Halogen comproportionation ↔ disproportionation
Halogens disproportionate in base (X₂ + OH⁻ → X⁻ + XO⁻) and comproportionate in acid (XO⁻ + X⁻ + H⁺ → X₂). Cold base stops at hypohalite (XO⁻); hot base pushes to halate (XO₃⁻). The equilibrium is pH-controlled: acid drives comproportionation, base drives disproportionation. This is why mixing bleach (NaClO) with acid produces toxic Cl₂.
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Amphoteric metals dissolving in base
Amphoteric metals (Al, Zn, Sn, Pb, and to some extent Cr) dissolve in both acids and excess strong base. In base, the metal is oxidized to a hydroxo complex (e.g., Al → [Al(OH)₄]⁻) while water is reduced to H₂. Balancing requires using OH⁻ and H₂O as reactants in basic solution. Their hydroxides also dissolve in excess base: Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻.
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Electrode discharge order
During electrolysis, the species discharged at each electrode follows a preferential order based on reduction/oxidation potentials. At the cathode, the easiest-to-reduce cation is deposited first. At the anode, the easiest-to-oxidize anion is discharged first. In aqueous NaCl: Cl⁻ is oxidized at the anode (not H₂O, due to overpotential) and H₂O is reduced at the cathode (not Na⁺, which has too negative a reduction potential).
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Dissolving post-hydrogen metals
Metals below hydrogen in the activity series (Cu, Ag, Hg, Au, Pt) cannot be dissolved by non-oxidizing acids (dilute HCl, dilute H₂SO₄). They require oxidizing acids where the anion acts as the oxidizer: dilute HNO₃ → NO, concentrated HNO₃ → NO₂, hot concentrated H₂SO₄ → SO₂. Gold and platinum require aqua regia (3HCl + HNO₃), where HNO₃ oxidizes and Cl⁻ complexes the metal ion.
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KMnO₄ products depend on pH
Permanganate (MnO₄⁻) reduction products vary with pH: in acid → Mn²⁺ (colorless, gains 5e⁻); in neutral/basic → MnO₂ (brown precipitate, gains 3e⁻); in very basic → MnO₄²⁻ (green, gains 1e⁻). Students who memorize only the acidic product will miss questions in neutral solution. Always check whether the question specifies "acidic," "acidified," or neither.
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Methods to produce Cl₂
Cl₂ can be produced by: (1) comproportionation of ClO⁻ + Cl⁻ in acid, (2) oxidation of Cl⁻ by MnO₂ in concentrated HCl with heat, (3) oxidation of Cl⁻ by KMnO₄ or K₂Cr₂O₇ in acid, (4) electrolysis of NaCl(aq) at the anode. The common thread: an oxidizing agent + Cl⁻ in acidic solution → Cl₂. Stronger oxidants require milder conditions.
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Boudouard equilibrium: CO vs CO₂ at high temperature
The Boudouard equilibrium (C + CO₂ ⇌ 2CO) explains why CO is the dominant carbon oxide at high temperatures: the reaction is endothermic and entropy-favored (1 mol gas → 2 mol gas), so high temperature shifts the equilibrium toward CO. This is why blast furnaces produce CO as the primary reducing agent for metal ores, and why carbon reduces metal oxides to the metal + CO (not CO₂) at high temperature.
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