§ C-9 Electrolytic Cell & Faraday’s Laws of Electrolysis (Grade 9–11 · Chemistry)
Cathodic copper plating, anodic O₂↑ evolution & mass law m = (M·I·t)/(z·F). The mass of substance liberated at an electrode during electrolysis is directly proportional to the total electric charge Q = I·t and the molar equivalent mass M/z.
Theory Summary & Scientific Mechanism
Electrolysis is a non-spontaneous redox process (Δ G > 0) driven by an external DC power supply across electrodes immersed in an electrolyte. Unlike a galvanic cell, in an electrolytic cell the cathode is the negative electrode (supplying electrons for reduction of cations), while the anode is the positive electrode (withdrawing electrons for oxidation).
During electrolysis of aqueous CuSO₄ with inert graphite or platinum electrodes, Cu²⁺ ions are preferentially reduced at the cathode (E^° = +0.34 V): Cu²⁺ + 2e⁻ → Cu⁰↓. At the inert anode, oxygenated SO₄²⁻ anions resist oxidation, so water molecules are oxidized instead, evolving oxygen gas bubbles: 2H₂O - 4e⁻ → O₂↑ + 4H⁺.
Quantitative electrode yields obey Faraday’s combined law of electrolysis: m = (M· I· t)/(z· F), where Q = I· t is total passed charge, z is the electron stoichiometry per ion, and F = e N_A ≈ 96 485 C/mol is Faraday’s constant (the charge of one mole of electrons).
Everyday Analogy
While a battery generates current from a spontaneous reaction, electrolysis uses an external power supply to drive a reaction in reverse, plating dissolved blue Cu²⁺ ions as shiny metallic copper onto the cathode! This is how jewelry is gold-plated and industrial copper is refined.
Interactive Question & Derivation
According to Faraday’s laws m = (M · I · t)/(z · F), how does the molar amount of metal plated on the cathode for a given charge Q = It change when switching from a monovalent ion Ag⁺ (z = 1) to a divalent ion Cu²⁺ (z = 2)?
- A: Reducing one mole of Cu²⁺ requires twice as many electrons (z = 2), so for the same charge Q = It the moles of deposited metal ν = (Q)/(zF) are halved
- B: The amount of plated metal is independent of ion charge z and current I
- C: The cathode loses mass because oxidation occurs at the cathode
Study Hint: By Faraday’s law of electrolysis m = (M · I · t)/(z · F), how does the mass of copper deposited on the cathode scale when current I or time t is doubled?
Proof & Explanation: By Faraday’s combined law of electrolysis, passing 1 F = 96 485 C of charge deposits exactly 1 mol of equivalents (ν = (I t)/(z F)). For aqueous CuSO₄ (z = 2), metallic copper plates onto the cathode while water oxidizes at the inert anode to evolve oxygen gas O₂↑ (z = 4).
Key Equations & 3D Simulation Parameters
m(Cu) = (M_Cu · I · t)/(z · F), V(O₂) = (I · t)/(4 F)· 22.4 L
- Direct electrolysis current I: 1 … 10 A (default: 5 A)
- Electrolysis duration t: 5 … 30 min (default: 15 min)
- Cation charge number z (1=Ag⁺, 2=Cu²⁺, 3=Au³⁺): 1 … 3 (default: 2)
- DC power supply voltage U: 2 … 8 V (default: 4.5 V)