CHEM1001 Worksheet 12 – Answers to Critical Thinking Questions

The worksheets are available in the tutorials and form an integral part of the learning outcomes and experience for this unit.

Model 1: Redox Reactions 1. Cu has oxidation number 0 in Cu and +2 in CuO.

2. O has oxidation number 0 in O2 and -2 in CuO. 3. The oxidation number of Cu is becoming more positive: it is losing electrons.

The oxidation number of O is becoming more negative: it is gaining electrons. 4. Cu is losing electrons: it is being oxidised. (OIL = “oxidation is loss”).

O is gaining electrons: it is being reduced. (RIG = “reduction is gain”). 5. If the oxidation number of an element becomes more positive (or less negative) in a reaction, the

element is being oxidised. If the oxidation number of an element becomes more negative (or less positive) in a reaction, the

element is being reduced. 6. 2 electrons are being transferred, from Cu to O.

7. Oxidising agent: reactant that does the oxidising. It takes electrons from something else and, in doing so, is itself reduced. O is the oxidising agent in this particular reaction.

Reducing agent: reactant that does the reducing. It gives electrons to something else and, in doing so, is itself oxidised. Cu is the reducing agent in this particular reaction.

8. See table below. Oxidised Reduced Oxidising

agent Reducing

Agent # electrons transferred

(a) FeO(s) + CO(g) ⇌ Fe(s) + CO!(g) +2 -2 +2 -2 0 +4 -2 C Fe Fe C 2

(b) FeCO!(s) ⇌ FeO(s) + CO!(g) +2 +4 -2 +2 -2 +4 -2 - - - - -

(c) 2H!(aq) +Mg(s) ⇌ Mg!!(aq) + H!(g) +1 0 +2 0 Mg H H Mg 2

(d) H!(aq) + OH!(aq) ⇌  H!O(l) +1 -2 +1 +1 -2 - - - - -

9. See below.

Cu ⇌ Cu!! +  2e! 2 × (2) + O! +  4e! ⇌ 2O!! (3)

2Cu  +  O! +  4e! ⇌ 2Cu!! +  4e! +  O!! 2 × (2) + (3)

2Cu  +  O! ⇌ 2Cu!! +  2O!! (Cross out electrons as they balance on each side.

10. 2Li   +  1/2O! ⇌ Li!O or 4Li   +  O! ⇌ 2Li!O 11. 2H+(aq) + 2e- à H2(g)

Mg(s) à Mg2+(aq) + 2e- 12. (a) 2Au3+(aq) + 6Cl-(aq) à 3Cl2 + 2Au

(b) 3Ag+(aq) + La(s) à 3Ag(s) + La3+(aq)

13. (a) Sn2+(aq) à Sn4+(aq) + 2e- Fe3+(aq) + e- à Fe2+(aq)

(b) Pb(s) + HSO4-(aq) à PbSO4(s) + H+(aq) + 2e-

PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e- à PbSO4(s) + 2H2O(l)

The ‘basic’ half cell reactions are Pb à Pb2+ + 2e- and Pb4+ + 2e- à Pb2+ but the actual reactions are the more complicated ones shown – these are important reaction as there form the basis of the lead storage battery used, for example, in cars.