Redox Worked ExamplesIGCSE Chemistry 0620

(a) The black solid turns pink-brown (copper forms), and droplets of colourless liquid (water) form at the cooler end of the tube.

(b) Copper(II) oxide loses oxygen, so it is reduced. Hydrogen gains oxygen (forming water), so it is oxidised.

(c) Oxidation and reduction happen at the same time in the same reaction — that is what redox means.

(d) The Roman numeral gives the oxidation number of copper in the compound: +2.

Use the rule that the oxidation numbers in a compound add up to zero. Chlorine in a chloride is −1 and oxygen in an oxide is −2.

(i) $3 \times (-1) = -3$, so Fe is +3: iron(III) chloride. (ii) O is −2, shared by two Cu atoms, so each Cu is +1: copper(I) oxide. (iii) $2 \times (-2) = -4$, so Mn is +4: manganese(IV) oxide.

(b) The (IV) shows that the oxidation number of nitrogen in the compound is +4.

(c) Sodium only ever has one oxidation number (+1) in its compounds, so no Roman numeral is needed. Transition elements such as iron and copper have variable oxidation numbers, so the numeral tells you which one.

Rules: O is −2 and H is +1 in these species; the oxidation numbers add up to zero in a compound and to the charge in an ion.

(a) $x + 3(-2) = -2$, so $x = +4$. (b) $x + 4(+1) = +1$, so $x = -3$.

(c) $x + 3(-2) = -1$, so $x = +5$.

(d) $4x + 10(-2) = 0$, so $4x = 20$ and $x = +5$. (e) $2x + 5(-2) = 0$, so $x = +5$ — which is why the catalyst in the Contact process is called vanadium(V) oxide.

(a)(i) Oxidation is an increase in oxidation number. (ii) Oxidation is loss of electrons.

(b) Iron: +2 in $\text{FeCl}_2$ → +3 in $\text{FeCl}_3$ — an increase, so iron is oxidised. Chlorine: 0 in $\text{Cl}_2$ (an element) → −1 in $\text{FeCl}_3$ — a decrease, so chlorine is reduced. Both happen together, so it is redox.

In terms of electrons, each iron(II) ion loses one electron:

(c) In reaction 2, Ag stays +1, Na stays +1, Cl stays −1, N stays +5 and O stays −2. No element changes its oxidation number, so it is not redox (it is a precipitation reaction).

Tin: +2 → +4, so $\text{SnCl}_2$ is oxidised. Iron: +3 → +2, so $\text{FeCl}_3$ is reduced.

An oxidising agent oxidises another substance and is itself reduced: $\text{FeCl}_3$. A reducing agent reduces another substance and is itself oxidised: $\text{SnCl}_2$. Answer: A.

B swaps the roles. C and D name products — the agents are always reactants.

(a) Chlorine displaces bromine and iodine; bromine displaces only iodine; iodine displaces neither. Order: chlorine > bromine > iodine.

(b) Potassium ions do not change — they are spectator ions and are left out. Chlorine molecules form chloride ions; bromide ions form bromine.

(c) Bromide ions lose electrons (they are oxidised to bromine) and chlorine gains electrons (it is reduced to chloride ions). Loss and gain of electrons happen together, so it is redox.

(d) Chlorine — it takes electrons from both bromide and iodide ions, so it oxidises the most other species.

(a) Acidified potassium manganate(VII) is purple.

(b) X turns the manganate(VII) from purple to colourless. Manganate(VII) is an oxidising agent, so it has been reduced by X — X is a reducing agent.

(c) Y turns potassium iodide from colourless to brown. Iodide ions have been oxidised to iodine, which is brown, so Y is an oxidising agent.

(d) Z causes no change in either test, so it is neither an oxidising agent nor a reducing agent (towards these reagents).

(a) Aluminium oxide is $\text{Al}_2\text{O}_3$ (Al is +3, O is −2). Oxygen atoms: 5 per $\text{V}_2\text{O}_5$ and 3 per $\text{Al}_2\text{O}_3$; the lowest common multiple is 15, so use $3\text{V}_2\text{O}_5$ and $5\text{Al}_2\text{O}_3$.

Then balance V ($3 \times 2 = 6$) and Al ($5 \times 2 = 10$).

(b)(i) Vanadium: +5 → 0 (decreases by 5; reduced). (ii) Aluminium: 0 → +3 (increases by 3; oxidised).

(c) A reducing agent is a substance that reduces another substance and is itself oxidised. (d) Aluminium.

(a) Each iron(III) ion gains one electron:

Two iodide ions each lose one electron to form one iodine molecule:

(b) Iodide ions are oxidised: they lose electrons (oxidation number −1 → 0). (c) The oxidising agent is iron(III) ions, $\text{Fe}^{3+}$ — they take the electrons and are reduced.

(d) Add aqueous sodium hydroxide (or aqueous ammonia): a green precipitate forms (insoluble in excess), showing $\text{Fe}^{2+}$.

(a) At the high temperature in the engine, nitrogen and oxygen from the air react together.

(b) Carbon: +2 in CO → +4 in $\text{CO}_2$. Nitrogen: +2 in NO → 0 in $\text{N}_2$ (an element). Oxygen stays −2.

(c) Carbon monoxide is oxidised (oxidation number increases; it also gains oxygen). Nitrogen monoxide is reduced (oxidation number decreases; it loses oxygen).

(d) The oxidising agent is the reactant that is reduced: nitrogen monoxide, NO.

(a) In ZnS, Zn is +2 so S is −2. In the element S: 0. In $\text{SO}_2$: $x + 2(-2) = 0$, +4. In $\text{SO}_3$: $x + 3(-2) = 0$, +6. In $\text{CaSO}_3$: $+2 + x + 3(-2) = 0$, +4.

(b) Reaction 4: sulfur stays at +4, calcium stays +2 and oxygen stays −2 — no oxidation numbers change. (It is an acid–base reaction, used in flue gas desulfurisation.)

(c) In reaction 1, sulfur goes from −2 to +4 (oxidised) and oxygen goes from 0 to −2 (reduced). The oxidising agent is oxygen, $\text{O}_2$. Zinc stays +2.

(d) In reaction 3, sulfur goes from +4 to +6 (oxidised), and oxygen in $\text{O}_2$ goes from 0 to −2, so oxygen is reduced.

Iodine in $\text{IO}_3^{\,-}$: $x + 3(-2) = -1$, so $x = +5$ (hence iodate(V)). It goes to 0 in $\text{I}_2$ — a decrease, so iodate(V) is reduced.

Iodide goes from −1 to 0 — an increase, so iodide is oxidised. A substance that is oxidised while reducing another is the reducing agent. Answer: B.

A and D reverse the directions. C is wrong because hydrogen stays at +1 in $\text{H}_2\text{O}$ — the hydrogen ions are not reduced.