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Electrolysis — Comprehensive E-Note (Detailed)
What you’ll learn — objectives & overview
Electrolysis uses electrical energy to force a non-spontaneous redox reaction. Passing current through an electrolyte causes ions to move and react at electrodes: oxidation at the anode and reduction at the cathode.
Use the navigation buttons below to move through pages. The quiz is at the end.
Electrolytes — substances that dissociate into ions in solution (or molten) and conduct electricity: strong electrolytes (e.g., NaCl, HCl, KOH) dissociate completely; weak electrolytes (e.g., acetic acid) partially dissociate.
Non-electrolytes — molecular compounds that do not form ions in solution (e.g., sugar, ethanol) and do not conduct current.
Example: NaCl dissolved in water → Na⁺ and Cl⁻ ions; apply current: Na⁺ moves toward cathode, Cl⁻ to anode.
First law: mass ∝ total charge (Q) passed. Second law: masses produced by the same charge are proportional to equivalent weights.
Useful relation: Q = I × t (Coulombs). Faraday constant: F ≈ 96485 C·mol⁻¹ (per mole of electrons).
Given: Q = 1.00×10⁵ C passed through Cu²⁺ solution. Equivalent weight (Cu) = 31.75 g per eq.
Mass = (Q / F) × equivalent weight = (1.00×10⁵ / 96485) × 31.75 ≈ 32.9 g.
If 0.5 g of Ag (molar mass 107.87 g/mol, Ag⁺ + e⁻ → Ag) is deposited in 10 minutes, what is the average current?
moles Ag = 0.5 / 107.87 = 0.004636 mol. moles e⁻ = 0.004636 (one e⁻ per Ag). Q = n·F = 0.004636×96485 ≈ 447.7 C. Time = 600 s. I = Q/t ≈ 0.746 A.
Reduction potentials: Na⁺/Na is very negative (~−2.71 V) while water reduction is easier under aqueous conditions (→ H₂). Therefore water is reduced and H₂ evolves at cathode.
Which ion is discharged depends on:
Example: in brine electrolysis, concentration and electrode type are chosen to favour Cl₂ production and NaOH formation.
The electrochemical series ranks half-reactions by E°. Cell EMF: E°cell = E°(cathode) − E°(anode). Positive E°cell → spontaneous galvanic cell.
E°(Cu²⁺/Cu)=+0.34 V, E°(Zn²⁺/Zn)=−0.76 V → E°cell = 0.34 − (−0.76) = +1.10 V (galvanic).
Major applications:
Impure copper anode dissolves to Cu²⁺; copper is deposited as pure metal at cathode; impurities either fall as anode slime or remain in solution.
Corrosion is a redox process where metal (anode area) oxidizes and electrons flow to cathodic areas. Methods of protection include:
Find mass of Cu deposited by a current of 2.0 A after 2 hours from Cu²⁺ solution. (Cu²⁺ + 2e⁻ → Cu)
Q = I t = 2 × (2×3600) = 14400 C. Moles e⁻ = Q/F = 14400/96485 ≈ 0.1493 mol e⁻. Moles Cu = 0.1493/2 = 0.07465 mol. Mass Cu = 0.07465 × 63.55 ≈ 4.74 g.
When you press Start Quiz you will see 30 questions and a 15:00 timer. Each question shows explanation after submission.
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