Gold is an unreactive metal that can only be oxidised under specific conditions. Some relevant half-equations and their standard electrode potentials are given. | | half-equation | Eº/V | |---|-------------------------------------------------|--------| | 1 | Au³⁺(aq) + 3e⁻ ⇌ Au(s) | +1.50 | | 2 | [AuCl₄]⁻(aq) + 3e⁻ ⇌ Au(s) + 4Cl⁻(aq) | +1.00 | | 3 | NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ ⇌ NO(g) + 2H₂O(l) | +0.96 |
- (a)(i)Define the term standard electrode potential.[2]
- (a)(ii)Draw a fully labelled diagram of the apparatus that should be used to measure the standard cell potential, E°cell, of Au³⁺(aq)/Au(s) and HNO₃(aq)/NO(g). Include all necessary chemicals.[4]
- (a)(iii)Write an ionic equation to show the spontaneous reaction that occurs when an electric current is drawn from the cell in (a)(ii).[1]
- (a)(iv)Calculate the E°cell of the reaction in (a)(iii). E°cell = V[1]
- (a)(v)Gold can be oxidised by a mixture of concentrated hydrochloric acid and concentrated nitric acid, known as aqua regia. Concentrated hydrochloric acid is 12 moldm⁻³. Concentrated nitric acid is 16 moldm⁻³. Explain why aqua regia is able to dissolve gold. In your answer, state and explain what effect the use of concentrated hydrochloric acid and concentrated nitric acid have on the E values of half-equations 2 and 3.[3]
- (b)(i)Write an equation for the reaction of AuCl₃ with H₂O₂.[1]
- (b)(ii)Determine the rate equation of the reaction. Show your reasoning, quoting data from the table. | experiment | [AuCl₃] /mol dm⁻³ | [H₂O₂] /mol dm⁻³ | rate of production of O₂(g) /dm³ minute⁻¹ | |------------|-------------------|-------------------|---------------------------------------| | 1 | 0.05 | 0.50 | 7.66 × 10⁻² | | 2 | 0.10 | 0.50 | 1.53 × 10⁻¹ | | 3 | 0.15 | 1.00 | 4.60 × 10⁻¹ |[3]
- (b)(iii)Use the results of experiment 2 to calculate the value of the rate constant, k, for this reaction. Include the units of k. rate constant, k = units =[2]
- (c)(i)Name the enthalpy changes labelled ∆H₄ and ∆H₆. ∆H₄ = ∆H₆ =[2]
- (c)(ii)Use the data in the table and data from the Data Booklet to calculate the lattice energy of AlF₃. | process | enthalpy change /kJ mol⁻¹ | |---------------------------------|---------------------------| | Al(s) → Al(g) | +326 | | Al(g) → Al³⁺(g) | +5137 | | F(g) → F⁻(g) | -328 | | Al(s) + 1.5F₂(g) → AlF₃(s) | -1504 | lattice energy of AlF₃ = kJ mol⁻¹[2]
- (c)(iii)Scandium fluoride, ScF₃, is an ionic compound. Use data from the Data Booklet to suggest how the lattice energy of AlF₃ compares with the lattice energy of ScF₃. Explain your answer.[2]
- (d)(i)AlF₃ is sparingly soluble in water. The concentration of its saturated solution at 298K is 6.5 × 10⁻² mol dm⁻³. Write an expression for the solubility product, Ksp, of AlF₃. Ksp =[1]
- (d)(ii)Calculate the numerical value of Ksp for AlF₃ at 298 K. Ksp =[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.
- [2] voltage of an electrode / half-cell
- compared/connected to (S)HE / hydrogen half-cell / electrode
- under standard conditions / 1 mol dm-³, 1 atm, 298 K
- [4] (diagram showing two half-cells, salt bridge, voltmeter, correct electrodes (Au(s) & Pt(s)), correct solutions (Au3+(aq) & HNO3(aq)), NO(g) delivery, 1 mol dm-3 concentrations, 1 atm pressure, 298K)
- [1] Au3+ + NO + 2H2O → Au + NO3- + 4H+
- [1] +1.50 -0.96 = + 0.54 (V)
- [3] M1 adding conc HNO3 shifts equilibrium 3 to the right
- M2 adding conc HCl shift equilibrium 2 to the left
- E for (half-equation 3) increases / more positive
- E for (half-equation 2) decreases / less positive
- M3 E(3) becomes greater than E(2)
- [1] 2AuCl3 + 3H2O2 → 2Au + 3O2 + 6HCl
- [3] M1 1st order w.r.t. AuCl3 because rate x2 / doubles when concentration ×2 / doubles
- M2 First order H2O2 × 2; AuCl3 × 3 rate × 6 so order = 1 for H2O2
- M3 rate = k [AuC13] [H2O2]
- [2] k = 1.53 x 10-1 / (0.10 × 0.50) = 3.06
- dm³ mol-2 minute-1
- [2] ∆H4 = (3 ×) electron affinity of fluorine / F
- ∆H6 = (enthalpy change of) formation of AlF3
- [2] M1 +326 + 1.5 × 158 + 5137 + 3x-328 + ∆Hlatt = -1504
- M2 ∆Hlatt = −6220 (kJ mol-1)
- [2] M1 lattice energy of ScF3 should be less exothermic ora
- M2 Sc ion / Sc3+ larger than Al ion / Al3+ AND lesser attraction between the ions / ionic bonds are weaker
- [1] Ksp = [Al3+][F-]3
- [1] Ksp = 6.5 × 10-2 × (3 × 6.5 × 10-2)3 = 4.8 × 10-4
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