Answers · Redox

Redox — return questions key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. ZR1 · Return

Model answer. The (II) in iron(II) oxide means iron is +2: one O²⁻ ion balances one Fe²⁺ ion. Iron(II) oxide becomes iron: it loses oxygen, so it is reduced. Carbon becomes carbon monoxide: it gains oxygen, so it is oxidised. Both happen at once: a redox reaction.

A full answer includes

(1) +2
(2) loses
(3) reduced
(4) gains
(5) oxidised

Also accept

+2; loses; reduced; gains; oxidised

Watch for

“The Roman numeral counts atoms, so copper(II) oxide has two copper atoms.” — The numeral gives the oxidation number of the element, here the charge on the metal ion: copper(II) means Cu²⁺, and copper(II) oxide, CuO, has one copper for each oxygen.
“Iron is oxidised, because it was in an oxide.” — Iron(II) oxide loses its oxygen and becomes iron, so it is reduced.

2. ZR2 · Return

Model answer. In SO₃²⁻ the three oxygens give −6 and the whole ion is −2, so sulfur is +4. In SO₄²⁻ sulfur is +6: it goes up, so the sulfite ions are oxidised. X oxidises them, so X is the oxidising agent: it takes their electrons, and gaining electrons is reduction, so X is reduced. Row A.

A full answer includes

I chose A.
Sulfur in SO₃²⁻: sulfur + 3 × (−2) = −2, so sulfur is +4.
X takes electrons from the sulfite ions, so X gains electrons and is reduced.

Also accept

Watch for

“B.” — +4 is right. But X does the oxidising: it takes electrons, so it is reduced. The sulfite ions are the ones oxidised.
“C.” — +6 would make the ion total 0, as in a compound. The ion is −2: sulfur + (−6) = −2, so sulfur is +4.
“D.” — Check both: the ion totals −2, so sulfur is +4; and the oxidising agent is itself reduced.
“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.

3. ZR3 · Return

Model answer. The brown is iodine, made from the iodide ions: iodine goes up from −1 to 0, so the iodide ions lose electrons. S took them, so S is an oxidising agent. The purple went because manganese went down from +7 to +2: the manganate(VII) ions gained electrons, which T gave. So T is a reducing agent. Row A.

A full answer includes

I chose A.
S is an oxidising agent: it took electrons from the iodide ions, which became brown iodine.
T is a reducing agent: it gave electrons to the manganate(VII) ions, so manganese went from +7 to +2.

Also accept

Watch for

“B.” — Both are the wrong way round. Iodide ions lose electrons to make iodine, so S, which took them, is an oxidising agent. Manganate(VII) gains electrons, so T, which gave them, is a reducing agent.
“C.” — S is right. But manganese went down from +7 to +2: it gained electrons from T, so T is a reducing agent.
“D.” — T is right. But iodide went up from −1 to 0: it lost electrons to S, so S is an oxidising agent.