Ionic Half Equations
Section: 4. Electrochemistry | Syllabus: Cambridge AS Level Physics 9702
What are Half Equations? Half equations show what happens at one electrode during electrolysis. They show the gain or loss of electrons by ions. We use half equations because: They show oxidation or reduction separately They clearly show electron transfer They help us understand the chemistry at each electrode They can be balanced to give the overall equation Key Features of Half Equations Include electrons (e⁻) to show charge transfer Must be balanced for atoms and charges Show state symbols (s, l, g, aq) Reduction (cathode): electrons on LEFT (reactants) Oxidation (anode): electrons on RIGHT (products) Remember: Electrons flow FROM anode TO cathode through the external circuit Cathode Half Equations (Reduction) General form: Positive ion + electrons → Neutral atom/molecule Ion⁺ + e⁻ → Atom Examples - Metal deposition: Cu²⁺(aq) + 2e⁻ → Cu(s) Ag⁺(aq) + e⁻ → Ag(s) Pb²⁺(l) + 2e⁻ → Pb(l) Al³⁺(l) + 3e⁻ → Al(l) Zn²⁺(aq) + 2e⁻ → Zn(s) Examples - Hydrogen formation: 2H⁺(aq) + 2e⁻ → H₂(g) Key point: The number of electrons equals the charge on the ion +1 ion needs 1 electron +2 ion needs 2 electrons +3 ion needs 3 electrons Anode Half Equations (Oxidation) General form: Negative ions → Neutral atom/molecule + electrons Ion⁻ → Atom + e⁻ Examples - Non-metal formation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ 2Br⁻(aq) → Br₂(l) + 2e⁻ 2I⁻(aq) → I₂(s) + 2e⁻ Examples - Oxygen formation from hydroxide: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ Alternative (from oxide ions in molten compounds): 2O²⁻(l) → O₂(g) + 4e⁻ Writing Half Equations - Step by Step Step 1: Identify the ion being discharged Step 2: Write the product (element or molecule) Step 3: Balance atoms (use coefficients if needed) Step 4: Add electrons to balance charges Step 5: Check: atoms balanced?
charges balanced? Step 6: Add state symbols Example 1: Copper at Cathode Step 1: Ion = Cu²⁺ Step 2: Product = Cu Step 3: Cu²⁺ → Cu (atoms balanced) Step 4: Charge on left = +2, on right = 0. Need 2e⁻ on left: Cu²⁺ + 2e⁻ → Cu Step 5: Check: Atoms ✓ (1 Cu each side), Charge ✓ (+2 -2 = 0 on left, 0 on right) Step 6: Cu²⁺(aq) + 2e⁻ → Cu(s) Example 2: Chlorine at Anode Step 1: Ion = Cl⁻ Step 2: Product = Cl₂ (chlorine is diatomic) Step 3: Need 2 Cl⁻ to make Cl₂: 2Cl⁻ → Cl₂ Step 4: Charge on left = -2, on right = 0.
Need 2e⁻ on right: 2Cl⁻ → Cl₂ + 2e⁻ Step 5: Check: Atoms ✓ (2 Cl each side), Charge ✓ (-2 on left, 0 -2 = -2 on right) Step 6: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ Example 3: Oxygen from Hydroxide at Anode Step 1: Ion = OH⁻ Step 2: Products = O₂ and H₂O Step 3: Need 4 OH⁻ to make 1 O₂ and 2 H₂O: 4OH⁻ → O₂ + 2H₂O Step 4: Charge on left = -4, on right = 0.
Need 4e⁻ on right: 4OH⁻ → O₂ + 2H₂O + 4e⁻ Step 5: Check: O atoms ✓ (4 each side), H atoms ✓ (4 each side), Charge ✓ (-4 on left, 0 -4 = -4 on right) Step 6: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ Common Half Equations - Quick Reference At Cathode (Reduction) Half Equation Hydrogen 2H⁺(aq) + 2e⁻ → H₂(g) Copper Cu²⁺(aq) + 2e⁻ → Cu(s) Silver Ag⁺(aq) + e⁻ → Ag(s) Lead Pb²⁺(l) + 2e⁻ → Pb(l) Aluminium Al³⁺(l) + 3e⁻ → Al(l) Zinc Zn²⁺(aq) + 2e⁻ → Zn(s) Iron(II) Fe²⁺(aq) + 2e⁻ → Fe(s) Sodium Na⁺(l) + e⁻ → Na(l) Common Half Equations - Anode At Anode (Oxidation) Half Equation Chlorine 2Cl⁻(aq) → Cl₂(g) + 2e⁻ Bromine 2Br⁻(aq) → Br₂(l) + 2e⁻ Iodine 2I⁻(aq) → I₂(s) + 2e⁻ Oxygen (from OH⁻) 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ Oxygen (from O²⁻) 2O²⁻(l) → O₂(g) + 4e⁻ Combining Half Equations To get the overall equation, combine the two half equations ensuring electrons cancel: Example: Electrolysis of molten lead bromide Cathode: Pb²⁺ + 2e⁻ → Pb Anode: 2Br⁻ → Br₂ + 2e⁻ Electrons balance (2 each), so add them: Overall: Pb²⁺ + 2Br⁻ → Pb + Br₂ Or simply: PbBr₂ → Pb + Br₂ Balancing Electrons When They Don't Match Example: Electrolysis of molten aluminium oxide Cathode: Al³⁺ + 3e⁻ → Al (needs 3 electrons) Anode: 2O²⁻ → O₂ + 4e⁻ (produces 4 electrons) Problem: 3 ≠ 4, electrons don't balance!
Solution: Find lowest common multiple (12) Multiply cathode equation by 4: 4Al³⁺ + 12e⁻ → 4Al Multiply anode equation by 3: 6O²⁻ → 3O₂ + 12e⁻ Now add them (12 electrons cancel): Overall: 4Al³⁺ + 6O²⁻ → 4Al + 3O₂ Or: 2Al₂O₃ → 4Al + 3O₂ Electron Flow Direction Important understanding: Electrons are RELEASED at the anode (oxidation) Electrons flow through the external circuit Electrons are GAINED at the cathode (reduction) Direction: Anode → External circuit → Cathode State Symbols in Half Equations Substance State symbol When used Ions in solution (aq) Aqueous electrolysis Ions in molten (l) Molten electrolysis Solid metal deposited (s) At cathode usually Molten metal (l) Very reactive metals, high temp Gas (g) H₂, O₂, Cl₂, etc.
Liquid bromine (l) or (aq) Br₂ dissolves in water Complete Example: Copper Sulfate Solution Cathode half equation: Cu²⁺(aq) + 2e⁻ → Cu(s) Anode half equation: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ Balancing electrons (LCM = 4): Multiply cathode by 2: 2Cu²⁺(aq) + 4e⁻ → 2Cu(s) Anode stays same: 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ Overall equation: 2Cu²⁺(aq) + 4OH⁻(aq) → 2Cu(s…
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