DP Chemistry (last assessment 2024)
Question 18M.2.sl.TZ2.3c.ii
Date | May 2018 | Marks available | [Maximum mark: 2] | Reference code | 18M.2.sl.TZ2.3c.ii |
Level | sl | Paper | 2 | Time zone | TZ2 |
Command term | Formulate | Question number | c.ii | Adapted from | N/A |
c.ii.
[Maximum mark: 2]
18M.2.sl.TZ2.3c.ii
The emission spectrum of an element can be used to identify it.
(c.ii)
Impure copper can be purified by electrolysis. In the electrolytic cell, impure copper is the anode (positive electrode), pure copper is the cathode (negative electrode) and the electrolyte is copper(II) sulfate solution.
Formulate the half-equation at each electrode.
[2]
Markscheme
Anode (positive electrode):
Cu(s) → Cu2+(aq) + 2e–
Cathode (negative electrode):
Cu2+(aq) + 2e– → Cu(s)
Accept Cu(s) – 2e– → Cu2+(aq).
Accept for →
Award [1 max] if the equations are at the wrong electrodes.
[2 marks]