Part 3: Hydroxide precipitates and redox interconversions

Distinguish acid–base precipitation, ligand substitution and electron transfer using the products, observations and balanced equations.

Aqueous ammonia has a base role as well as a ligand role

OH⁻ removes protons from coordinated water and forms metal hydroxide precipitates. Simplified equations such as Fe³⁺ + 3OH⁻ → Fe(OH)₃(s) are acceptable descriptions of the precipitation stoichiometry. The oxidation state does not change.

Aqueous ammonia produces the same initial hydroxides because NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. For example Fe³⁺ + 3NH₃ + 3H₂O → Fe(OH)₃(s) + 3NH₄⁺. It is not necessary to put NH₃ ligands into every initial precipitate formula.

Excess reagent distinguishes the ions

Of these five hydroxides, Cr(OH)₃ dissolves in excess NaOH to form green [Cr(OH)₆]³⁻. It is amphoteric: it also reacts with acid to give an aqueous chromium(III) species. Do not infer that all transition-metal hydroxides dissolve in excess alkali.

In excess ammonia, the copper precipitate dissolves to its deep-blue complex and chromium forms its purple hexaammine complex under suitable conditions. Fe²⁺, Fe³⁺ and Mn²⁺ hydroxides do not dissolve in excess ammonia in the required test scheme. Record a precipitate dissolving and the resulting solution colour; colour alone omits a major observation.

Changing the oxidation state is a separate process

Acidified permanganate oxidises Fe²⁺ to Fe³⁺; the permanganate purple colour disappears while Fe(III) species form. Iodide can reduce Fe³⁺ to Fe²⁺ while being oxidised to brown aqueous iodine. Check the oxidation numbers: Fe falls from +3 to +2, so it gains electrons.

The green-to-brown change of an Fe(OH)₂ precipitate on standing is associated with oxidation by air. It is not evidence that hydroxide was an oxidising agent in the initial precipitation. The order and timing of observations matter.

Separate oxidation from the chromate–dichromate equilibrium

In alkaline solution, hydrogen peroxide can oxidise chromium(III) to yellow chromate(VI). One balanced equation is 2[Cr(OH)₆]³⁻ + 3H₂O₂ → 2CrO₄²⁻ + 2OH⁻ + 8H₂O. Chromium rises from +3 to +6 while peroxide is reduced.

Acidifying chromate shifts 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O towards orange dichromate. Chromium stays +6, so this colour change is not redox. Reduction of acidified dichromate produces Cr³⁺, commonly observed as green under the reaction conditions. Zinc with acid can cause further reduction to blue Cr²⁺ if conditions allow; avoid claiming every zinc experiment stops at +3.

Build an unfamiliar equation from supplied half-equations rather than memorising an unbalanced colour sequence. For reduction specifically to +3, Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O supplies the electron balance.

Copper(I) illustrates precipitation and disproportionation

Adding iodide to Cu²⁺ forms a white CuI precipitate and iodine, whose brown colour may mask the white solid. The equation 2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂ shows copper reduction from +2 to +1 and iodide oxidation. Formation of insoluble CuI stabilises the Cu(I) product in this reaction.

Free aqueous Cu⁺ readily disproportionates: 2Cu⁺ → Cu²⁺ + Cu. Copper in the same starting oxidation state is both oxidised and reduced. This does not mean that every Cu(I) compound immediately disproportionates; precipitation and ligand environment affect the species present.

Use this decision process for an unfamiliar change: identify a solid forming or dissolving; compare ligands; calculate metal oxidation numbers; then select precipitation, substitution or redox. More than one type may occur in an overall sequence.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources