2.3. Formulas
A subsection of Chemistry, 9701, through 2. Atoms, molecules and stoichiometry
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Data should be selected from Table3.1 in order to answer some parts of this question. Table 3.1 electrode reaction E o / V Mg2+ + 2e– Mg –2.38 Mn2+ + 2e– Mn –1.18 Mn3+ + e– Mn2+ +1.49 MnO2 + 4H+ + 2e– Mn2+ + 2H2O +1.23 MnO4 – + e– MnO4 2– +0.56 MnO4 – + 4H+ + 3e– MnO2 + 2H2O +1.67 MnO4 – + 8H+ + 5e– Mn2+ + 4H2O +1.52 An electrochemical cell can be constructed from a Mg2+ / Mg half-cell and a MnO4 – / Mn2+ half‑cell. The standard cell potential of this cell can be calculated using the standard electrode potentials of the two half-cells. Define standard electrode potential. Include details of the standard conditions used. Complete the diagram below to show an electrochemical cell constructed from a Mg2+ / Mg half‑cell and a MnO4 – / Mn2+ half-cell. Label your diagram.  Use a positive (+) sign and a negative (–) sign to identify the polarity of each of the two electrodes in your diagram. Use an arrow and the symbol ‘e’ to show the direction of electron flow in the external circuit. Calculate the standard cell potential, , of this cell.  = V Construct an equation for the cell reaction. Predict how the cell reaction will change, if at all, when the solution in the Mg2+ / Mg half‑cell is diluted by the addition of a large volume of water. Explain your answer. A molten magnesium salt is electrolysed for 15.0minutes by a constant current. 4.75×1022 magnesium atoms are produced at the cathode. Calculate the value of the current used.  current = A 
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THEORY
2022
Paper 4, Variant 2
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Some electrode potentials are shown in Table 3.1. Table 3.1 electrode reaction E o– / V V2+ + 2e– V –1.20 V3+ + e– V2+ –0.26 VO2+ + 2H+ + e– V3+ + H2O +0.34 VO2 + + 2H+ + e– VO2+ + H2O +1.00 Fe2+ + 2e– Fe –0.44 Fe3+ + 3e– Fe –0.04 Fe3+ + e– Fe2+ +0.77 2H+ + 2e– H2 0.00 Cl O– + H2O + 2e– Cl – + 2OH– +0.89 Complete the diagram to show a standard hydrogen electrode. Label your diagram. Identify all substances. You do not need to state standard conditions. An electrochemical cell is set up using an Fe3+ / Fe2+ electrode and a standard hydrogen electrode. Identify the positive electrode in the electrochemical cell and the direction of electron flow in the external circuit. positive electrode Electrons flow from the electrode to the electrode. The vanadium-containing species in the electrode reactions given in Table 3.1 are V, V2+, V3+, VO2+ and VO2 +. Identify one vanadium-containing species that does not react with Fe2+ ions under standard conditions. Use data from Table 3.1 to explain your answer. Identify all the vanadium-containing species that will react with Fe2+ ions under standard conditions. Write an equation for one of the possible reactions identified in . Another electrochemical cell is set up using an Fe3+ / Fe2+ electrode and an alkaline Cl O– / Cl – electrode. The concentration of Fe3+ is 1000 times greater than the concentration of Fe2+ in the Fe3+ / Fe2+ electrode. All other conditions are standard. Use the Nernst equation to calculate the E value of the Fe3+ / Fe2+ electrode. Show your working. E = V Write an equation for the reaction that occurs in the cell, under these conditions. Another electrochemical cell is set up using an Fe2+ / Fe electrode and an alkaline Cl O– / Cl – electrode under standard conditions. Calculate the value of ΔG o– for the cell. ΔGo – = kJ mol–1 A solution of iron(sulfate, FeSO4is electrolysed with iron electrodes. Under the conditions used, no gas is evolved at the cathode. A current of 0.640 A is passed for 17.0 minutes. The mass of the cathode increases by 0.185 g. Use these results to calculate an experimental value for the Avogadro constant, L. Show your working. L = mol–1 Iron(chloride, FeCl2, is oxidised by chlorine to form iron(chloride, FeCl 3, under standard conditions. 2FeCl 2+ Cl 22FeCl 3ΔH o– = –128 kJ mol–1 Table 3.2 species S o– / J K–1 mol–1 Cl 2223 FeCl 2120 FeCl 3142 Use Table 3.2 and other data to calculate the Gibbs free energy change, ΔG o–, for this reaction. Show your working. ΔG o– = kJ mol–1 Predict whether this reaction becomes more or less feasible at a higher temperature. Explain your answer. The reaction becomes feasible. explanation
9701_w23_qp_42
THEORY
2023
Paper 4, Variant 2
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The structure of the polydentate ligand, EDTA4–, is shown in . NCH2CH2N CCH2 O O O CCH2 O CH2C O O O CH2C O The stability constants, at 298 K, of five octahedral complexes are given in Table 4.1. Table 4.1 complex Kstab [Cu(EDTA)]2– 6.31 × 1019 [Cr(EDTA)]2– 1.00 × 1013 [Cr(EDTA)]– 1.00 × 1024 [Fe(EDTA)]2– 2.00 × 1014 [Fe(EDTA)] – 1.26 × 1025 Define stability constant. Calculate the oxidation states of Cu in [Cu(EDTA)]2– and Cr in [Cr(EDTA)]–. Cu Cr Deduce the number of lone pairs donated by each EDTA4– ligand in a single [Fe(EDTA)]2– complex ion. Identify the most stable complex in Table 4.1. Explain your choice. In a solution at equilibrium at 298 K, [[Cu(H2O)6]2+] = 3.00 × 10–10 mol dm–3 and [EDTA4–] = 5.00 × 10–12 mol dm–3. Use the expression for Kstab to calculate the concentration of [Cu(EDTA)]2– in this solution. Show your working. [[Cu(EDTA)]2–] = mol dm–3 A solution of [Cu(EDTA)]2– ions is pale blue while a solution of [Cu(NH3)4(H2O)2]2+ ions is deep blue. Explain this difference in colour.
9701_w23_qp_42
THEORY
2023
Paper 4, Variant 2
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Transition metal atoms and transition metal ions form complexes by combining with ligands. Explain why transition elements form complex ions. Co2+ ions form complex ion G. Each G ion contains two Co2+ ions, both of which are octahedrally coordinated. Each G ion contains one O2 molecule, which donates one pair of electrons to each Co2+ ion, and one NH2 – ion, which donates one pair of electrons to each Co2+ ion. The remaining ligands are NH3 molecules. Deduce the formula of complex ion G. Include its overall charge. formula of G The d-orbitals of the Co2+ ions present in complex ion G are split. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level in each Co2+ ion. number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level Co2+ ions form a different complex ion, M. Each M ion contains two Co2+ ions, both of which are octahedrally coordinated, but the ligands are different from the ligands in G. Explain why G and M have different colours. Cadmium forms complex ion X, [Cd(NH3)4]2+. When a solution containing CN– ions is added to an aqueous solution of X, a ligand exchange reaction takes place, forming complex ion Y. Y contains no NH3 ligands and no H2O ligands. Y is in a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium do not change in this reaction. Write an ionic equation for this reaction, using the formulae of the complex ions. Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl – ions. When NaCl is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, Kstab. smallest value of Kstab largest value of Kstab Ethanedioate ions, C2O4 2–, form complexes with transition element ions. The concentration of C2O4 2– ions can be found by reaction with acidified Cr2O7 2– ions. C2O4 2– ions are protonated and form HOOCCOOH molecules which are oxidised by Cr2O7 2–. The half-equations are shown. Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O 2CO2 + 2H+ + 2e– HOOCCOOH Construct an equation for the reaction between acidified Cr2O7 2– and HOOCCOOH. A 25.0 cm3 sample of a solution of Na2C2O4 reacts with exactly 16.20 cm3 of an acidified solution of 0.0500 mol dm–3 K2Cr2O7. Calculate the concentration of the solution of Na2C2O4. = mol dm–3
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THEORY
2024
Paper 4, Variant 1
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Transition metal atoms and transition metal ions form complexes by combining with species called ligands. When NaOHis added to an aqueous solution containing [Co(H2O)6]2+ a precipitation reaction occurs accompanied by a colour change. In this reaction two of the water ligands each lose one H+ ion. The H+ ions are gained by OH– ions from the NaOH. State the colour change seen in this precipitation reaction. from to Complete the ionic equation for this precipitation reaction. [Co(H2O)6]2+ + + This precipitation reaction can also be described as a different type of reaction. Name this type of reaction. L is an uncharged tridentate ligand. L donates three lone pairs to a metal atom or ion. Cobalt forms an octahedral complex ion, E, with L. Complex ion E has a 2+ charge. Give the formula of E. Identify the oxidation state of cobalt in E. The d-orbitals of the cobalt atom or ion present in E are split in energy. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level. number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level Define the term non-degenerate d-orbitals. The mineral chromite contains a compound which has the formula FeCrnO4. The oxidation state of iron in FeCrnO4 is +2. A sample of 4.18 g of FeCrnO4 is dissolved in an excess of sulfuric acid. The resulting solution is made up to 250 cm3. This is solution F. All the Fe2+ ions in 25.0 cm3 of solution F are oxidised to Fe3+ ions by exactly 18.7 cm3 of 0.0200 mol dm–3 KMnO4. One MnO4 – ion reacts with five Fe2+ ions. Assume no other oxidation reaction occurs. Write an equation for the reaction of Fe2+ ions with MnO4 – ions in acid solution. Calculate the number of moles of Fe2+ ions in 25.0 cm3 of solution F. number of moles of Fe2+ ions = Calculate the Mr of FeCrnO4 and use your answer to deduce the value of n. Mr of FeCrnO4 = value of n =
9701_w24_qp_42
THEORY
2024
Paper 4, Variant 2
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Transition metal atoms and transition metal ions form complexes by combining with ligands. Explain why transition elements form complex ions. Co2+ ions form complex ion G. Each G ion contains two Co2+ ions, both of which are octahedrally coordinated. Each G ion contains one O2 molecule, which donates one pair of electrons to each Co2+ ion, and one NH2 – ion, which donates one pair of electrons to each Co2+ ion. The remaining ligands are NH3 molecules. Deduce the formula of complex ion G. Include its overall charge. formula of G The d-orbitals of the Co2+ ions present in complex ion G are split. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level in each Co2+ ion. number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level Co2+ ions form a different complex ion, M. Each M ion contains two Co2+ ions, both of which are octahedrally coordinated, but the ligands are different from the ligands in G. Explain why G and M have different colours. Cadmium forms complex ion X, [Cd(NH3)4]2+. When a solution containing CN– ions is added to an aqueous solution of X, a ligand exchange reaction takes place, forming complex ion Y. Y contains no NH3 ligands and no H2O ligands. Y is in a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium do not change in this reaction. Write an ionic equation for this reaction, using the formulae of the complex ions. Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl – ions. When NaCl is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, Kstab. smallest value of Kstab largest value of Kstab Ethanedioate ions, C2O4 2–, form complexes with transition element ions. The concentration of C2O4 2– ions can be found by reaction with acidified Cr2O7 2– ions. C2O4 2– ions are protonated and form HOOCCOOH molecules which are oxidised by Cr2O7 2–. The half-equations are shown. Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O 2CO2 + 2H+ + 2e– HOOCCOOH Construct an equation for the reaction between acidified Cr2O7 2– and HOOCCOOH. A 25.0 cm3 sample of a solution of Na2C2O4 reacts with exactly 16.20 cm3 of an acidified solution of 0.0500 mol dm–3 K2Cr2O7. Calculate the concentration of the solution of Na2C2O4. = mol dm–3
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THEORY
2024
Paper 4, Variant 3
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An aqueous solution of cobalt(chloride is a pink colour due to the presence of [Co(H2O)6]2+ complex ions. Complete Table6.1 to describe what is observed when the named reagent is added to an aqueous solution of cobalt(chloride. Table 6.1 reagent colour of cobalt-containing product state of cobalt-containing product NaOHan excess of conc. HCl  Write an equation for the reaction between [Co(H2O)6]2+ ions and NaOH. Write an equation for the reaction between [Co(H2O)6]2+ ions and an excess of conc.HCl. Define stability constant. Write an expression for the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion. Kstab =  Give the units of the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion.  units = The numerical value of the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion is 7.7×104. In an aqueous solution the concentration of the [Co(NH3)6]2+ complex ion is 0.0740 mol dm–3 and the concentration of NH3 is 0.480 mol dm–3 at equilibrium. Calculate the concentration of [Co(H2O)6]2+ in this solution.  concentration = mol dm–3 
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THEORY
2022
Paper 4, Variant 2
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