DP Chemistry (last assessment 2024)

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Question 22N.2.sl.TZ0.3c.ii

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Date November 2022 Marks available [Maximum mark: 1] Reference code 22N.2.sl.TZ0.3c.ii
Level sl Paper 2 Time zone TZ0
Command term State, Write Question number c.ii Adapted from N/A
c.ii.
[Maximum mark: 1]
22N.2.sl.TZ0.3c.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)

(c.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

This was a well answered question. 60% of the candidates deduced the half-equation for the reaction occurring at the anode. The question also discriminated well between high-scoring and low-scoring candidates.

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