Two colour tests — answer key

1. ZQ16 · Practice

Options as numbered on the pupil sheet: 1 — brown; 2 — purple; 3 — colourless.

Model answer. Acidified potassium manganate(VII) is purple; a reducing agent turns it colourless, as manganese goes from +7 to +2. Aqueous potassium iodide is colourless; an oxidising agent turns it brown, as the iodide ions are oxidised to iodine.

A full answer includes

• 1. Purple.
• 2. Colourless.
• 3. Colourless.
• 4. Brown.

Also accept

• purple; colourless; colourless; brown

Watch for

• “The manganate(VII) test turns brown.” — Brown is the iodide test, where iodine forms. Acidified manganate(VII) goes from purple to colourless.
• “A paler purple shows a reducing agent.” — A paler purple alone proves nothing: water would make it paler too, by spreading it out. The test needs the purple to go completely: colourless.

2. ZQ17 · Practice

Model answer. The purple MnO₄⁻ ion becomes the almost colourless Mn²⁺ ion: manganese goes down from +7 to +2, so it gains electrons and is reduced. Those electrons came from P, so P is oxidised: P is a reducing agent.

A full answer includes

• P is a reducing agent: it gives electrons, and manganese is reduced from +7 to +2.

Also accept

• P is a reducing agent; manganese is reduced

Watch for

• “P is an oxidising agent, because the purple is oxidised.” — Manganese goes down from +7 to +2: it is reduced. Something that reduces another substance is a reducing agent.
• “P is a reducing agent, and manganese is oxidised.” — A reducing agent reduces the other substance: manganese goes down, so it is reduced.
• “P is an oxidising agent that gives electrons to the manganese.” — Giving electrons is what a reducing agent does. P gives electrons and reduces the manganese, so P is a reducing agent.

3. ZQ18 · Practice

Model answer. The iodide ion, I⁻, is −1; iodine, I₂, is an element on its own, 0. Going up from −1 to 0, each iodide ion loses an electron: it is oxidised. Hydrogen peroxide took those electrons, so it was reduced: it is an oxidising agent. The brown is iodine, made from the colourless iodide ions.

A full answer includes

• (1) −1
• (2) 0
• (3) loses
• (4) oxidised
• (5) an oxidising agent

Also accept

• −1; 0; loses; oxidised; an oxidising agent

Watch for

• “The brown is the colour of the substance added.” — The substance added was colourless, and so was the potassium iodide. The brown is iodine, made in the tube from the iodide ions.
• “Hydrogen peroxide is oxidised, so it is the oxidising agent.” — The iodide ions are oxidised. Hydrogen peroxide takes their electrons, so it is reduced: that makes it the oxidising agent.

4. ZQ19 · Practice

Model answer. An oxidising agent takes electrons from iodide ions and makes brown iodine: only X turns the potassium iodide brown. W turns the manganate(VII) colourless, so W is a reducing agent. Y only makes the purple paler: water would do that by dilution, so the test has not shown a reducing agent in Y. Z changes neither. Neither test shows an oxidising agent in W, Y or Z.

A full answer includes

• I chose X.
• I used the iodide test: brown shows an oxidising agent.
• I did not choose Y: a paler purple alone does not show an agent, and the manganate(VII) test shows reducing agents, not oxidising agents.

Also accept

• X

Watch for

• “W.” — W turns acidified potassium manganate(VII) colourless: it gives electrons, so it is a reducing agent.
• “Y.” — A paler purple alone does not show an agent: dilution could cause it. And the manganate(VII) test shows reducing agents, not oxidising agents.
• “Z.” — Z shows no change in either test. An oxidising agent turns potassium iodide brown.

5. ZQ20 · Practice

Model answer. (a) A change from purple to colourless would show a reducing agent. No change only shows that this test has not shown T acting as a reducing agent. T may be an oxidising agent, and it may react in ways in which no electrons move, such as forming a precipitate or neutralising an acid. (b) Add a few drops of T to aqueous potassium iodide. If T is an oxidising agent, it oxidises the iodide ions to iodine, and the colourless solution turns brown.

A full answer includes

• (a) No change only shows that this test has not shown T acting as a reducing agent.
• (a) T may still be an oxidising agent, or react in other ways, such as forming a precipitate.
• (b) Add T to aqueous potassium iodide.
• (b) If T is an oxidising agent, the solution turns from colourless to brown.

Also accept

• The test has not shown T acting as a reducing agent; potassium iodide turns brown

Watch for

• “No change with acidified potassium manganate(VII) means the substance is unreactive.” — It only means this test has not shown the substance acting as a reducing agent. It may still be an oxidising agent, or react in other ways, such as forming a precipitate.
• “Test T again with more acidified potassium manganate(VII).” — Manganate(VII) is itself an oxidising agent: it tests for reducing agents. To find an oxidising agent, use potassium iodide.

6. ZT3 · Apply

Model answer. (a) In SO₂ two oxygens give −4, so sulfur is +4. In SO₄²⁻ four oxygens give −8 and the total is −2, so sulfur is +6. (b) Sulfur goes up from +4 to +6, an increase in oxidation number, so sulfur dioxide is oxidised: it is a reducing agent. (c) The manganate(VII) ion takes the electrons: manganese goes down from +7 to +2, so it is reduced, and acidified potassium manganate(VII) is the oxidising agent. This colour change is used as the test for sulfur dioxide.

A full answer includes

• (a) Sulfur in SO₂: +4.
• (a) Sulfur in SO₄²⁻: +6.
• (b) Oxidised: the oxidation number of sulfur increases.
• (c) The oxidising agent is acidified potassium manganate(VII).
• (c) Manganese goes from +7 to +2.
• I did not call sulfur dioxide the oxidising agent.

Also accept

• +4 and +6; oxidised; manganate(VII), +7 → +2

Watch for

• “Sulfur dioxide is the oxidising agent, because it contains oxygen.” — Sulfur goes up from +4 to +6: sulfur dioxide is oxidised, so it is the reducing agent.
• “The oxidising agent is the substance that is oxidised.” — It is the other way round. The oxidising agent oxidises the other substance: it takes electrons, so it is itself reduced. The substance oxidised is the reducing agent.