Strong and weak acids and bases, neutralisation and its net ionic equation, oxidation numbers, what is oxidised and what is reduced, and balancing by half-reactions in acid and in base.
Chapter 4, part 2
Acid-base reactions and redox reactions, from slides 16 to 29 of the Chapter 4 partial notes and the second lecture video.
Homework: Classifying Chemical Reactions, in the Chapter 4 folder, covers both.
Two more kinds of reaction in water. In an acid-base reaction an H⁺ moves from the acid to the base. In a redox reaction electrons move from one substance to another, and oxidation numbers keep track of where they went.
Gives H⁺ ions (H₃O⁺) in water.
Gives OH⁻ ions in water.
A proton donor: it gives away an H⁺.
A proton acceptor: it takes an H⁺. Ammonia is one.
A monoprotic acid gives up one H⁺ (HCl, HNO₃); a diprotic acid two (H₂SO₄). H₃PO₄ can give three.
Any acid not on the strong list is weak.
acid + base → salt + water
Water forms. H⁺ and OH⁻ join into H₂O molecules, which take the ions out of the solution.
The acid’s H and the base’s metal swap partners.
Its cation comes from the base, its anion from the acid: NaOH and HCl give NaCl.
In an ionic equation, split strong acids, strong bases and soluble salts into ions. Weak acids, ammonia, water and solids stay whole.
Loss of electrons. The oxidation number goes up.
Gain of electrons. The oxidation number goes down.
LEO the lion says GER: Lose Electrons Oxidation, Gain Electrons Reduction. Or OIL RIG: Oxidation Is Loss, Reduction Is Gain.
The substance oxidized is always the reducing agent. The substance reduced is always the oxidizing agent.
Each atom gets the charge it would have if every electron were transferred completely. A reaction is redox if the oxidation number of any atom changes: usually single displacements, combinations, decompositions and combustions.
General rules
Particular atoms
Everything else comes from the sum. The notes mark this table “LEARN”.
The oxidation number of sulfur in sulfuric acid. Hydrogen is +1 and oxygen −2, and the compound is neutral:
2(+1) + x + 4(−2) = 0, so x = +6
Phosphorus burning: P₄(s) + 5O₂(g) → P₄O₁₀(s). Phosphorus goes from 0 to +5, so it is oxidized and P₄ is the reducing agent. Oxygen goes from 0 to −2, so it is reduced and O₂ is the oxidizing agent.
4x + 10(−2) = 0, so x = +5 in P₄O₁₀
A redox reaction can be split into the oxidation and the reduction, each written with its electrons so that its charge balances.
1. Divide the reaction into two half-reactions.
2. Balance each: atoms other than O and H first, then O by adding H₂O, then H by adding H⁺, then charge by adding electrons.
3. Multiply so the electrons lost equal the electrons gained.
4. Add the half-reactions and simplify.
5. Check the atoms and the charges.
The solution has OH⁻, not H⁺. Balance exactly as in acid, then add as many OH⁻ to both sides as there are H⁺. Each H⁺ + OH⁻ becomes H₂O. Simplify the water and check again.
Balance under basic conditions, slide 29. Nitrogen goes from +3 to −3 (reduced), aluminum from 0 to +3 (oxidized).
NO₂⁻(aq) + Al(s) → NH₃(g) + AlO₂⁻(aq)
Reduction. Two O on the left, so 2H₂O on the right; that makes 7 H on the right, so 7H⁺ on the left; the charge is then +6 on the left and 0 on the right, so 6e⁻ on the left.
NO₂⁻ + 7H⁺ + 6e⁻ → NH₃ + 2H₂O
Oxidation. Two H₂O on the left supply the oxygen, 4H⁺ on the right balance the hydrogen, and 3e⁻ on the right balance the charge.
Al + 2H₂O → AlO₂⁻ + 4H⁺ + 3e⁻
Six electrons against three: double the oxidation, then add the two halves.
NO₂⁻ + 7H⁺ + 2Al + 4H₂O → NH₃ + 2H₂O + 2AlO₂⁻ + 8H⁺
Simplify: 7H⁺ cancels against 8H⁺, leaving one H⁺ on the right, and 2H₂O against 4H₂O, leaving two on the left.
NO₂⁻ + 2Al + 2H₂O → NH₃ + 2AlO₂⁻ + H⁺
Basic solution: add one OH⁻ to each side. H⁺ + OH⁻ makes one more water on the right, which cancels one on the left.
NO₂⁻(aq) + 2Al(s) + H₂O(l) + OH⁻(aq) → NH₃(g) + 2AlO₂⁻(aq)
Check: 1 N, 2 Al, 4 O and 3 H on each side, and a charge of −2 on each side.
The questions from the Acid-Base Reactions & Redox Reactions in-class sheet, checked one part at a time. Write charges with a caret: OH^- reads as OH⁻.
A fresh problem every time. Wrong answers get a reason; the working shows when you finish.
Start another round for a fresh set of ten.
Helpers for the homework. They add up what you write and check it; the numbers themselves stay yours.
Type a formula or an ion, such as K2Cr2O7 or SO4^2-. Give each element its oxidation number; the line underneath adds them up against the charge, and Check marks each one.
Write each half-reaction with its H₂O, H⁺ (or OH⁻) and electrons as e^-, then the overall equation. Each table counts atoms and charge on both sides as you set the coefficients.