Topic 6: Electrochemistry

Question 1

Definitions I:
(a) Redox
(b) Oxidation
(c) Reduction

Answer 1

Definitions I:
(a) Reaction involving both oxidation and reduction by electron transfer.
(b) Loss of electrons, increase in oxidation number.
(c) Gain of electrons, decrease in oxidation number.

Question 2

Definitions II:
(a) Oxidising agent
(b) Reducing agent
(c) Disproportionation

Answer 2

Definitions II:
(a) Substance that accepts electrons, causes oxidation of another species.
(b) Substance that donates electrons, causes reduction of another species.
(c) Reaction where a species is both oxidised and reduced.

Question 3

What are the oxidation numbers of:
(a) S in SO₄²⁻? [1 Mark]
(b) Mn in MnO₄⁻? [1 Mark]

Answer 3

(a) S: +6 (O = -2, total -8, S balances to -2 charge) (1 mark)
(b) Mn: +7 (O = -8, Mn balances to -1 charge) (1 mark)

Question 4

Balance using oxidation numbers: MnO₄⁻ + H⁺ + Fe²⁺ → Mn²⁺ + Fe³⁺ + H₂O [3 Marks]

Answer 4

Mn: +7 to +2, -5 (1 mark)
Fe: +2 to +3, +1, ×5 Fe (1 mark)
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O (1 mark)

Question 5

Identify the redox process in:
(a) 2H₂ + O₂ → 2H₂O [1 Mark]
(b) Cl₂ + 2NaOH → NaCl + NaClO + H₂O [1 Mark]

Answer 5

(a) H₂ oxidised (0 to +1), O₂ reduced (0 to -2) (1 mark)
(b) Cl₂ disproportionates (0 to -1 and +1) (1 mark)

Question 6

Calculate oxidation numbers in:
(a) Cr₂O₇²⁻ [1 Mark]
(b) H₂O₂ [1 Mark]

Answer 6

(a) Cr: +6 (O = -14, 2Cr = +12, each +6) (1 mark)
(b) O: -1 (H = +2, 2O = -2, each -1) (1 mark)

Question 7

Is Fe²⁺ oxidised or reduced in Fe²⁺ → Fe³⁺ + e⁻, and what is the agent? [2 Marks]

Answer 7

Oxidised, loses electron (+2 to +3) (1 mark)
Oxidising agent accepts electron (e.g., O₂) (1 mark)

Question 8

Balance the disproportionation of Cu⁺ → Cu + Cu²⁺ using oxidation numbers. [2 Marks]

Answer 8

Cu⁺: +1, to 0 (Cu) and +2 (Cu²⁺) (1 mark)
2Cu⁺ → Cu + Cu²⁺ (1 mark)

Question 9

Why is MnO₄⁻ a strong oxidising agent in acidic solution? [2 Marks]

Answer 9

Mn +7 gains 5 electrons to +2 (1 mark)
High reduction potential in acid (1 mark)

Question 10

Assign Roman numerals to:
(a) Fe in Fe₂O₃ [1 Mark]
(b) Cu in CuCl₂ [1 Mark]

Answer 10

(a) Fe(III), +3 (O = -6, 2Fe = +6) (1 mark)
(b) Cu(II), +2 (Cl = -2, Cu = +2) (1 mark)

Question 11

Identify oxidation and reduction in Zn + Cu²⁺ → Zn²⁺ + Cu [2 Marks]

Answer 11

Zn oxidised (0 to +2), loses electrons (1 mark)
Cu²⁺ reduced (+2 to 0), gains electrons (1 mark)

Question 12

Balance Fe²⁺ + H₂O₂ → Fe³⁺ + H₂O in acidic conditions using oxidation numbers. [3 Marks]

Answer 12

Fe: +2 to +3, ×2 (1 mark)
H₂O₂: -2 to 0 (O: -1 to -2 in H₂O) (1 mark)
2Fe²⁺ + H₂O₂ + 2H⁺ → 2Fe³⁺ + 2H₂O (1 mark)