge Q = It, with current I in amperes, time t in seconds and Q in coulombs. Here Q denotes electric charge, not the reaction quotient used in equilibrium. The amount of electrons is n(e⁻) = Q/F, using F = 96 500 C mol⁻¹ from the 2026 data book. A half-equation then converts electrons to reactant or product. If z electrons make one formula unit of product, n(product) = Q/(zF). Convert to mass with m = nM or to ideal gas volume at SLC with V = n × 24.8 L mol⁻¹.
Worked example 1: copper deposition
Original calculation: an idealised copper-plating cell carries 1.20 A for 600 s, at 100% current efficiency. Cu²⁺(aq) + 2e⁻ → Cu(s). Q = 1.20 × 600 = 720 C; n(e⁻) = 720/96 500 = 0.00746114 mol. n(Cu) = 0.00373057 mol and m(Cu) = 0.00373057 × 63.5 = 0.237 g. The plated object is the cathode. A soluble copper anode can replenish Cu²⁺. If hydrogen formation diverts some charge, the deposited copper mass is smaller.
Worked example 2: work backwards to time
How long is required to deposit 0.500 g Cu at 2.00 A if 80.0% of charge deposits copper? n(Cu) = 0.500/63.5 = 0.00787402 mol; required product-forming n(e⁻) = 0.0157480 mol. Useful charge = 1519.69 C. Total charge = 1519.69/0.800 = 1899.61 C, hence t = 1899.61/2.00 = 950 s = 15.8 min. Dividing by efficiency correctly makes the required time longer.
5. Products can change as electrolysis proceeds
Make a species inventory at the relevant time, not only at the start. Reduction requires an oxidising agent actually present at the cathode; an electrode made of a metal does not automatically mean its dissolved cation is present. A copper electrode and Cu²⁺ ions are different species. A reactive anode can introduce new ions into solution, and those ions may later reach the cathode and compete for reduction. Depletion of one species can also make a previously minor reaction important.
For aqueous copper(II) sulfate with inert electrodes, an idealised process deposits copper at the cathode while oxidising water at the anode. Combining the electron-balanced half-equations gives 2Cu²⁺(aq) + 2H₂O(l) → 2Cu(s) + O₂(g) + 4H⁺(aq). Sulfate is a spectator in this model. Cu²⁺ concentration falls and acidity increases. With a soluble copper anode instead, copper oxidation replenishes Cu²⁺ at the same ideal rate at which cathode deposition removes it; electrolyte concentration can remain approximately constant if volume stays constant.
Faraday calculations therefore need explicit operating assumptions: constant or known current, a specified electron-to-product ratio, enough reactant, and stated current efficiency. Charge cannot make more copper than the available supply allows. Beyond depletion, a different cathode reaction can carry current; you must not continue extrapolating a copper-deposition formula indefinitely. Changing composition can affect product selectivity before mathematical depletion is reached, so idealised thresholds are models, not universal switch points.
Worked example 3: electrode choice changes the material balance
Transfer, original idealised dataset. A 250.0 mL solution initially contains 0.120 mol L⁻¹ Cu²⁺. Inert electrodes operate at 0.965 A for 50.0 min. Assume all cathodic charge deposits copper, all anodic charge forms oxygen, constant solution volume, and no competing reactions during this interval. Find copper mass, final [Cu²⁺] and oxygen volume at SLC.
Q = 0.965 × (50.0 × 60) = 2895 C; n(e⁻) = 2895/96 500 = 0.0300 mol. Cu²⁺(aq) + 2e⁻ → Cu(s), so n(Cu) = 0.0150 mol and m(Cu) = 0.0150 × 63.5 = 0.953 g. Initial n(Cu²⁺) = 0.120 × 0.2500 = 0.0300 mol, so 0.0150 mol remains and final concentration is 0.0600 mol L⁻¹. The reactant-availability check passes.
Anode oxidation transfers four electrons per O₂, giving n(O₂) = 0.0300/4 = 0.00750 mol and V(O₂) = 0.186 L at SLC. The ideal charge required to remove all initial Cu²⁺ is 2 × 0.0300 × 96 500 = 5790 C, corresponding to 100 min at this current. Real selectivity need not remain unchanged up to that endpoint. If a soluble copper anode replaces the inert anode, the ideal cathode deposit is still 0.953 g, but the anode loses that mass, [Cu²⁺] remains 0.120 mol L⁻¹, and oxygen is not the specified anode product.
6. Use charge as a shared accounting quantity
Two cells in series carry the same charge when the same current passes for the same time. They need not deposit the same mass or amount. For a metal-ion reduction requiring z electrons, m = QM/(zF). Equal charge therefore compares the ratio M/z, provided efficiencies are equal. For example, one electron forms one Ag atom but two electrons form one Cu atom; ignoring z loses the chemical information.
An inverse problem can determine an unknown ion charge from z = n(e⁻)/n(metal), but only when the deposited substance, molar mass and current efficiency are known. A non-integer experimental ratio does not prove fractional ionic charge: gas production, incomplete deposit recovery or current measurement error can affect it. A deposit measured while wet may appear too heavy; lost flakes may make it too light. One anomalous result alone does not establish whether the problem is systematic or random. Seek repeats and a mechanism.
For changing current, total charge is the area under the current–time graph. For constant-current intervals, add each It contribution. Do not use the final current multiplied by the whole time. This arithmetic extension uses the same Faraday principle and is useful for variable renewable electricity: the total electrons, not the highest instantaneous current, constrain ideal production.
Practice
- Why does adding water change the prediction for sodium production?
- Give oxidation and reduction equations for transfer of copper from a soluble anode to a cathode.
- At 100% efficiency, calculate H₂ volume at SLC from 0.500 A for 1930 s.
- What current deposits 0.635 g Cu in 965 s at 100% efficiency?
- A cell runs at 0.400 A for 300 s, then 0.700 A for 600 s. Assuming all cathodic charge reduces Cu²⁺, find the deposited mass.
- The same 1930 C passes through separate ideal silver and copper deposition cells. Calculate each mass using Ag = 107.9 and Cu = 63.5 g mol⁻¹.
- An ideal experiment deposits 0.5935 g of tin when 965 C passes. Given M(Sn) = 118.7 g mol⁻¹, infer the ion charge. What assumption would gas bubbles at the cathode call into question?
Worked answers
- Water supplies a competing cathode reduction and is reduced more readily than Na⁺ under ordinary aqueous conditions; hydrogen forms instead. Molten salt has no water competitor.
- Anode: Cu(s) → Cu²⁺(aq) + 2e⁻. Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s). Net effect is copper transfer; label different electrodes so cancellation does not hide that physical change.
- Q = 965 C; n(e⁻) = 0.0100 mol; n(H₂) = 0.00500 mol from two electrons per molecule. V = 0.00500 × 24.8 = 0.124 L.
- n(Cu) = 0.635/63.5 = 0.0100 mol. Q = 2 × 0.0100 × 96 500 = 1930 C. I = 1930/965 = 2.00 A.
- Q = 0.400 × 300 + 0.700 × 600 = 540 C. n(e⁻) = 540/96 500 = 0.00559585 mol; n(Cu) = 0.00279793 mol. m = 0.00279793 × 63.5 = 0.178 g. The final current alone would overestimate the first interval’s charge.
- n(e⁻) = 0.0200 mol in each. Ag⁺(aq) + e⁻ → Ag(s) gives 0.0200 mol Ag and 2.16 g. Cu²⁺(aq) + 2e⁻ → Cu(s) gives 0.0100 mol Cu and 0.635 g. Equal charge means equal electrons, not equal metal amounts or masses.
- n(Sn) = 0.5935/118.7 = 0.005000 mol; n(e⁻) = 965/96 500 = 0.0100 mol; ratio = 2, so Sn²⁺(aq) + 2e⁻ → Sn(s). Gas suggests a competing cathode reaction, calling the 100% metal-deposition current-efficiency assumption into question.
