Redox Reactions & Electrochemistry

Electrochemistry bridges thermodynamics and redox chemistry through quantitative relationships. Oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons (decrease in oxidation number). Key oxidation number rules: free element = 0, F always = −1, O = −2 (except peroxides −1, OF₂ +2), H = +1 (except metal hydrides −1), and the sum of oxidation numbers equals the charge on the species.

Galvanic (voltaic) cells are spontaneous (ΔG < 0, E°cell > 0). The anode (oxidation) is the negative terminal, and the cathode (reduction) is the positive terminal. Electron flow: anode → external circuit → cathode. A salt bridge (KCl in agar) maintains electrical neutrality.

Electrolytic cells are non-spontaneous (require external power). Here, the anode is positive and the cathode is negative — opposite sign convention from galvanic cells.

Standard electrode potential E° is measured against the SHE (E° = 0.00 V). More positive E° = stronger oxidizing agent; more negative E° = stronger reducing agent.

Cell EMF: E°cell = E°cathode − E°anode (always cathode minus anode).

Solved Example 1: Daniell cell: Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s). E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V

Nernst equation: E = E° − (RT/nF) ln Q. At 25°C: E = E° − (0.0592/n) log Q. At equilibrium, E = 0, so E° = (0.0592/n) log K (derivation: setting E = 0 in the Nernst equation and Q = K at equilibrium).

Solved Example 2: EMF with [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.01 M at 25°C. E = 1.10 − (0.$\frac{0592}{2}$) log(0.$\frac{1}{0}$.01) = 1.10 − 0.0296 × log(10) = 1.10 − 0.0296 = 1.0704 V

Gibbs energy and EMF: ΔG° = −nFE° (F = 96485 C/mol ≈ 96500 C/mol).

Conductance: Specific conductance κ = 1/ρ (S/cm). Molar conductivity Λm = κ × 1000/M (S·cm²/mol, M = molarity). For strong electrolytes: Λm = Λ°m − A√C (Debye-Huckel-Onsager; slight increase on dilution). For weak electrolytes: Λm increases sharply at high dilution; degree of dissociation α = Λm/Λ°m.

Kohlrausch's law: Λ°m = ν₊λ°₊ + ν₋λ°₋. Application: Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl) (adding and canceling common ions).

Faraday's laws of electrolysis: First law: w = ZIt, where Z = M/(nF) (electrochemical equivalent). Combined: w = MIt/(nF), where M = molar mass, I = current (A), t = time (s), n = electrons transferred. Dimensional analysis: (g/mol × A × s) / (mol⁻¹ × C/mol) → g/mol × C / (C/mol) = g ✓ (since A·s = C) 1 Faraday = 96485 C = charge of 1 mol electrons.

Solved Example 3: Copper deposited by 2 A for 1 hour through CuSO₄ solution. w = MIt/(nF) = (63.5 × 2 × 3600) / (2 × 96500) = $\frac{457200}{193000}$ = 2.37 g

Batteries: Dry cell (Leclanché, 1.5 V, non-rechargeable), lead storage battery (Pb/PbO₂, 2 V/cell, rechargeable), fuel cells (H₂-O₂, continuous feed, ~70% efficiency). Corrosion is an electrochemical process: Fe oxidizes at anodic spots, O₂ reduces at cathodic spots, forming rust (Fe₂O₃·xH₂O). Prevention: galvanization (Zn coating), cathodic protection, painting.

The key testable concept is Nernst equation calculations for non-standard conditions and Faraday's law numericals for mass deposition in electrolysis.