DP Chemistry (last assessment 2024)
Question 22N.2.hl.TZ0.3f.ii
| Date | November 2022 | Marks available | [Maximum mark: 1] | Reference code | 22N.2.hl.TZ0.3f.ii |
| Level | hl | Paper | 2 | Time zone | TZ0 |
| Command term | State, Write | Question number | f.ii | Adapted from | N/A |
f.ii.
[Maximum mark: 1]
22N.2.hl.TZ0.3f.ii
Consider the following reaction:
Cu2+ (aq) + Fe (s) → Fe2+ (aq) + Cu (s)
The diagram shows an unlabelled voltaic cell for the reaction:
Cu2+ (aq) + Fe (s) → Fe2+ (aq) + Cu (s)
(f.ii)
Write the half-equation for the reaction occurring at the anode (negative electrode).
[1]
Markscheme
Fe (s) → Fe2+ (aq) + 2e− ✔
Accept equilibrium arrow.
Do not award ECF for Cu (s) → Cu2+ (aq) + 2e−.
Examiners report
Most, almost irrespective of the annotation of the diagram, managed to write the correct half equation for the reaction at the iron anode.
