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Date May 2017 Marks available 1 Reference code 17M.2.sl.TZ2.2
Level SL Paper 2 Time zone TZ2
Command term Deduce Question number 2 Adapted from N/A

Question

An acidic sample of a waste solution containing Sn2+(aq) reacted completely with K2Cr2O7 solution to form Sn4+(aq).

State the oxidation half-equation.

[1]
a.i.

Deduce the overall redox equation for the reaction between acidic Sn2+(aq) and Cr2O72–(aq), using section 24 of the data booklet.

[1]
a.ii.

Calculate the percentage uncertainty for the mass of K2Cr2O7(s) from the given data.

[1]
b.i.

The sample of K2Cr2O7(s) in (i) was dissolved in distilled water to form 0.100 dmsolution. Calculate its molar concentration.

[1]
b.ii.

10.0 cm3 of the waste sample required 13.24 cm3 of the K2Cr2O7 solution. Calculate the molar concentration of Sn2+(aq) in the waste sample.

[2]
b.iii.

Markscheme

Sn2+(aq) → Sn4+(aq) + 2e

 

Accept equilibrium sign.

Accept Sn2+(aq) – 2e → Sn4+(aq).

[1 mark]

a.i.

Cr2O72–(aq) + 14H+(aq) + 3Sn2+(aq) → 2Cr3+(aq) + 7H2O(l) + 3Sn4+(aq)

 

Accept equilibrium sign.

[1 mark]

a.ii.

«13.239 g ± 0.002 g so percentage uncertainty» 0.02 «%»

 

Accept answers given to greater precision, such as 0.0151%.

[1 mark]

b.i.

« [K2Cr2O7] = \(\frac{{13.239{\text{ g}}}}{{294.20{\text{ g}}\,{\text{mo}}{{\text{l}}^{ - 1}} \times 0.100{\text{ d}}{{\text{m}}^3}}}\) =» 0.450 «mol\(\,\)dm–3»

[1 mark]

b.ii.

n(Sn2+) = «0.450 mol\(\,\)dm–3 x 0.01324 dm3 x \(\frac{{3\,mol}}{{1\,mol}}\) =» 0.0179 «mol»

«[Sn2+] = \(\frac{{0.0179\,mol}}{{0.0100\,mol}}\) =» 1.79 «mol\(\,\)dm–3»

 

Award [2] for correct final answer.

[2 marks]

b.iii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
b.iii.

Syllabus sections

Core » Topic 9: Redox processes » 9.1 Oxidation and reduction
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