Part 3: Halides with concentrated sulfuric acid

Separate proton transfer from redox. Track sulfur oxidation states, identify the observations and construct balanced equations for each reduction product.

The halide-ion trend goes the opposite way

Halide ions become stronger reducing agents down the group: F⁻ < Cl⁻ < Br⁻ < I⁻. Their outer electrons are farther from the nucleus and more shielded, so they are donated more readily. This is electron loss by X⁻, not electron gain by X₂.

For solid sodium halides with concentrated H₂SO₄, begin with an acid–base step that generates HX. Bromide and iodide can then reduce the sulfuric acid. H₂SO₄ acts first as a proton donor; in the later redox step it acts as an oxidising agent.

The acid–base step: no oxidation-state change

The equations below use sodium hydrogensulfate for the usual concentrated-acid representation. Chloride and fluoride do not reduce H₂SO₄ under these test conditions. White steamy/misty fumes are associated with hydrogen halides in moist air; the pure gases themselves are colourless.

Bromide and iodide also have this initial step, but subsequent redox chemistry makes additional products. White fumes therefore do not prove the absence of redox.

Bromide reduces sulfur from +6 to +4

Bromide loses electrons to form Br₂. Sulfur in H₂SO₄ gains two electrons per sulfur atom and forms SO₂. Look for orange/red-brown bromine vapour, alongside the initial hydrogen-halide fumes. SO₂ is colourless; its presence is not shown by the orange colour.

Construct the equation from one oxidation half-equation and one reduction half-equation. In the combined equation all electrons must cancel.

Iodide can produce SO₂, sulfur or H₂S

Iodide is a stronger reducing agent and can reduce sulfur further. Starting from sulfur at +6, SO₂ contains S at +4, elemental S is 0, and H₂S contains S at −2. The reductions require respectively 2, 6 and 8 electrons per sulfur atom.

Because 2I⁻ → I₂ + 2e⁻, multiply that half-equation by 1, 3 or 4. Different products can occur in the mixture; a question naming one product needs the equation for that product. Do not force all possible sulfur products into one unbalanced equation.

Worked construction of the hydrogen-sulfide equation

Balance sulfur first, then oxygen with four H₂O on the right. With H₂SO₄ on the left, eight additional H⁺ balance hydrogen. Add eight electrons on the left to balance charge. Multiply the iodide oxidation by four and add.

Using SO₄²⁻ instead of H₂SO₄ requires ten H⁺, not eight, in the reduction half-equation. Both forms can describe the same redox bookkeeping if they are balanced consistently.

The 2023 Q06.3 report highlights errors in this reduction half-equation and failure to cancel electrons. Check: one S, four O, ten H and charge zero on both sides of the final H₂SO₄-based equation.

Write what is observed, then explain it

Iodine may appear as a grey-black solid or purple vapour. These observations identify oxidised iodine, not the reduction product of sulfur. Yellow sulfur is a different solid. Name each observation with the corresponding product.

These reactions involve corrosive concentrated acid and toxic gases. They belong in a controlled, supervised laboratory procedure with a fume cupboard; never identify H₂S or SO₂ by deliberately smelling the reaction mixture.

Quick checks

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

Sources