Oxidation and reduction
5 marks · HardThe following ionic equation represents a reaction between zinc metal and copper(II) ions:
State which species is oxidised and which is reduced. Give your answers as chemical symbols or formulae.
Write a half-equation for the oxidation process.
Explain, in terms of electron transfer, why zinc is considered a reducing agent in this reaction.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Zn is oxidised; Cu^{2+} is reduced
(b) Zn → Zn^{2+} + 2e^-
(c) Zinc acts as a reducing agent because it loses electrons to copper(II) ions. By donating electrons, zinc reduces Cu^{2+} to Cu.
Common mistakes: Mixing up oxidation and reduction processes; Omitting electron transfer in explanations
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Identifies Zn as being oxidised.
B2 (a) · 1 mark · Identifies Cu^{2+} as being reduced.
B3 (b) · 1 mark · Writes the correct half-equation for oxidation: Zn → Zn^{2+} + 2e^-
B4 (c) · 1 mark · Mentions that zinc loses electrons.
B5 (c) · 1 mark · States that zinc reduces Cu^{2+} (by electron transfer) or explains reduction of Cu^{2+} to Cu by Zn.
Worked solution
(a)
Zn is oxidised (it loses electrons); Cu^{2+} is reduced (gains electrons).
(b)
(c)
Zinc is a reducing agent because it loses two electrons to copper(II) ions. The electrons from Zn convert Cu^{2+} ions to metallic Cu.