DP Chemistry (last assessment 2024)

Question 20N.3.hl.TZ0.b

Date	November 2020	Marks available	[Maximum mark: 3]	Reference code	20N.3.hl.TZ0.b
Level	hl	Paper	3	Time zone	TZ0
Command term	Calculate	Question number	b	Adapted from	N/A

[Maximum mark: 3]

20N.3.hl.TZ0.b

A voltaic cell is made up of nickel and magnesium half-cells.

$Mg (s) | {Mg}^{2 +} (aq) | | {Ni}^{2 +} (aq) | Ni (s)$

Calculate the cell potential for $0.0100 mol {dm}^{- 3} {Mg}^{2 +} (aq)$ and $0.800 mol {dm}^{- 3} {Ni}^{2 +} (aq)$ at $298 K$ . Use sections 1, 2 and 24 of the data booklet.

[3]

Markscheme

$E^{Ɵ} = « 2.37 - 0.26 = » (+) 2.11 « V »$ ✔

$« Q = (\frac{0.0100}{0.800}) = » 0.0125$ AND $« n = » 2$ ✔

$« E = E^{Ɵ} - (\frac{R T}{n F}) \ln Q = 2.11 - (\frac{8.31 \times 298}{2 \times 96 500}) \ln 0.0125 = » (+) 2.17 « V »$ ✔

Award [3] for correct final answer.

Examiners report

The cell potential numerical problem was well executed and many scored all three marks, by calculating the final answer as +2.17 V. The most common errors related to either incorrectly calculating Q or identifying n = 2 in the intermediate, numerical step.

Syllabus sections

Additional higher level (AHL)

Additional higher level (AHL) » Topic 19: Redox processes » 19.1 Electrochemical cells

Additional higher level (AHL) » Topic 19: Redox processes

Question 20N.3.hl.TZ0.b

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