DP Chemistry (first assessment 2025)

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Question 23M.2.HL.TZ1.4

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Date May 2023 Marks available [Maximum mark: 7] Reference code 23M.2.HL.TZ1.4
Level HL Paper 2 Time zone TZ1
Command term Calculate, Formulate, Outline, State Question number 4 Adapted from N/A
4.
[Maximum mark: 7]
23M.2.HL.TZ1.4

Redox reactions can be used to produce electricity.

(a)

State the oxidation state of sulfur in copper (II) sulfate.

[1]

Markscheme

+6/VI ✓

 

Do not accept 6/6+.

(b)

A voltaic cell was constructed using a copper (II) sulfate/copper half-cell and a zinc sulfate/zinc half-cell.

(b.i)

Outline why electrons flow from zinc to copper when these half cells are connected with a wire. Use section 25 of the data booklet.

[1]

Markscheme

Zinc more reactive/ «better» reducing agent/ «more» easily oxidized/loses electrons «more» easily. ✓
 

Accept “zinc higher in the activity «series»”.
Accept “zinc has a negative electrode potential/Cu has a positive electrode potential”.

(b.ii)

Formulate equations for the reactions taking place at each electrode.

 

Anode (negative electrode):  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 

Cathode (positive electrode): . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 

[2]

Markscheme

Anode (negative electrode):
Zn(s) → Zn2+(aq) + 2e−  ✓

Cathode (positive electrode):
Cu2+(aq) + 2e− → Cu(s) ✓

 

Award [1 max] for equilibria.
Award [1 max] for equations at the wrong electrodes.
State symbols not required for mark.

(c.i)

Calculate the standard cell potential for the voltaic cell in part (b). Use section 24 of the data booklet.

[1]

Markscheme

«Eθcell = +0.34−(−0.76) = +» 1.10 «V» ✓

 

Accept ECF from 4 (b) (ii).

(c.ii)

Calculate the standard Gibbs free energy change, ΔG⦵, in kJ mol−1, for this reaction. Use section 1 of the data booklet. (If you did not answer part (c)(i) use 1.05 V, but this is not the correct value.)

[2]

Markscheme

«ΔG⦵ = −nFE⦵ =» −2 × 9.65 × 104 × 1.10 ✓
−212.3 «kJ mol−1» ✓

Alternate:
«ΔG⦵ =» −2 × 9.65 × 104 × 1.05 ✓
−202.7 «kJ mol−1» ✓

 

Award [2] for the correct final answer.