A-LevelChemistryElectrochemistry9701/41 · Oct/Nov 2021 · Q610 Marks

An excess of sodium iodide is added to a solution of copper(II) sulfate. Iodine and a white precipitate of copper(I) iodide are formed.

(a)Write an equation for the reaction that occurs.[1]
(b)(i)Explain why the copper(II) sulfate solution is coloured.[4]
(b)(ii)Suggest why the precipitate of copper(I) iodide is white.[1]
(c)Use suitable E° values from the Data Booklet to predict whether iodide ions can reduce Cu²⁺ to Cu⁺ under standard conditions. Explain your answer.[2]
(d)An excess of sodium iodide is added to copper(II) sulfate solution. Copper(I) iodide forms as a precipitate. After precipitation, [Cu⁺] is much lower than 1.0 moldm⁻³. Use this information and your answer to (c) to explain how the relevant electrode potentials change and hence why I⁻ ions can reduce Cu²⁺ ions.[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

[1] 2Cu2+ + 4I- → 2Cul + I2 OR 2CuSO4 + 4Nal → 2CuI + I2 + 2Na2SO4
[1] M1: d-d orbital splitting occurs
[1] M2: electron(s) promoted / excited
[1] M3: wavelength / frequency of light is absorbed
[1] M4: colour seen is complementary OR wavelength / frequency of light not absorbed is seen
[1] (for Cu+) 3d10 OR 3d subshell full
[1] M1: (Cu2+ / Cu+) Eº = (+)0.15 V AND (I2 / I¯) Eº = (+)0.54 V
[1] M2: No, since (Eºcell) negative / −0.39 V OR No, since (I2 / I¯) is more positive than (Cu2+ / Cu+) OR No, I2 is more easily reduced OR No, I2 stronger oxidant ORA
[1] M1: Cu2+ / Cu+ E becomes more positive as equilibrium shifts to the right
[1] M2: The new E for Cu2+ / Cu⁺ is more positive than 0.54 / Eº (I2 / I¯)

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