Electrochemistry

2(a)(i) M1 the only number extracted: 762, 1560, 496 2 M2 correct multiplier, other four numbers used and calculation to the answer –272 = +416 + ½(496) + 762 + 1560 –141 + 798 + ΔH lattice ∴ ΔH = –3915 (kJ mol–1) ecf lattice 2(a)(ii) 20 × [0.9(+2) + 0.1(+3)] –2x = 0 ∴ x = 21 1 2(a)(iii) • FeO more exothermic/more negative 2 • Fe2+ has smaller radius/higher charge density (also same charge) • greater attraction/ stronger ionic bonds (between Fe2+ and O2–) All three for two marks © UCLES 2021 Page 5 of 13 2(b)(i) • Fe2+ reduced to Fe OR oxid no. Fe +2 → 0 1 • Fe2+ oxidised to Fe3+ (in Fe O ) OR oxid no. Fe +2 → +3 3 4 BOTH bullets required 2(b)(ii) 2O2– → O + 4e– 1 2 2(b)(iii) M1: coulombs and correct use of ÷ 96500 3 M2: correct use of 3 and 8 M3: correct use of 55.8 and answer M1: Q = It = 50 × 6 × 602 OR 1.08 × 106 C AND no. of faraday = 1.08 × 106 ÷ 96500 OR 11.2 / 11.19 mol e– M2: Fe2+ + 2Fe3+ + 8e– → 3Fe ∴ moles of Fe = 3 / 8 × M1 = 4.20 mol Fe ecf M3: mass of Fe = 55.8 × M2 = 234.2 g ecf 3sf min 2(c)(i) Any one of: 1 small size / compact, low mass, high voltage OWTTE 2(c)(ii) Li from +1 to +1 1 Fe from +3 to +2 2(c)(iii) LiC + FePO → LiFePO + 6C 1 6 4 4 Question Answer Marks

3(a)(i) (an element) forming stable ion / ions / compound(s) / oxidation state(s) 1 AND with partially filled / incomplete AND d orbitals / d subshell / d shell 3(a)(ii) (melting point) higher 1 AND (density) higher 3(b)(i) M1: emf / potential difference / difference in electrode potential between two half-cells / two electrodes (in a cell) 2 M2: (all solutions being) 1 mol dm–3 AND either 1 atm OR 298 K © UCLES 2021 Page 8 of 17 3(b)(ii) salt bridge, voltmeter, Cu(s), Cu2+(aq), Pt(s), Fe2+ and Fe3+(aq) 3 two for one mark, four for two marks, six for three marks 3(c)(i) M1: 2I– + 2Fe3+ → I + 2Fe2+ 2 2 M2: S O 2– + 2Fe2+ → 2SO 2– + 2Fe3+ 2 8 4 3(c)(ii) M1: I / I– +0.54 V AND Fe3+ / Fe2+ + 0.77 V AND [Fe(CN) ]3– / [Fe(CN) ]4– +0.36 V 2 2 6 6 M2: Eo of I / I– is more positive / greater than Eo of [Fe(CN) ]3– / [Fe(CN) ]4– 2 6 6 OR Eo = –0.18 V so no reaction occurs cell OR Eo of Fe3+ / Fe2+ is more positive / greater than Eo of I / I– 2 OR Eo = 0.23 V so reaction occurs [1] cell 3(d)(i) S O 2– and tartrate ions are both negatively charged / both reactants same charge 1 2 8 AND so repel each other OR have a high E a 3(d)(ii) C H O 2– + 2H O ⇌ 2CO + 2HCO – + 6H+ + 6e– 1 4 4 6 2 2 2 © UCLES 2021 Page 9 of 17 3(e)(i) 3 reagent structure of organic product type of reaction an excess of LiAlH reduction 4 an excess of CH COCl condensation 3 M1: product with LiAlH 4 M2: product with CH COCl 3 M3: both types of reaction 3(e)(ii) 1 OR dianion of tartrate with two cations present © UCLES 2021 Page 10 of 17

3(a)(i) M1: voltage of an electrode / a half-cell compared to / connected to (standard) hydrogen electrode / half-cell 2 M2: (at concentration of) 1 mol / dm3 AND (pressure of) 1 atm / 101 kPa (or in Pa) AND 298 K / 25°C 3(a)(ii) 1 3(a)(iii) 3 OR M1 / M2: two 3D isomers of [Ru(en) ]3+ 3 M3: optical / enantiomerism 3(b)(i) Eo = 1.07 – 0.80 = (+)0.27 V 1 cell AND direction of electron flow = Ag+ / Ag to Br / Br– 2 3(b)(ii) M1: Eo 3rd box ticked 2 cell M2: [Ag+] decreases AND so (Ag+ / Ag) equilibrium shifts to the left OR [Ag+] decreases AND E for (Ag+ / Ag) becomes less positive / more negative © UCLES 2021 Page 7 of 16 3(c)(i) (a species) that uses / shares a lone pair of electrons to form a coordinate bond to a metal atom / ion 1 3(c)(ii) K = [Ag(S O ) 3–] [Br–] / [S O 2–]2 1 c 2 3 2 2 3 3(c)(iii) M1: K = K × K = 15.7 2 c stab sp M2: 1 OR none / no units 3(d) M1: highest [Ag(CN) ]– [Ag(S O ) ]3– [Ag(NH ) ]+ lowest 2 2 2 3 2 3 2 M2: K [Ag(CN) ]– is highest / [Ag(CN) ]– is the most stable stab 2 2 OR higher K forms the more stable complex stab Question Answer Marks

3(a)(i) (an element) forming stable ion / ions / compound(s) / oxidation state(s) 1 AND with partially filled / incomplete AND d orbitals / d subshell / d shell 3(a)(ii) (melting point) higher 1 AND (density) higher 3(b)(i) M1: emf / potential difference / difference in electrode potential between two half-cells / two electrodes (in a cell) 2 M2: (all solutions being) 1 mol dm–3 AND either 1 atm OR 298 K © UCLES 2021 Page 8 of 17 3(b)(ii) salt bridge, voltmeter, Cu(s), Cu2+(aq), Pt(s), Fe2+ and Fe3+(aq) 3 two for one mark, four for two marks, six for three marks 3(c)(i) M1: 2I– + 2Fe3+ → I + 2Fe2+ 2 2 M2: S O 2– + 2Fe2+ → 2SO 2– + 2Fe3+ 2 8 4 3(c)(ii) M1: I / I– +0.54 V AND Fe3+ / Fe2+ + 0.77 V AND [Fe(CN) ]3– / [Fe(CN) ]4– +0.36 V 2 2 6 6 M2: Eo of I / I– is more positive / greater than Eo of [Fe(CN) ]3– / [Fe(CN) ]4– 2 6 6 OR Eo = –0.18 V so no reaction occurs cell OR Eo of Fe3+ / Fe2+ is more positive / greater than Eo of I / I– 2 OR Eo = 0.23 V so reaction occurs [1] cell 3(d)(i) S O 2– and tartrate ions are both negatively charged / both reactants same charge 1 2 8 AND so repel each other OR have a high E a 3(d)(ii) C H O 2– + 2H O ⇌ 2CO + 2HCO – + 6H+ + 6e– 1 4 4 6 2 2 2 © UCLES 2021 Page 9 of 17 3(e)(i) 3 reagent structure of organic product type of reaction an excess of LiAlH reduction 4 an excess of CH COCl condensation 3 M1: product with LiAlH 4 M2: product with CH COCl 3 M3: both types of reaction 3(e)(ii) 1 OR dianion of tartrate with two cations present © UCLES 2021 Page 10 of 17

1(a) moles of H = 462 / 24 000 = 0.01925 [1] 2 2 molecules of H = 0.019 × 6.02 × 1023 2 = 1.16 × 1022 (1.1×1022 / 1.2 × 1022) [1] min 2sf ecf M1 1(b) number of electrons = 1.16 × 1022 × 2 1 = 2.32 × 1022 [1] min 2sf ecf 1a 1(c) Q = 2.32 ×1022 × 1.6 × 10–19 1 = 3.71 × 103 [1] min 2sf ecf 1b 1(d) x = 3.71 × 103 / (14 × 60) 1 = 4.4 (A) [1] min 2sf ecf 1c 1(e)(i) ΔS = 262 + 205 – 140 = (+) 327 (J K–1 mol–1) [1] 1 1(e)(ii) ΔG = ΔH – TΔS OR use of Gibbs equation [1] 2 ΔG = 572 – (298 × 0.327) = (+)474.6 (kJ mol–1) [1] min 3sf ecf 1e(i) 1(e)(iii) becomes more feasible / spontaneous 1 as TΔS is more positive / –TΔS becomes more negative   Question Answer Marks

3(a) 0.351 / 24 = 0.015 (mol) [1] 2 0.015 × 6.02 × 1023 = 9.0 × 1021 / 8.8 × 1021 [1] 3(b) 1.76 × 1022 / 1.8 × 1022 [1] 1 3(c) 2817 / 2816 / 2820 / 2800 C [1] 1 3(d) 15 / 15.1 / 15.05 / 15.15 minutes [1] 1 3(e)(i) –20 [1] 1 © UCLES 2021 Page 6 of 12 3(e)(ii) states / uses correct Gibbs equation [1] 2 answer = 190 / 191 / 190.0 [1] 3(e)(iii) Becomes less feasible / less spontaneous / 1 AND because ΔS is negative / TΔS becomes more negative / –TΔS becomes more positive [1] Question Answer Marks

1(a) moles of H = 462 / 24 000 = 0.01925 [1] 2 2 molecules of H = 0.019 × 6.02 × 1023 2 = 1.16 × 1022 (1.1×1022 / 1.2 × 1022) [1] min 2sf ecf M1 1(b) number of electrons = 1.16 × 1022 × 2 1 = 2.32 × 1022 [1] min 2sf ecf 1a 1(c) Q = 2.32 ×1022 × 1.6 × 10–19 1 = 3.71 × 103 [1] min 2sf ecf 1b 1(d) x = 3.71 × 103 / (14 × 60) 1 = 4.4 (A) [1] min 2sf ecf 1c 1(e)(i) ΔS = 262 + 205 – 140 = (+) 327 (J K–1 mol–1) [1] 1 1(e)(ii) ΔG = ΔH – TΔS OR use of Gibbs equation [1] 2 ΔG = 572 – (298 × 0.327) = (+)474.6 (kJ mol–1) [1] min 3sf ecf 1e(i) 1(e)(iii) becomes more feasible / spontaneous 1 as TΔS is more positive / –TΔS becomes more negative   Question Answer Marks

3(a)(i) evidence of tangent drawn at t = 40 s and calculation of gradient 1 = 0.000170 (mol dm–3 s–1) min 2sf 3(a)(ii) M1 evidence of construction lines and calculation of two t 2 1/2 OR evidence of construction lines and times for halving of concentration M2 deduction: constant half-life / constant time between halving of concentration  1st order 3(b) M1 two half-lives in 320 s so t = 160 s 2 1 / 2 M2 k = 0.693 / 160 = 0.00433 s–1 ecf min 2sf 3(c)(i) voltage / EMF / potential difference when a half-cell is connected to a (standard) hydrogen electrode under standard 1 conditions 3(c)(ii) ions move (from the salt bridge) to maintain charge balance / complete the circuit 1 © UCLES 2022 Page 7 of 15 3(c)(iii) 3 Pt, H (g), good delivery system, 298 K, 1 atm 2 Pt, Cr O 2-(aq), H+(aq) / Cr3+(aq), 1 mol dm–3 seen once, voltmeter 2 7 three for one mark, six for two marks, nine for three marks 3(c)(iv) SHE labelled negative AND arrow in the external circuit moving away from this electrode 1 3(d)(i) M1 Cr O 2– + 3CH CHO + 8H+ ⇌ 2Cr3+ + 3CH COOH + 4H O 2 2 7 3 3 2 ALLOW Cr O 2– + 3CH CHO + 5H+ ⇌ 2Cr3+ + 3CH COO– + 4H O 2 7 3 3 2 M1 E o = +2.27 (V) cell 3(d)(ii) M1 Go = -nFE o 2 cell M2 Go = –4  96 500  2.01 = –775 860 J mol–1 Go = –776 kJ mol–1 min 3sf © UCLES 2022 Page 8 of 15

5(a) 3 substance liberated substance liberated electrolyte at the anode at the cathode PbBr (l) Br / bromine Pb / lead 2 2 concentrated NaCl(aq) Cl / chlorine H / hydrogen 2 2 Cu(NO ) (aq) O / oxygen (+ H O) Cu / copper 3 2 2 2 two for one mark, four for two marks, six for three marks © UCLES 2022 Page 9 of 16 5(b)(i) F = Le OR F is directly proportional to L 1 5(b)(ii) number of Cu2+ formed = 0.35 / 63.5 = 5.51  10–3 4 Q = I  t = 0.60  30  60 = 1080 C number of electrons = 1080 / 1.610-19 = 6.75  1021 ecf number of Cu2+ ions = 6.75  1021 / 2 = 3.375  1021 ecf number of Cu2+ ions per mole (L) = 3.375  1021 / 5.51  10-3 = 6.12  1023 ecf min 2sf all five points for four marks ALLOW valid alternate calculations of L Question Answer Marks

3(a)(i) evidence of tangent drawn at t = 40 s and calculation of gradient 1 = 0.000170 (mol dm–3 s–1) min 2sf 3(a)(ii) M1 evidence of construction lines and calculation of two t 2 1/2 OR evidence of construction lines and times for halving of concentration M2 deduction: constant half-life / constant time between halving of concentration  1st order 3(b) M1 two half-lives in 320 s so t = 160 s 2 1 / 2 M2 k = 0.693 / 160 = 0.00433 s–1 ecf min 2sf 3(c)(i) voltage / EMF / potential difference when a half-cell is connected to a (standard) hydrogen electrode under standard 1 conditions 3(c)(ii) ions move (from the salt bridge) to maintain charge balance / complete the circuit 1 © UCLES 2022 Page 7 of 15 3(c)(iii) 3 Pt, H (g), good delivery system, 298 K, 1 atm 2 Pt, Cr O 2-(aq), H+(aq) / Cr3+(aq), 1 mol dm–3 seen once, voltmeter 2 7 three for one mark, six for two marks, nine for three marks 3(c)(iv) SHE labelled negative AND arrow in the external circuit moving away from this electrode 1 3(d)(i) M1 Cr O 2– + 3CH CHO + 8H+ ⇌ 2Cr3+ + 3CH COOH + 4H O 2 2 7 3 3 2 ALLOW Cr O 2– + 3CH CHO + 5H+ ⇌ 2Cr3+ + 3CH COO– + 4H O 2 7 3 3 2 M1 E o = +2.27 (V) cell 3(d)(ii) M1 Go = -nFE o 2 cell M2 Go = –4  96 500  2.01 = –775 860 J mol–1 Go = –776 kJ mol–1 min 3sf © UCLES 2022 Page 8 of 15

3(a)(i) (Ion (Sn2+/Sn4+) concentration) 1 mol dm–3 1 AND 298 K (25 °C) 3(a)(ii) • both half-cells have Pt or C electrode 3 • Sn2+ / Sn4+ AND H+ solutions • feasible gas delivery system • H label 2 • V/voltmeter AND correct circuit AND salt bridge touching solution • salt bridge labelled 3(a)(iii) no (reaction) AND 2 both Eo values (Sn2+ / Sn) –0.14 and (Cl / Cl –) +1.36 [1] 2 E is –1.5 V / E is negative cell cell OR Eo of Sn4+ / Sn2+ is more negative/smaller than Eo of Cl / Cl –[1] 2 3(a)(iv) Sn2+→Sn4+ and VO2+→V3+ [1] 2 Sn2++ 2VO2+ + 4H+ → Sn4+ + 2V3+ +2H O [1] 2 3(b) moles of Sn = 2.95 / 118.7 = 0.0249 moles 2 moles of Al (is 2 / 3 moles of Sn) = 0.0166 moles [1] mass of Al = 0.0166  27 = 0.447 / 0.448 g to 3sf [1] ecf Question Answer Marks © UCLES 2022 Page 7 of 16

3(a)(i) the voltage produced by a half-cell compared with a standard hydrogen electrode [1] 2 1 mol dm–3, 298 K, 1 atm [1] 3(a)(ii) Mg wire and Pt wire [1] 3 voltmeter, salt bridge, complete circuit [1] solutes Mg2+ and MnO –, Mn2+, H+ [1] 4 3(a)(iii) Mg is minus, Pt is plus 1 arrow points towards MnO – / Mn2+ half-cell 4 3(a)(iv) 3.90 V 1 3(a)(v) 5Mg + 2MnO – +16H+ → 5Mg2+ + 2Mn2+ + 8H O 1 4 2 3(a)(vi) no change and 1 dilution will make Mg2+ / Mg potential even more negative 3(b) either: 2 4.75  1022  2  1.60  10–19 = 15 200 C OR 2  96 500  (4.75  1022) / 6.02  1023 = 15 228 C [1] 15 200 / (15  60) = 16.9 A OR 15 228 / (15  60) = 16.9 A [1] © UCLES 2022 Page 6 of 11

3(a)(i) (Ion (Sn2+/Sn4+) concentration) 1 mol dm–3 1 AND 298 K (25 °C) 3(a)(ii) • both half-cells have Pt or C electrode 3 • Sn2+ / Sn4+ AND H+ solutions • feasible gas delivery system • H label 2 • V/voltmeter AND correct circuit AND salt bridge touching solution • salt bridge labelled 3(a)(iii) no (reaction) AND 2 both Eo values (Sn2+ / Sn) –0.14 and (Cl / Cl –) +1.36 [1] 2 E is –1.5 V / E is negative cell cell OR Eo of Sn4+ / Sn2+ is more negative/smaller than Eo of Cl / Cl –[1] 2 3(a)(iv) Sn2+→Sn4+ and VO2+→V3+ [1] 2 Sn2++ 2VO2+ + 4H+ → Sn4+ + 2V3+ +2H O [1] 2 3(b) moles of Sn = 2.95 / 118.7 = 0.0249 moles 2 moles of Al (is 2 / 3 moles of Sn) = 0.0166 moles [1] mass of Al = 0.0166  27 = 0.447 / 0.448 g to 3sf [1] ecf Question Answer Marks © UCLES 2022 Page 7 of 16

2(a) H PO 1 3 2 2(b)(i) electrode potential E would become more positive / less negative (than E⦵) 1 lower [H PO –] AND shifts equilibrium to the right-hand side 1 2 2 2(b)(ii) +1.57 – 0.74 = (+)0.83 (V) 1 © UCLES 2023 Page 6 of 17 2(b)(iii) 3 • Pt(s) / C / graphite • Cr(s) • H+(aq) / any strong acid • Cr3+(aq) • H (g) 2 • voltmeter / V • salt bridge labelled • conditions of 1 atm AND 1 mol dm–3 / 1 M / 1 mol / dm3 • other liquid level (salt bridge) and wire to electrode any three [1] any six [2] all nine [3] 2(b)(iv) Pt electrode positive AND flow of electrons anticlockwise (to the SHE) 1 2(b)(v) H PO – + 3OH– + Ni2+ → HPO 2– + 2H O + Ni 1 2 2 3 2 2(c)(i) (6.4 / 24000) ÷ 60 = 4.44  10–6 (mol dm-3 s–1) min 2sf 1 2(c)(ii) [H PO –] doubles / 2 from experiments 1 to 2 ORA 1 2 2 volume of H produced doubles / 2 (∴ first order wrt [H PO –]) 2 2 2 [H PO –]  3 and [OH–]  ½ from experiments 1 to 3 ORA 1 2 2 volume of H produced falls to ¾ original 2 (if first order wrt [H PO –] then must be second order wrt [OH–]) 2 2 ALLOW input data into rate equation and show k is the same k =2.8  10–6 / k = 6.7  10–2 (1 / 15) / k = 4 for all experiments [2] © UCLES 2023 Page 7 of 17 2(c)(iii) mol–2 dm6 s–1 1 2(c)(iv) t = 0.693 / 8.25  10–5 = 8400 (s) OR t = In2 / 8.25  10–5 = 8401.8 (s) 1 ½ ½ 2(c)(v) (k ) increases (with temperature) 1 1 2(d) • reactants adsorb (to surface of catalyst) 2 • bonds (in reactant) weaken • (reaction occurs and the) products are desorbed OR reaction occurs and substances are desorbed any two [1] all three [2] Question Answer Marks

9(a) potential difference / voltage between the two half-cells / two electrodes (in a cell) under standard conditions 1 9(b) 3 salt bridge voltmeter complete circuit [wires/salt bridge in/touching both solutions] Cl Cl – good delivery system (no arrow required) 2 Pt Pt Fe2+ and Fe3+ Any three [1], any six [2], all nine [3] 9(c) M1 Go ) = – nFE o OR –2  96500  0.59 2 cell  M2 Go = –2  96500  0.59 = –113870 J mol-1 Go = –114 kJ mol-1 min 3sf ecf 9(d)(i) E = 0.77 + (0.059 / 1) log (1/0.15) use of z = 1 1 E = 0.82 V 9(d)(ii) E = 0.59 + 0.77 – (answer to (d)(i)) = 0.54 V ecf 1 cell © UCLES 2023 Page 19 of 19

3(a)(i) hydrogen, delivery system, H+, platinum, [1] 1 3(a)(ii) iron . . . . . hydrogen . . . . . iron [1] 1 3(b)(i) (for specified V2+, V3+ or VO2+) Eo is more positive than / above –0.44 AND more negative than / below 0.77 V [1] 1 3(b)(ii) V and VO + [1] 1 2 3(b)(iii) V + Fe2+ → V2+ + Fe OR 1 VO + + 2H+ + Fe2+ → VO2+ + H O + Fe3+ [1] 2 2 3(c)(i) Nernst: E = 0.77 + (0.059 / z)log[ox] / [red] [1] 2 0.947 [1] 3(c)(ii) 2Fe3+ + Cl– + 2OH– → 2Fe2+ + ClO– + H O [1] 1 2 3(d) Eo = 1.33 V [1] 3 cell Go = –nEo F [1] cell –257 [1] © UCLES 2023 Page 6 of 14 3(e) 0.64  17  60 = 653 / 652.8 Coulombs [1] 3 652.8 ÷ 1.6x10–19 = 4.08  1021 (number of electrons) 4.08  1021 ÷ 2 = 2.04  1021 (number of atoms Fe) [1] 0.185 ÷ 55.8 = 3.31  10–3 (number of moles Fe atoms 2.04  1021 ÷ 3.31  10–3 = L = 6.153  1023 [1] 3(f)(i) S = –179 [1] 3 G = H –TS [1] –74.7 [1] 3(f)(ii) less AND G becomes more positive [1] 1 Question Answer Marks

2(a)(i) 2LiNO → Li O + 2NO + ½O 1 3 2 2 2 2(a)(ii) M1 radius / size of cation / M+ increases OR charge density of ion decreases 2 M2 less polarisation / distortion of anion / nitrate ion / NO – / less weakening of NO bond 3 2(b) M1 M2 any two for one mark or all four for two marks: 3  mol total MnO – = 0.125  0.0500 = 6.25  10–3 4  mol Fe2+ = 0.0400  0.0225 = 9.00  10–4  mol unreacted MnO – = 9.00  10–4 ÷ 5 = 1.80  10–4 ecf 4  mol reacted MnO – = 6.25  10–3 - 1.80  10–4 = 6.07  10–3 ecf 4 M3 mol NO – = 2.5  6.07  10–3 = 1.5175  10–2 2 conc NaNO = 4  1.5175  10–2 = 6.07-6.08  10–2 mol dm–3 ecf min 2sf 2 2(c)(i) 3MnO 2– + 4H+ → 2MnO – + MnO + 2H O 2 4 4 2 2 M1 MnO 2– as a reactant and MnO – + MnO products identified 4 4 2 M2 correct equation 2(c)(ii) (E ) decreases / becomes less positive AND as [H+] decreases AND equilibrium shifts to the left 1 cell OR in alkali the E = 0.60 – 0.56 = 0.04 V (working required) cell © Cambridge University Press & Assessment 2024 Page 6 of 15

5(a)(i) (Ar) 3d10 AND (Ar) 3d9 1 5(a)(ii) similar energy of 3d and 4s subshells 1 5(b)(i) pale blue precipitate 1 5(b)(ii) [Cu(H O) ]2+ + 2OH– → Cu(OH) (H O) + 2H O 1 2 6 2 2 4 2 5(c)(i) deep blue solution 1 5(c)(ii) [Cu(H O) ]2+ + 4NH → [Cu(NH ) (H O) ]2+ + 4H O 1 2 6 3 3 4 2 2 2 5(c)(iii) ligand exchange 1 5(d)(i) M1: E = Eo +(0.059 / z) log(ox/red) 2 M2: E = 1.52 + (0.059 / 5)  log (0.002) © Cambridge University Press & Assessment 2025 Page 9 of 16 5(d)(ii) 1.15 1 5(d)(iii) 5Cu + 2MnO – + 16H+ → 5Cu2+ + 2Mn2+ + 8H O 1 4 2 5(d)(iv) Cu2+ / Cu, Cu2+ / Cu, MnO – / Mn2+ / platinum 1 4 5(e) M1: moles of copper = 0.0120 mol 3 M2: coulombs required = 2322 C M3: current required = 0.129 A Question Answer Marks

2(a)(i) A = [Co(H O) ]2+ pink 4 2 6 B = [Co(H O) (OH) ] / Co(OH) blue 2 4 2 2 C = [Co(NH ) ]2+ straw / yellow-brown 3 6 B = precipitation / deprotonation / acid-base C = ligand exchange Any two [1] any four [2] any six [3] all eight [4] 2(a)(ii) concentrated HCl 1 2(b)(i) • five (3d) orbitals (lines, boxes) in the isolated Co2+ ion of same energy 2 • splitting three higher and two lower d orbitals • energy of all d orbitals in the complex clearly higher than all d orbitals in isolated ion any two [1] all three [2] © Cambridge University Press & Assessment 2025 Page 7 of 17 2(b)(ii) 1 2(c) M1: H O acting as oxidising agent with Mn(OH) / Mn2+ 4 2 2 2 M2: Mn2+ + H O → MnO + 2H+ AND E o = +0.56 2 2 2 cell OR Mn(OH) + H O → MnO + 2H O AND E o = +0.92 2 2 2 2 2 cell M3: H O acting as reducing agent with MnO 2 2 2 M4: MnO + 2H+ + H O → Mn2+ + 2H O + O AND Eo = +0.54 2 2 2 2 2 cell 2(d)(i) colourless to pale pink 1 2(d)(ii) • moles MnO – = 0.0500  18.60 / 1000 = 9.30  10-4 in 25.0 cm3 4 4 • moles Fe2+ = 5  9.30  10-4 = 4.65  10-3 in 25.0 cm3 • moles Fe2+ = 6 x 4.65  10-3 = 2.79  10-2in 150.0 cm3 M1 / M2: any two of the above bullets [1] all three [2] OR • moles MnO – = 0.0500  18.60 / 1000 = 9.30  10-4 in 25.0 cm3 4 • moles MnO – = 6  9.30  10-4 = 5.58  10-3 in 150.0 cm3 4 • moles Fe2+ = 5  5.58  10-3 = 2.79  10-2in 150.0 cm3 M1 / M2: any two of the above bullets [1] all three [2] M3: mass Fe = 55.8  2.79  10-2 = 1.55682 (g) in two tablets M4: mass Fe = 0.5  1.55682 = 0.77841 (g) in one tablet AND mass Fe = 7.7841  10–2  1000 = 778.4 © Cambridge University Press & Assessment 2025 Page 8 of 17

5(a)(i) (Ar) 3d10 AND (Ar) 3d9 1 5(a)(ii) similar energy of 3d and 4s subshells 1 5(b)(i) pale blue precipitate 1 5(b)(ii) [Cu(H O) ]2+ + 2OH– → Cu(OH) (H O) + 2H O 1 2 6 2 2 4 2 5(c)(i) deep blue solution 1 5(c)(ii) [Cu(H O) ]2+ + 4NH → [Cu(NH ) (H O) ]2+ + 4H O 1 2 6 3 3 4 2 2 2 5(c)(iii) ligand exchange 1 5(d)(i) M1: E = Eo +(0.059 / z) log(ox/red) 2 M2: E = 1.52 + (0.059 / 5)  log (0.002) © Cambridge University Press & Assessment 2025 Page 9 of 16 5(d)(ii) 1.15 1 5(d)(iii) 5Cu + 2MnO – + 16H+ → 5Cu2+ + 2Mn2+ + 8H O 1 4 2 5(d)(iv) Cu2+ / Cu, Cu2+ / Cu, MnO – / Mn2+ / platinum 1 4 5(e) M1: moles of copper = 0.0120 mol 3 M2: coulombs required = 2322 C M3: current required = 0.129 A Question Answer Marks

4(a) M1: potential difference / voltage / EMF AND between two half-cells / two electrodes (in a cell) 2 M2: (at concentration of) 1 mol dm–3 AND (pressure of) 1 atm / 101 kPa AND (temperature of) 298 K / 25 °C © Cambridge University Press & Assessment 2025 Page 9 of 19 4(b)(i) 3 M1: salt bridge labelled AND voltmeter / V AND wire to electrode and liquid level shown [complete circuit] M2: Cu(s) AND Cu2+(aq) M3: Zn(s) AND Zn2+(aq) 4(b)(ii) ions AND electrons 1 4(b)(iii) M1: Nernst equation E = Eo + (0.059/z) log(Zn2+/Zn) 2 M2: E = –0.76 + (0.059 / 2) log(0.25) = –0.778 V OR E = –0.76 + (8.31  298) / (96 500  2) ln(0.25) = –0.778 V min 2sf 4(c)(i) (from) +4 (to) +3 1 4(c)(ii) Zn + 2MnO 2 → ZnO + Mn 2 O 3 1 4(c)(iii) Eo (MnO 2 / Mn 2 O 3 ) = 1.47 – 1.28 = (+)0.19 V 1 © Cambridge University Press & Assessment 2025 Page 10 of 19

5(a)(i) Co2+ = [Ar] 3d7 AND Co3+ = [Ar] 3d6 1 5(a)(ii) has vacant d orbitals that are energetically accessible 1 5(a)(iii) • pink 2 • blue • aqueous Any two [1], all three [2] 5(a)(iv) [Co(H 2 O) 6 ]2+ + 4HCl → [CoCl4 ]2– + 6H 2 O + 4H+ OR [Co(H 2 O) 6 ]2+ + 4Cl– → [CoCl4 ]2– + 6H 2 O 1 © Cambridge University Press & Assessment 2025 Page 9 of 15 5(a)(v) ligand exchange 1 5(a)(vi) [Co(H O) ]2+ + 2OH– → Co(OH) (H O) + 2H O OR 1 2 6 2 2 4 2 [Co(H O) ]2+ + 2OH– → Co(OH) + 6H O 2 6 2 2 5(b)(i) [oxidised] 2 M1: Nernst equation E = Eo + (0.059 / z) × log( ) [reduced] M2: E = –0.28 + (0.059 / 2) × log(0.02) 5(b)(ii) 1.66 V 1 5(b)(iii) 3Co + Cr O 2– + 14H+ → 3Co2+ + 2Cr3+ + 7H O 1 2 7 2 5(b)(iv) Co2+ / Co, Co2+ / Co, Cr O 2– / Cr3+ 1 2 7 5(c) M1: moles of cobalt = 0.547 / 58.9 = 0.00929 3 M2: coulombs required = 0.00929  96500  2 = 1792 M3: time required = 1792 / 0.50 = 3584.8 s = 59.7 minutes 3sf Question Answer Marks

4(a) M1: potential difference / voltage / EMF AND between two half-cells / two electrodes (in a cell) 2 M2: (at concentration of) 1 mol dm–3 AND (pressure of) 1 atm / 101 kPa AND (temperature of) 298 K / 25 °C © Cambridge University Press & Assessment 2025 Page 9 of 19 4(b)(i) 3 M1: salt bridge labelled AND voltmeter / V AND wire to electrode and liquid level shown [complete circuit] M2: Cu(s) AND Cu2+(aq) M3: Zn(s) AND Zn2+(aq) 4(b)(ii) ions AND electrons 1 4(b)(iii) M1: Nernst equation E = Eo + (0.059/z) log(Zn2+/Zn) 2 M2: E = –0.76 + (0.059 / 2) log(0.25) = –0.778 V OR E = –0.76 + (8.31  298) / (96 500  2) ln(0.25) = –0.778 V min 2sf 4(c)(i) (from) +4 (to) +3 1 4(c)(ii) Zn + 2MnO 2 → ZnO + Mn 2 O 3 1 4(c)(iii) Eo (MnO 2 / Mn 2 O 3 ) = 1.47 – 1.28 = (+)0.19 V 1 © Cambridge University Press & Assessment 2025 Page 10 of 19

4(a) Go = –nF Eo 1 A Go = – nF E . cell cell 4(b) Cu2+ + 2V2+ → Cu + 2V3+ 2 CON cancellable species / electrons If ⇌ read forwards (+) 0.60 I 0.6 + not needed but – is CON. 4(c) M1 Cu 2 or in words if unambiguous. M2 V2+ 4(d) M1 Cl / chlorine 2 CON if both oxidised and reduced form 2 given. M2 V / vanadium CON if both oxidised and reduced form given. 4(e) All candidates get three marks 3 © Cambridge University Press & Assessment 2026 Page 11 of 21 Trace ID: PC-F9EA0E463D Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

5(a) M1 emf / potential difference / voltage 2 A half-cell for electrode. and between / compared to and a standard hydrogen electrode and another electrode CON hydrogen cell. M2 under standard conditions of • 1 atm / 101 kPa / 100 kPa • 298 K / 25 °C, • (all solutions being) 1M / 1 mol dm–3 u / c 5(b)(i) M1M2 3 A M3 two calculated E cell • Zn + 2Fe3+ → Zn2+ + 2Fe2+ • bullet 1 E = 1.53 (V) cell • NO – + 10H+ + 8Fe2+ → NH + + 3H O + 8Fe3+ • bullet 2 E = +0.10 (V) 3 4 2 cell • NO – + 10H+ + 4Zn → NH + + 3H O + 4Zn2+ • bullet 3 E = +1.63 (V). 3 4 2 cell ● ✓✓ M3 E is positive (to be feasible) I M3 use of ‘greater’ for more positive. cell or Eo of Fe3+ / Fe2+ and NO – / NH + is more positive 3 4 or Eo of redox system 1 and 2 are more positive ora u / c 5(b)(ii) high E for the reaction OWTTE 1 I rate is too slow. a I non-standard conditions are used. © Cambridge University Press & Assessment 2026 Page 16 of 25 Trace ID: PC-9D76E91B48 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(b)(iii) 3 I inclusion negative ion with SHE like Cl–, NO –, 3 SO 2–. 4 CON bullet 2 & 9 battery in the circuit. A bullet 6 / 7 graphite / C for Pt. State symbols not required. • labelled salt bridge • voltmeter / V / potentiometer • H 2 • good delivery system (of H ) 2 • H+(aq) / HCl(aq) solution labelled [state symbol not required] • Pt electrode • Pt electrode • NO and NH + and H+ solution labelled 3 4 or HNO and NH NO 3 4 3 • complete circuit (salt bridge touches the solution & wires from V go to the electrodes) ● ● ✓ ● ● ✓● ● ✓ 5(c)  Go = -nE F 2 A M1 use of Go = –nEF and use of –645.6. cell –645.6  1000 = –3  E  96 500 M2 E = (+)2.23 (V) min 2sf A ecf [1] from cell Go = nE F → E = –2.23. cell cell © Cambridge University Press & Assessment 2026 Page 17 of 25 Trace ID: PC-9D76E91B48 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

4(a) Go = –nF Eo 1 A Go = – nF E . cell cell 4(b) Cu2+ + 2V2+ → Cu + 2V3+ 2 CON cancellable species / electrons If ⇌ read forwards (+) 0.60 I 0.6 + not needed but – is CON. 4(c) M1 Cu 2 or in words if unambiguous. M2 V2+ 4(d) M1 Cl / chlorine 2 CON if both oxidised and reduced form 2 given. M2 V / vanadium CON if both oxidised and reduced form given. 4(e) All candidates get three marks 3 © Cambridge University Press & Assessment 2026 Page 11 of 21 Trace ID: PC-F2CDA313A1 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

1(a) M1 Oxygen / O 2 I ‘O’ alone. 2 M2 4OH– → 2H O + O + 4e– I state symbols. 2 2 A multiples, use of ‘⇌’. A 4OH– – 4e– → 2H O + O . 2 2 1(b)(i) M1 2.4  12  60 or 1728 (C) 2 A 0.00896 or 0.008955. M2 1728 / (2  96 500) = 0.00895 / 0.0090 mol min 2sf 1.4945  10–4 (no  60) [1]. 0.0179 [1] (no ÷ 2). or Correct answer [2]. 1728 0.009 [1] (1sf). =0.00897 min 2sf ecf 2 ( 1.610−196.0221023) 1(b)(ii) © Cambridge University Press & Assessment 2026 Page 7 of 21 V = n R p T → 1 ( b ) ( i) 1  0 8 1 . 0 3 0 1 0  6 5 3 = 4 . 8 1  1 0 − 4 m 3 1 ECF from 1(b)(i), e.g. if mol in 1(b)(i) = 0.0179, volume = 0.962. 4.81×10–4 m3 × 1000 = 0.481 (dm3) 0.482 if 0.00897 mol. answer must be 3sf 0.484 if 0.0090 mol. 0.962 if 0.0179 mol. A use of their unrounded previous value, e.g. 0.962 if 0.018 mol (i.e. they are using 0.0179). correct answer [1]. Trace ID: PC-0F078E40FE Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 1(c) Full expression: M1 © Cambridge University Press & Assessment 2026 Page 8 of 21 1 . 6 3 6  0 1 0 0 0 1 −  9  2 3 8 . 5 1 = 6 . 0 8  1 0 2 3 ( m o −l 1 ) 2 I 6  1023, 6.1  1023 as not 3 sf 6.08  1023 scores [2]. A – ve sign if seen in working, but 36 000 ÷ 1.6  10–19 or 2.25  1023 electrons / atoms Na –6.08  1023 scores max [1]. M2 ans ÷ (8.51 / 23) = 6.08  1023 must be 3sf ECF from use of incorrect A for M2. r M for NaCl used instead of Na gives or (using F = –Le) r L = 1.55  1024 [1] M1 8.51 ÷ 23.0 = 0.370 moles of electrons / atoms Na and use of atomic number instead of A r gives 36 000 ÷ 0.370 or 97 297 (value for F) L = 2.91  1023 [1] M2 ans ÷ 1.6  10–19 = 6.08  1023 must be 3sf  rather than ÷ A r gives 8.33  1022 [1]. Question Answer Marks Guidance