Electrochemical series, conductors & electrolytes, electrolysis of molten and aqueous solutions, Faraday calculations, and four industrial applications.
Section 1 — The Electrochemical Series
The series ranks metals by ease of losing electrons (ionising). A higher metal is a stronger reducing agent and displaces any metal below it from a salt solution.
Metals — Reactivity Series
| Metal | Ion | |
|---|---|---|
| Potassium K | K⁺ | ↑ Strongest reducer |
| Calcium Ca | Ca²⁺ | ↑ Easier to ionise |
| Sodium Na | Na⁺ | |
| Magnesium Mg | Mg²⁺ | |
| Aluminium Al | Al³⁺ | |
| Zinc Zn | Zn²⁺ | |
| Iron Fe | Fe²⁺ | |
| Lead Pb | Pb²⁺ | |
| Hydrogen H | H⁺ | Reference |
| Copper Cu | Cu²⁺ | |
| Silver Ag | Ag⁺ | ↓ Weakest reducer |
Observations: Mg gets smaller · Blue fades · Pink Cu deposits · Temperature rises.
Non-Metals (Oxidising Power)
Ranked by ease of gaining electrons. Higher = stronger oxidising agent and displaces those below from their compounds.
Section 2 — Electrical Conduction & Electrolytes
| Type | Ionisation | Bulb | Examples |
|---|---|---|---|
| Strong electrolyte | Fully ionised → single arrow | Bright | HCl(aq), H₂SO₄(aq), NaOH(aq), NaCl(aq) |
| Weak electrolyte | Partially ionised ⇌ | Dim | CH₃COOH(aq), NH₃(aq), H₂CO₃(aq) |
| Non-electrolyte | Does not ionise | No glow | Ethanol, glucose, sugar solution, wax |
Section 3 — Electrolysis
Molten PbBr₂ — Worked Example
Aqueous Solutions — Which Ion Is Discharged?
| Electrode | Factor | Rule |
|---|---|---|
| ANODE | Anion series position | Lower in series = easier to discharge. OH⁻ > I⁻ > Br⁻ > Cl⁻ > NO₃⁻ > SO₄²⁻ > F⁻ |
| ANODE | Concentration | Concentrated Cl⁻, Br⁻, I⁻ overrides position. Dilute NaCl → O₂; Conc. NaCl (brine) → Cl₂ |
| ANODE | Electrode type | INERT (graphite/Pt): normal discharge. ACTIVE (e.g. Cu): anode metal ionises instead |
| CATHODE | Cation series position | Lower in series = easier to discharge. Cu²⁺ below H⁺ → Cu deposits. Na⁺ above H⁺ → H₂ instead |
| Electrolyte | Electrodes | Anode | Cathode | Electrolyte Change |
|---|---|---|---|---|
| Dilute H₂SO₄ | Inert | O₂ | H₂ | More concentrated |
| Dilute NaCl | Inert | O₂ | H₂ | More concentrated |
| Conc. NaCl (brine) | Inert | Cl₂ | H₂ | Becomes alkaline (NaOH) |
| CuSO₄ | Inert graphite | O₂ | Cu (pink) | Paler blue, more acidic |
| CuSO₄ | Active copper | Cu dissolved | Cu (pink) | Unchanged |
Section 4 — Quantitative Electrolysis (The Faraday)
5-Step Method
- 1Calculate Q = I × t — convert minutes to seconds first!
- 2Write the half-equation at the relevant electrode.
- 3Find moles of electrons per mole of product from the half-equation.
- 4n(product) = Q ÷ (n(e⁻) × 96,500)
- 5Convert: mass = n × M, or volume = n × Vₘ.
Section 5 — Industrial Applications
Al is too high in the series for carbon reduction. Extracted by electrolysis of molten Al₂O₃ dissolved in cryolite.
Cathode: Al³⁺(l) + 3e⁻ → Al(l)
Anode: 2O²⁻(l) → O₂(g) + 4e⁻
Graphite anodes corrode and must be replaced regularly.
Purifies impure Cu: impure metal = anode, pure metal = cathode, CuSO₄ = electrolyte.
Anode: Cu(s) → Cu²⁺(aq) + 2e⁻
Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)
Impurities sink as sludge. Electrolyte unchanged.
Deposits a thin metal coat on an object (corrosion protection, appearance).
Object = cathode; plating metal = anode; metal salt solution = electrolyte.
Cr³⁺(aq) + 3e⁻ → Cr(s) [on cathode]Thickens Al₂O₃ protective layer. Al article = anode; dilute H₂SO₄ = electrolyte.
Al(s) → Al³⁺ + 3e⁻ [anode]
Al³⁺ + O²⁻ → Al₂O₃ [thick layer]
More corrosion-resistant; can be dyed attractive colours.
Section 6 — Resources
Section 7 — CSEC Practice Questions
1 Fe is above Cu in the electrochemical series → Fe is a stronger reducing agent → reaction occurs.
2 Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
3 Ionic: Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)
a Ions present: Na⁺(aq), Cl⁻(aq), H⁺(aq), OH⁻(aq)
b Cathode: 2H⁺(aq) + 2e⁻ → H₂(g)
Anode: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
c Although OH⁻ is lower in the anion series, in concentrated NaCl the Cl⁻ concentration is very high. The concentration factor overrides position — Cl⁻ is preferentially discharged.
a t = 40 × 60 = 2400 s. Q = 3.0 × 2400 = 7200 C
b Half-equation: Cu²⁺ + 2e⁻ → Cu (2 mol e⁻ per mol Cu)
n(Cu) = 7200 ÷ (2 × 96500) = 0.0373 mol
c m(Cu) = 0.0373 × 63.5 = 2.37 g
d Active copper anode dissolves: Cu(s) → Cu²⁺(aq) + 2e⁻. Solution colour stays constant as Cu²⁺ is replenished.
1 Al is very high in the electrochemical series. Its Al³⁺ ions are so stable that carbon (coke) cannot reduce them — C is not a strong enough reducing agent.
2 Al is extracted by electrolysis of molten Al₂O₃ (bauxite) dissolved in molten cryolite (which lowers the melting point from ~2000°C to ~950°C).
3 Cathode: Al³⁺(l) + 3e⁻ → Al(l) [molten Al sinks]
Anode: 2O²⁻(l) → O₂(g) + 4e⁻ [O₂ attacks graphite → CO₂ → anodes corrode and must be replaced]
a Graphite (inert): Anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻ [O₂ evolved]. Cathode: Cu²⁺ + 2e⁻ → Cu [pink deposits]. Cu²⁺ removed but not replaced → solution gets paler blue.
b Copper (active): Anode: Cu → Cu²⁺ + 2e⁻ [anode dissolves]. Cathode: Cu²⁺ + 2e⁻ → Cu [pink deposits]. Cu²⁺ consumed = Cu²⁺ produced → solution colour stays the same.