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A-Level · Chemistry

AC11 Chemistry: Electrode Potentials and Transition Metals

AQA 7405 · Calculator allowed · about 165 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question is about the electron configurations of d-block elements and the definition of a transition element.

(a)Define, in terms of electron configuration, what is meant by the term transition element.(2)

(b)Chromium, Cr (Z = 24), has an anomalous ground-state electron configuration. Give the full electron configuration of a chromium atom, and explain why this differs from the configuration predicted by the simple aufbau filling order.(2)

(c)Iron, Fe, has atomic number 26. Give the electron configurations of the Fe2+ and Fe3+ ions.(2)

(d)Zinc, Zn (Z = 30), is a d-block element. Explain why zinc is not classified as a transition element.(2)

(Total for Question 1 is 8 marks)

This question is about ligands, complex ions and the multidentate ligand EDTA4-.

(a)Define the term ligand.(1)

(b)State the coordination number and name the shape of each of the following complex ions: [Cu(H2O)6]2+, [CuCl4]2-, [Ag(NH3)2]+.(3)

(c)Explain, in terms of bonding, how a ligand such as water bonds to the central metal ion in a complex ion such as [Cu(H2O)6]2+.(3)

(d)EDTA4- is described as a hexadentate ligand. Explain what is meant by the term hexadentate, and state the coordination number of the complex ion formed between EDTA4- and a single Ca2+ ion.(2)

(e)Complexes formed with EDTA4- have a much larger stability constant than the equivalent complex formed with six monodentate ligands of similar donor strength (e.g. six water molecules). Explain this observation in terms of entropy.(3)

(Total for Question 2 is 12 marks)

Explain why many transition metal ion complexes are coloured, but the corresponding complexes of calcium (Ca2+) and zinc (Zn2+) are colourless. Your answer should refer to d-orbitals, ligands and electronic transitions.

(Total for Question 3 is 6 marks)

This question is about ligand substitution reactions of copper(II) complexes.

(a)Aqueous ammonia is added, a little at a time and then in excess, to a solution of aqueous copper(II) sulfate. Describe what would be observed, and give equations for the two stages of this reaction.(4)

(b)When excess concentrated hydrochloric acid is added instead to aqueous copper(II) sulfate, the solution turns from pale blue to yellow-green and the coordination number of copper changes from 6 to 4. Write an equation for this ligand substitution, and explain, in terms of ligand size, why the coordination number decreases.(3)

(c)The stability constant, Kstab, for the reaction [Cu(H2O)6]2+(aq) + 4NH3(aq) ≤> [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l) is Kstab = 2.0 x 10¹³ mol^-4 dm12 at 298 K. Write the expression for Kstab, and state what this very large value indicates about the position of equilibrium and the relative stability of the two complexes.(3)

(Total for Question 4 is 10 marks)

Ammonium vanadate(V) solution (containing VO2+ ions, yellow) is reduced by an excess of zinc metal in dilute sulfuric acid. Standard electrode potentials: VO2+ + 2H+ + e- -> VO2+ + H2O, E = +1.00 V; VO2+ + 2H+ + e- -> V3+ + H2O, E = +0.34 V; V3+ + e- -> V2+, E = -0.26 V; Zn2+ + 2e- -> Zn, E = -0.76 V.

(a)State the sequence of colour changes observed as reduction proceeds from vanadium(V) to vanadium(II), naming the oxidation state and colour present at each stage.(4)

(b)Using the standard electrode potentials given, show by calculation that zinc metal is able to reduce VO2+ all the way down to V2+, i.e. that each of the three reduction steps is thermodynamically feasible.(4)

(c)Write the overall balanced ionic equation, including state symbols, for the direct reduction of VO2+ to V2+ by zinc metal in acidic solution.(4)

(Total for Question 5 is 12 marks)

Copper(I) ions are unstable in aqueous solution but can be stabilised as insoluble compounds or as complex ions. Standard electrode potentials: E(Cu2+/Cu+) = +0.15 V; E(Cu+/Cu) = +0.52 V.

(a)Use an appropriate calculation to show that aqueous Cu+ ions disproportionate to form Cu2+ and Cu, and state which copper species is oxidised and which is reduced in this reaction.(3)

(b)Despite this, solid copper(I) oxide, Cu2O, and copper(I) chloride, CuCl, are both stable compounds that do not spontaneously disproportionate. Suggest why Cu+ is stable in these solids but not in aqueous solution.(3)

(c)In the presence of excess copper metal and a source of ammonia, Cu+ can be stabilised in solution as the complex ion [Cu(NH3)2]+, which does not readily disproportionate. Explain, in terms of the relative stabilisation of the Cu+ and Cu2+ oxidation states by ligands, how forming this complex ion helps to prevent disproportionation.(4)

(Total for Question 6 is 10 marks)

The standard electrode potential for [Fe(H2O)6]3+/[Fe(H2O)6]2+ is +0.77 V. When cyanide ligands replace the water ligands, the standard electrode potential for [Fe(CN)6]3-/[Fe(CN)6]4- falls to +0.36 V.

(a)State what this fall in E value indicates about the relative oxidising strength of [Fe(CN)6]3- compared with [Fe(H2O)6]3+.(2)

(b)Explain, in terms of the relative stabilisation of the Fe3+ and Fe2+ oxidation states by the CN- ligand, why complexing with cyanide causes this decrease in electrode potential.(3)

(c)Given E(I2/I-) = +0.54 V, use the E value of +0.36 V for [Fe(CN)6]3-/[Fe(CN)6]4- to determine whether [Fe(CN)6]3- can oxidise iodide ions to iodine, and contrast this with the behaviour of [Fe(H2O)6]3+.(3)

(Total for Question 7 is 8 marks)

This question is about catalysis by transition metals and their ions.

(a)Iron is used as a heterogeneous catalyst in the Haber process, and platinum, palladium and rhodium are used in catalytic converters. Explain, using the terms adsorption and active site, how a heterogeneous catalyst increases the rate of a reaction, and explain why catalytic converters use a fine mesh or honeycomb structure.(4)

(b)The reaction between C2O4²-(aq) and MnO4-(aq) in acidic solution is autocatalysed by the Mn2+ ions produced. Explain what is meant by autocatalysis, why the initial rate of this reaction is slow, and describe the shape of a graph of [MnO4-] against time that you would expect.(4)

(c)The reaction between peroxodisulfate ions, S2O8²-, and iodide ions, I-, is slow when the ions are mixed directly, but is catalysed homogeneously by Fe2+ ions. Standard electrode potentials: E(S2O8²-/SO4²-) = +2.01 V; E(Fe3+/Fe2+) = +0.77 V; E(I2/I-) = +0.54 V. Show, using these E values, that Fe2+ can catalyse this reaction via two separate feasible steps, and explain why this two-step route avoids the main reason the direct, uncatalysed reaction is slow.(5)

(Total for Question 8 is 13 marks)

This question uses practical techniques from the required practical on titration. EDTA4- forms a 1:1 complex ion with Ca2+, [Ca(EDTA)]2-, and this reaction is used to determine the total hardness of a water sample. A 100.0 cm3 sample of tap water is titrated with 0.0100 mol dm-3 EDTA solution, using an indicator that changes colour once all the free Ca2+ (and Mg2+) has been complexed. The mean titre required is 24.60 cm3.

(a)State the coordination number of Ca2+ in the [Ca(EDTA)]2- complex ion, and explain why EDTA4- can form this complex with only one EDTA4- ion per Ca2+ ion.(2)

(b)Describe the colour change that would be observed at the end point of this titration, in general terms, given that the indicator used binds to free Ca2+ (giving one colour) but is a different colour when it is not bound to a metal ion.(2)

(c)Calculate the amount, in mol, of EDTA4- used in the titration, and hence the concentration, in mol dm-3, of Ca2+ in the water sample.(4)

(d)Express this hardness as a concentration in mg dm-3 of CaCO3 equivalent (Mr of CaCO3 = 100.1), and use the scale below to classify this water sample: soft, less than 60 mg dm-3; moderately hard, 60 to 120 mg dm-3; hard, 120 to 180 mg dm-3; very hard, greater than 180 mg dm-3 (all as CaCO3).(3)

(Total for Question 9 is 11 marks)

This question uses techniques from the required practical on measuring the emf of an electrochemical cell. A student sets up a cell using a platinum electrode dipping into a solution containing both Fe3+(aq) and Fe2+(aq) ions, connected by a salt bridge to a second platinum electrode dipping into a solution containing both Cr3+(aq) and Cr2+(aq) ions, all under standard conditions. Standard electrode potentials: E(Fe3+/Fe2+) = +0.77 V; E(Cr3+/Cr2+) = -0.41 V.

(a)Explain why a platinum electrode, rather than a strip of metal, must be used for each of these two half-cells.(2)

(b)Calculate the standard emf of this cell.(1)

(c)State which half-cell forms the positive electrode of this cell, and explain your reasoning.(2)

(d)The student's measured emf for this cell was 1.16 V, slightly below the theoretical value. Suggest two reasons for this discrepancy that relate specifically to this cell using platinum electrodes with ion-ion redox couples in solution (rather than solid metal electrodes).(3)

(Total for Question 10 is 8 marks)

This question is about general properties of transition elements and a calculation on a hydrated transition metal salt.

(a)Explain, referring to the energies of the 3d and 4s orbitals, why transition metals commonly show variable oxidation states.(2)

(b)State two characteristic properties, other than variable oxidation states, that are shared by transition metals and/or their compounds.(2)

(c)A sample of hydrated cobalt(II) chloride, CoCl2.xH2O, has a molar mass of 237.9 g mol-1. Given that the molar mass of anhydrous CoCl2 is 129.9 g mol-1, calculate x, the number of moles of water of crystallisation per mole of CoCl2.(4)

(Total for Question 11 is 8 marks)

This question is about precipitation reactions of transition metal ions with aqueous sodium hydroxide.

(a)State the colour of the precipitate formed, and identify the species present, when aqueous sodium hydroxide is added separately to solutions of Fe2+(aq), Fe3+(aq), and Cu2+(aq).(3)

(b)Chromium(III) hydroxide, Cr(OH)3, is described as amphoteric. Using ionic equations, explain what this means with reference to its reactions with excess sodium hydroxide and with a strong acid.(3)

(c)The green precipitate of Fe(OH)2 slowly turns to a rust-orange colour on standing in air. Explain this observation, giving an equation.(2)

(Total for Question 12 is 8 marks)

Standard electrode potentials: E(MnO4-/Mn2+, acidic) = +1.51 V; E(O2/H2O) = +1.23 V, for the half-equation O2 + 4H+ + 4e- -> 2H2O.

(a)Use these data to show, by calculation, that acidified potassium manganate(VII) solution is thermodynamically capable of oxidising water to oxygen gas.(3)

(b)Despite this, acidified KMnO4(aq) solutions can be stored and used as titrants over the timescale of a normal titration without significant decomposition. Explain this observation in terms of reaction kinetics.(3)

(c)Suggest why KMnO4(aq) solutions are typically stored in dark (brown glass) bottles, away from direct sunlight.(2)

(Total for Question 13 is 8 marks)

A colorimeter is calibrated using standard solutions of hydrated copper(II) sulfate, plotting absorbance against concentration to give a straight-line calibration graph through the origin, with gradient (calibration constant) k = 42.5 dm3 mol-1 (absorbance = k x concentration). An unknown copper(II) sulfate solution gives an absorbance of 0.638 using the same colorimeter and filter.

(a)Explain, in terms of the colour of the solution, why a red/orange filter (rather than a blue filter) should be selected for this colorimeter measurement.(2)

(b)Using the calibration data given, calculate the concentration, in mol dm-3, of Cu2+ in the unknown solution.(3)

(c)This solution was prepared by diluting 10.0 cm3 of the original stock solution to a total of 250 cm3 with distilled water before measuring its absorbance. Calculate the concentration, in mol dm-3, of Cu2+ in the original (undiluted) stock solution.(2)

(Total for Question 14 is 7 marks)

Mark scheme · AC11 Chemistry: Electrode Potentials and Transition Metals

Question 1

(a) B1 a d-block element (oe)
(a) B1 that forms at least one stable ion with an incomplete (partially-filled) d sub-shell
(a) Answer: A transition element is a d-block element that forms at least one ion with an incomplete d sub-shell.
(b) B1 1s2 2s2 2p6 3s2 3p6 3d5 4s1 (oe, allow [Ar]3d5 4s1)
(b) B1 a half-filled 3d sub-shell (3d5) with a single 4s electron is more stable/lower in energy than the predicted 3d4 4s2 arrangement (oe)
(b) Answer: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 ([Ar]3d5 4s1); one electron moves from 4s into 3d to give the extra stability of a half-filled 3d sub-shell.
(c) B1 Fe2+: [Ar]3d6 (oe)
(c) B1 Fe3+: [Ar]3d5 (oe)
(c) Answer: Fe2+: [Ar]3d6. Fe3+: [Ar]3d5.
(d) M1 zinc atoms, [Ar]3d10 4s2, only ever form Zn2+ ions, by loss of the two 4s electrons
(d) A1 giving [Ar]3d10, a complete (full) d sub-shell, so zinc never forms an ion with an incomplete d sub-shell (oe)
(d) Answer: Zinc only ever forms Zn2+, with configuration [Ar]3d10 - a full d sub-shell - so it does not meet the 'incomplete d sub-shell' requirement and is not classed as a transition element.

Question 2

(a) B1 a molecule or ion that donates a lone pair of electrons to a (central) metal ion/atom, forming a dative covalent (coordinate) bond (oe)
(a) Answer: A ligand is a molecule or ion that can donate a lone pair of electrons to a central metal ion, forming a dative covalent bond.
(b) B1 [Cu(H2O)6]2+: coordination number 6, octahedral
(b) B1 [CuCl4]2-: coordination number 4, tetrahedral
(b) B1 [Ag(NH3)2]+: coordination number 2, linear
(b) Answer: [Cu(H2O)6]2+: 6, octahedral. [CuCl4]2-: 4, tetrahedral. [Ag(NH3)2]+: 2, linear.
(c) M1 each water ligand has a lone pair of electrons on the oxygen (donor) atom
(c) A1 this lone pair is donated into an empty (vacant) orbital on the central Cu2+ ion, forming a dative covalent (coordinate) bond, in which both electrons of the shared pair originate from the ligand
(c) B1 six such bonds form around the Cu2+ ion (matching the coordination number), arranged octahedrally in space, giving the shape of the complex ion
(c) Answer: Each water molecule donates a lone pair from its oxygen atom into an empty orbital on Cu2+, forming a dative covalent bond; six such bonds arranged octahedrally give the [Cu(H2O)6]2+ complex ion.
(d) B1 hexadentate means the ligand has six donor atoms (lone pairs) and so can form six dative covalent bonds to the same central metal ion simultaneously (oe)
(d) B1 coordination number of the Ca-EDTA complex = 6
(d) Answer: Hexadentate means the ligand can form six dative bonds to one central ion at once (it has six donor atoms); the coordination number of [Ca(EDTA)]2- is 6.
(e) M1 when EDTA4- replaces six monodentate H2O ligands, the total number of free/independent particles in solution increases: 2 reactant particles (the hydrated ion and EDTA4-) become 7 product particles (the new complex plus six released water molecules) (oe)
(e) M1 this contrasts with monodentate ligand substitution, where the number of particles stays roughly the same (6 ligands replace 6 water molecules), giving little entropy change (oe)
(e) A1 the large increase in the number of particles gives a large positive entropy change, making delta G more negative (via delta G = delta H - T delta S), so the chelate complex is thermodynamically more stable, giving a much larger stability constant (the chelate effect) (oe)
(e) Answer: Substituting one EDTA4- for six water ligands increases the total number of particles in solution (2 to 7), giving a large positive entropy change; this makes delta G more negative than for an equivalent monodentate substitution (where particle numbers barely change), so the chelate complex is much more stable (the chelate effect).

Question 3

Level 3 (5-6): A coherent, logically structured explanation covering: the five d orbitals of a free/isolated metal ion are degenerate; ligands bonding to the metal ion cause the d orbitals to split into two groups of different energy; an electron can be promoted (absorbing energy) from the lower to the higher set of d orbitals (a d-d transition), and the energy absorbed corresponds to a frequency of visible light; the complementary colour (the light not absorbed) is seen; the size of the splitting (and hence the colour observed) depends on the ligand and the metal ion; Ca2+ and Zn2+ have either an empty (d0) or full (d10) d sub-shell, so no d-d transition is possible, hence they are colourless.
Level 2 (3-4): Some relevant chemistry given, e.g. describing d-orbital splitting by ligands and d-d electron transitions, but with incomplete or partly inaccurate linkage to why Ca2+/Zn2+ complexes are specifically colourless.
Level 1 (1-2): Basic or isolated statement(s), e.g. 'transition metals have partly-filled d orbitals' or 'electrons absorb light', with little or no development of the mechanism.
Level 0 (0): No relevant chemistry given, or answer entirely incorrect/irrelevant.
Indicative content:
- In a free (isolated, gaseous) transition metal ion, the five d orbitals are degenerate (all the same energy).
- When ligands bond to the central metal ion, they cause the d orbitals to split into two groups (sets) of different energy (e.g. a lower set of three and a higher set of two in an octahedral complex).
- An electron in the lower-energy set of d orbitals can absorb a specific quantity (quantum) of energy and be promoted into the higher-energy set; this is called a d-d transition.
- The energy absorbed corresponds to a particular frequency/wavelength of light, which for transition metal complexes typically falls within the visible region of the spectrum.
- The colour observed is the complementary colour to the light absorbed (the wavelengths not absorbed are transmitted or reflected, and this is what is seen).
- The size of the energy gap between the two sets of d orbitals (and therefore the frequency of light absorbed, and the colour seen) depends on the identity of both the ligand and the metal ion/its oxidation state.
- Ca2+ has an empty d sub-shell (d0), so there are no d electrons available to be promoted; Zn2+ has a full d sub-shell (d10), so there is no vacant d orbital for an electron to be promoted into.
- Because neither Ca2+ nor Zn2+ can undergo a d-d transition, no visible light is absorbed and their complexes are colourless.

Question 4

(a) B1 with a small volume of aqueous ammonia, a pale blue precipitate forms
(a) B1 Cu2+(aq) + 2NH3(aq) + 2H2O(l) -> Cu(OH)2(s) + 2NH4+(aq)
(a) B1 in excess ammonia the precipitate dissolves to give a deep (royal) blue solution
(a) B1 Cu(OH)2(s) + 4NH3(aq) + 2H2O(l) -> [Cu(NH3)4(H2O)2]2+(aq) + 2OH-(aq)
(a) Answer: A pale blue precipitate of Cu(OH)2 forms first (Cu2+ + 2NH3 + 2H2O -> Cu(OH)2 + 2NH4+); in excess ammonia this dissolves to give a deep blue solution of [Cu(NH3)4(H2O)2]2+ (Cu(OH)2 + 4NH3 + 2H2O -> [Cu(NH3)4(H2O)2]2+ + 2OH-).
(b) B1 [Cu(H2O)6]2+(aq) + 4Cl-(aq) -> [CuCl4]2-(aq) + 6H2O(l)
(b) M1 chloride ions are larger (have a bigger ionic radius) than water molecules
(b) A1 so fewer Cl- ions than H2O molecules can fit (pack) around the central Cu2+ ion, reducing the coordination number from 6 to 4 (oe)
(b) Answer: [Cu(H2O)6]2+ + 4Cl- -> [CuCl4]2- + 6H2O; chloride ions are larger than water molecules, so fewer can pack around the Cu2+ ion, giving a lower coordination number (4 instead of 6).
(c) B1 Kstab = [Cu(NH3)4(H2O)2]2+ / ([Cu(H2O)6]2+ x [NH3]⁴)
(c) M1 a very large Kstab means the equilibrium position lies very far to the right (almost entirely towards the ammine complex) oe
(c) A1 so the ammine complex, [Cu(NH3)4(H2O)2]2+, is much more thermodynamically stable than the aqua complex, [Cu(H2O)6]2+ oe
(c) Answer: Kstab = [Cu(NH3)4(H2O)2]2+ / ([Cu(H2O)6]2+ x [NH3]⁴); the very large value shows the equilibrium lies almost completely to the right, meaning the ammine complex is far more stable than the aqua complex.

Question 5

(a) B1 vanadium(V), VO2+, yellow
(a) B1 vanadium(IV), VO2+, blue
(a) B1 vanadium(III), V3+, green
(a) B1 vanadium(II), V2+, violet (lilac)
(a) Answer: Yellow VO2+ (V, +5) -> blue VO2+ (V, +4) -> green V3+ (V, +3) -> violet/lilac V2+ (V, +2).
(b) M1 step 1: E cell = E(VO2+/VO2+) - E(Zn2+/Zn) = 1.00 - (-0.76) = +1.76 V, positive so feasible
(b) M1 step 2: E cell = E(VO2+/V3+) - E(Zn2+/Zn) = 0.34 - (-0.76) = +1.10 V, positive so feasible
(b) M1 step 3: E cell = E(V3+/V2+) - E(Zn2+/Zn) = -0.26 - (-0.76) = +0.50 V, positive so feasible
(b) A1 since E(Zn2+/Zn) = -0.76 V is more negative than all three vanadium E values, E cell is positive at every stage, so zinc can reduce V(+5) all the way to V(+2) cao
(b) Answer: E cell = +1.76 V, +1.10 V and +0.50 V for the three steps respectively; all positive, so zinc reduces VO2+ fully to V2+.
(c) M1 combined vanadium reduction half-equation: VO2+ + 4H+ + 3e- -> V2+ + 2H2O
(c) M1 zinc oxidation half-equation: Zn -> Zn2+ + 2e-
(c) M1 half-equations correctly scaled to balance electrons transferred (vanadium x2, zinc x3, giving 6 electrons)
(c) A1 2VO2+(aq) + 8H+(aq) + 3Zn(s) -> 2V2+(aq) + 4H2O(l) + 3Zn2+(aq) fully balanced with state symbols cao
(c) Answer: 2VO2+(aq) + 8H+(aq) + 3Zn(s) -> 2V2+(aq) + 4H2O(l) + 3Zn2+(aq)

Question 6

(a) M1 E cell = E(Cu+/Cu) - E(Cu2+/Cu+) = 0.52 - 0.15 = +0.37 V
(a) A1 positive E cell, so 2Cu+(aq) -> Cu2+(aq) + Cu(s) is thermodynamically feasible cao
(a) B1 some Cu+ is reduced to Cu (gains an electron) while the rest of the Cu+ is oxidised to Cu2+ (loses an electron); i.e. Cu+ acts as both the oxidising and reducing agent
(a) Answer: E cell = +0.37 V, positive, so 2Cu+ -> Cu2+ + Cu is feasible; Cu+ is simultaneously reduced (to Cu) and oxidised (to Cu2+) - it disproportionates.
(b) M1 in the solid state the Cu+ ions are held in a fixed lattice, strongly bonded to the oxide/chloride ions (by ionic/some covalent-character bonding), rather than existing as free, mobile hydrated ions
(b) A1 there is no mechanism for two neighbouring Cu+ ions in a rigid solid lattice to come together and exchange an electron in the way that mobile Cu+(aq) ions can in solution (oe)
(b) A1 so, although disproportionation may still be thermodynamically favourable overall, these solids are kinetically stable (do not disproportionate at a measurable rate) under normal conditions (oe)
(b) Answer: In Cu2O and CuCl the Cu+ ions are fixed in a rigid lattice rather than existing as free, mobile hydrated ions, so there is no route for neighbouring Cu+ ions to exchange electrons; these solids are kinetically stable even though Cu+(aq) disproportionates readily.
(c) M1 ammonia ligands form complexes with both Cu+ and Cu2+, but the two oxidation states are not stabilised to the same extent (oe)
(c) M1 if ammonia complexation stabilises Cu+ (lowers its energy/makes [Cu(NH3)2]+ have a much larger stability constant) more than it stabilises Cu2+, then converting Cu+ to Cu2+ (as required for disproportionation) becomes energetically less favourable than in the uncomplexed aqua-ion system (oe)
(c) A1 this effectively raises the relevant electrode potential for the Cu2+/Cu+ couple, reducing (or reversing the sign of) E cell for the disproportionation reaction, so it is no longer thermodynamically favourable, and free Cu+(aq) is removed from solution as the stable complex (oe)
(c) A1 this implies that the stability constant of [Cu(NH3)2]+ (the Cu+ ammine complex) is much larger than that of the corresponding Cu2+ ammine complex, relative to their aqua ions (oe)
(c) Answer: Ammonia stabilises Cu+ (as [Cu(NH3)2]+) much more strongly than it stabilises Cu2+, so oxidising Cu+ to Cu2+ becomes less favourable once both are complexed; this effectively changes the relevant electrode potential enough that disproportionation is no longer feasible, implying the Cu+ ammine complex has a disproportionately large stability constant compared with the Cu2+ ammine complex.

Question 7

(a) B1 [Fe(CN)6]3- is a weaker oxidising agent than [Fe(H2O)6]3+ (oe)
(a) B1 because it has a lower (less positive) tendency to be reduced/to gain an electron (oe)
(a) Answer: [Fe(CN)6]3- is a weaker oxidising agent than [Fe(H2O)6]3+, since its (less positive) E value shows a smaller tendency to be reduced.
(b) M1 CN- ligands stabilise the Fe3+ ion (with its higher charge density) more strongly than they stabilise the Fe2+ ion (oe)
(b) M1 so complexation with CN- lowers the energy of the Fe3+ complex, [Fe(CN)6]3-, by more than it lowers the energy of the Fe2+ complex, [Fe(CN)6]4- (oe)
(b) A1 this differential stabilisation makes the Fe3+ oxidation state relatively more stable (harder to reduce) within the cyanide complex than within the aqua complex, so the electrode potential for the couple decreases (oe)
(b) Answer: CN- ligands stabilise Fe3+ more strongly than Fe2+ (because of its higher charge/greater polarising power), lowering the energy of [Fe(CN)6]3- relatively more; this makes Fe3+ harder to reduce within the complex, so the electrode potential of the couple falls from +0.77 V to +0.36 V.
(c) M1 for [Fe(CN)6]3-: E cell = 0.36 - 0.54 = -0.18 V, negative, so [Fe(CN)6]3- cannot oxidise I- to I2
(c) M1 for [Fe(H2O)6]3+: E cell = 0.77 - 0.54 = +0.23 V, positive, so [Fe(H2O)6]3+ can oxidise I- to I2
(c) A1 so complexing Fe3+/Fe2+ with cyanide changes the feasibility of this reaction: the aqua ion oxidises iodide but the cyanide complex does not cao
(c) Answer: [Fe(CN)6]3-: E cell = -0.18 V, not feasible. [Fe(H2O)6]3+: E cell = +0.23 V, feasible. Complexation with cyanide therefore prevents the oxidation of iodide that the aqua ion can carry out.

Question 8

(a) B1 reactant molecules are adsorbed onto active sites on the surface of the catalyst
(a) B1 adsorption weakens the bonds within the reactant molecules (or holds them in a favourable orientation/proximity), providing an alternative reaction pathway of lower activation energy
(a) B1 product molecules desorb from the surface, freeing the active site for further reactant molecules
(a) B1 a fine mesh/honeycomb structure maximises the surface area (and so the number of exposed active sites) for a given mass of the expensive precious metal, increasing the rate of reaction while minimising cost
(a) Answer: Reactants adsorb onto active sites, weakening bonds and lowering activation energy via an alternative pathway; products desorb, freeing the site for reuse. A honeycomb/fine mesh structure maximises surface area (and active sites) per unit mass of catalyst, increasing rate while minimising the amount of costly metal needed.
(b) B1 autocatalysis is where a reaction is catalysed by one of its own products (here, Mn2+) (oe)
(b) B1 at the very start of the reaction there is no Mn2+ present, so the reaction can only proceed by the slow, uncatalysed pathway (partly because C2O4²- and MnO4- are both negatively-charged ions, so electrostatic repulsion between them contributes to a high activation energy) (oe)
(b) M1 as the reaction proceeds, Mn2+ is produced and catalyses further reaction, so the rate increases (the reaction accelerates) over the early part of the reaction (oe)
(b) A1 the graph of [MnO4-] against time starts with a shallow (slow) negative slope, becomes steeper (faster) as [Mn2+] builds up, then levels off towards zero as the reactants are used up towards completion (an S-shaped/sigmoidal decay curve) (oe)
(b) Answer: Autocatalysis is catalysis of a reaction by one of its own products; here Mn2+ catalyses the reaction. Initially, with no Mn2+ present and both reactants negatively charged, the rate is slow; as Mn2+ builds up the rate increases, giving a concentration-time graph for MnO4- that starts shallow, steepens, then levels off near zero (an S-shaped/sigmoidal curve).
(c) M1 step 1: 2Fe2+ + S2O8²- -> 2Fe3+ + 2SO4²-; E cell = 2.01 - 0.77 = +1.24 V, feasible
(c) M1 step 2: 2Fe3+ + 2I- -> 2Fe2+ + I2; E cell = 0.77 - 0.54 = +0.23 V, feasible
(c) A1 Fe2+ is regenerated in step 2, so it acts as a true catalyst overall (net equation: S2O8²- + 2I- -> 2SO4²- + I2) cao
(c) M1 the direct, uncatalysed reaction between S2O8²- and I- is slow mainly because both ions are negatively charged, so there is strong electrostatic repulsion between them, giving a high activation energy (oe)
(c) A1 in each catalysed step, one of the two reacting species is a cation (Fe2+ or Fe3+), so the anion-anion repulsion is avoided, lowering the activation energy of each step and increasing the overall rate (oe)
(c) Answer: Step 1: E cell = +1.24 V (feasible); Step 2: E cell = +0.23 V (feasible); Fe2+ is regenerated, so it is a true catalyst. The uncatalysed reaction is slow due to repulsion between the two anions; the catalysed route avoids this because each step involves a cation (Fe2+ or Fe3+) reacting with an anion.

Question 9

(a) B1 coordination number = 6
(a) B1 because EDTA4- is hexadentate, providing all six donor atoms (bonds) needed by Ca2+, so only one EDTA4- ion is required per Ca2+ ion (oe)
(a) Answer: Coordination number = 6; since EDTA4- is hexadentate it alone can supply all six dative bonds to Ca2+, so a 1:1 complex forms.
(b) B1 before the end point, free Ca2+ ions are still present in solution, bound to the indicator, giving the indicator's 'metal-bound' colour
(b) B1 at the end point, the last free Ca2+ ions are complexed by EDTA4- (which binds Ca2+ more strongly than the indicator does), releasing the indicator, which changes to its own (free/uncomplexed) colour (oe)
(b) Answer: While free Ca2+ remains, the indicator stays bound to it, showing its metal-bound colour; at the end point, EDTA4- (which binds Ca2+ more strongly) displaces the last of the indicator from Ca2+, and the colour changes to the indicator's free (unbound) colour.
(c) M1 n(EDTA4-) = 0.0100 x (24.60/1000) = 2.460 x 10^-4 mol
(c) M1 n(Ca2+) = n(EDTA4-) = 2.460 x 10^-4 mol (1:1 ratio)
(c) M1 concentration = n/V = 2.460e-4 / 0.1000
(c) A1 2.46 x 10^-3 mol dm-3 cao (awrt 2.46e-3)
(c) Answer: 2.46 x 10^-3 mol dm-3
(d) M1 mass of CaCO3 equivalent per dm3 = 2.460 x 10^-3 x 100.1
(d) A1 246 mg dm-3 cao (awrt 246)
(d) B1 classified as very hard water (since 246 mg dm-3 is greater than 180 mg dm-3) ft
(d) Answer: 246 mg dm-3 as CaCO3; this is classified as very hard water.

Question 10

(a) M1 neither half-cell involves a solid metal in equilibrium with its own ion (both couples are between two different oxidation states of ions in solution), so there is no solid metal electrode available to use (oe)
(a) A1 platinum is used because it is inert/unreactive: it does not itself take part in the reaction, but simply provides a conducting surface at which electrons can be transferred to or from the ions in solution (oe)
(a) Answer: Both couples involve only dissolved ions (no solid metal), so an inert platinum electrode is used to conduct electrons to/from the solution without itself reacting.
(b) B1 +1.18 V cao (E cell = 0.77 - (-0.41))
(b) Answer: +1.18 V
(c) B1 the Fe3+/Fe2+ half-cell is the positive electrode
(c) B1 because it has the more positive (higher) standard electrode potential, so it is the couple that is reduced (accepts electrons) when the cell operates (oe)
(c) Answer: The Fe3+/Fe2+ half-cell; it has the more positive E value, so it is reduced (accepts electrons from the Cr3+/Cr2+ half-cell) when the cell operates.
(d) B1 Cr2+ is a very strong reducing agent and is readily oxidised by dissolved oxygen in the air/solution before or during the measurement, reducing the effective concentration of Cr2+ (moving that half-cell away from standard conditions and altering its potential) (oe)
(d) B1 surface contamination, incomplete platinisation, or slow electron-transfer kinetics at the inert platinum surface can mean true equilibrium (and hence the full theoretical potential) has not yet been established when the reading is taken (oe)
(d) B1 small amounts of current drawn by the voltmeter (not perfectly infinite resistance) reduce the reading slightly below the true zero-current emf (oe); allow any two valid, distinct reasons specific to this system
(d) Answer: For example: Cr2+ is readily air-oxidised by dissolved oxygen, shifting that half-cell from standard conditions; and/or slow electron-transfer kinetics or surface contamination at the platinum electrodes mean equilibrium is not fully established when the reading is taken.

Question 11

(a) M1 the 3d and 4s sub-shells are very close together in energy (oe)
(a) A1 so successive ionisation energies (for removing electrons beyond the first two 4s electrons, from the 3d sub-shell) increase relatively gradually, without one very large jump, allowing a variable number of electrons to be lost to form ions of different, comparably-stable oxidation states (oe)
(a) Answer: The 3d and 4s orbitals are close in energy, so successive ionisation energies increase gradually rather than showing one large jump, allowing several oxidation states to be reasonably stable.
(b) B1 forming coloured compounds/ions (in solution or as solids)
(b) B1 acting as catalysts (as elements or as compounds/ions); allow forming complex ions with ligands as an alternative second property
(b) Answer: For example: forming coloured compounds/ions, and acting as catalysts (or forming complex ions with ligands).
(c) M1 mass attributable to water = 237.9 - 129.9 = 108.0
(c) M1 x = 108.0 / 18.0
(c) A1 x = 6.00 cao
(c) B1 so the formula is CoCl2.6H2O
(c) Answer: x = 6; the formula is CoCl2.6H2O.

Question 12

(a) B1 Fe2+(aq): green precipitate of Fe(OH)2
(a) B1 Fe3+(aq): orange-brown (rust-coloured) precipitate of Fe(OH)3
(a) B1 Cu2+(aq): pale blue precipitate of Cu(OH)2
(a) Answer: Fe2+: green Fe(OH)2. Fe3+: orange-brown Fe(OH)3. Cu2+: pale blue Cu(OH)2.
(b) B1 amphoteric means Cr(OH)3 can react with both a base and an acid (oe)
(b) B1 with excess OH-(aq): Cr(OH)3(s) + 3OH-(aq) -> [Cr(OH)6]3-(aq), the precipitate dissolves to give a (dark) green solution
(b) B1 with a strong acid: Cr(OH)3(s) + 3H+(aq) -> Cr3+(aq) + 3H2O(l), the precipitate dissolves to give a green Cr3+ solution
(b) Answer: Amphoteric means it reacts with both acids and bases. With excess OH-: Cr(OH)3 + 3OH- -> [Cr(OH)6]3- (dissolves, green solution). With acid: Cr(OH)3 + 3H+ -> Cr3+ + 3H2O (dissolves, green solution).
(c) B1 Fe(OH)2 is oxidised by dissolved oxygen (and water) in the air to Fe(OH)3, which is the rust-orange colour observed (oe)
(c) B1 4Fe(OH)2(s) + O2(g) + 2H2O(l) -> 4Fe(OH)3(s)
(c) Answer: Fe(OH)2 is oxidised by atmospheric oxygen to Fe(OH)3: 4Fe(OH)2 + O2 + 2H2O -> 4Fe(OH)3.

Question 13

(a) M1 E cell = E(MnO4-/Mn2+) - E(O2/H2O) = 1.51 - 1.23
(a) A1 = +0.28 V cao
(a) B1 positive E cell, so the reaction (MnO4- oxidising H2O to O2) is thermodynamically feasible
(a) Answer: E cell = +0.28 V, positive, so acidified KMnO4 is thermodynamically capable of oxidising water to oxygen.
(b) M1 a positive E cell value only shows that a reaction is thermodynamically feasible (energetically favourable); it gives no information about the rate at which the reaction occurs (oe)
(b) M1 in the absence of a catalyst, this decomposition reaction has a high activation energy, so its rate at room temperature is extremely slow (oe)
(b) A1 so, although KMnO4(aq) does very slowly decompose over long-term storage (which is why standard solutions are periodically re-standardised), this occurs far too slowly to have any measurable effect during the timescale of a single titration (oe)
(b) Answer: A positive E cell only shows the reaction is feasible, not that it is fast; with no catalyst present, the reaction has a high activation energy and is extremely slow at room temperature, so KMnO4(aq) remains usable as a titrant over the timescale of a titration, even though it very slowly decomposes over long-term storage.
(c) M1 light (particularly UV/visible light) can provide energy that catalyses (accelerates) the decomposition reaction, e.g. by a photochemical pathway (oe)
(c) A1 storing the solution in the dark minimises exposure to light, slowing this decomposition and keeping the concentration of the standard solution constant (accurate) for longer (oe)
(c) Answer: Light can catalyse (speed up) the decomposition of KMnO4(aq); storing it in the dark minimises this, keeping its concentration reliable for longer.

Question 14

(a) B1 aqueous Cu2+ solutions ([Cu(H2O)6]2+) are pale blue, meaning they strongly absorb light from the complementary (red/orange) part of the visible spectrum (oe)
(a) B1 using a red/orange filter means the light passing through the solution is strongly absorbed, giving the largest, most sensitive change in absorbance reading with concentration (oe)
(a) Answer: The blue solution strongly absorbs red/orange light (its complementary colour); using a red/orange filter means absorbance changes most sensitively with concentration.
(b) M1 concentration = absorbance / k = 0.638 / 42.5
(b) A1 0.0150 mol dm-3 cao (awrt 0.0150)
(b) B1 correct rearrangement of absorbance = k x concentration shown
(b) Answer: 0.0150 mol dm-3
(c) M1 c1 = c2 x V2 / V1 = 0.0150 x 250 / 10.0
(c) A1 0.375 mol dm-3 cao ft from (b)
(c) Answer: 0.375 mol dm-3

Chemistry: Electrode Potentials and Transition Metals - Worksheets, Questions and Revision