IB DP Chemistry Reactivity 3.2 Electron transfer reactions HL Paper 2- Exam Style Questions - New Syllabus
Question
Potassium, K, and potassium chloride, KCl, both form lattice structures in the solid state.
(a) Predict, with a reason, the electrical conductivity of K (s) and KCl (s). [2]
(b) Two electrolytic cells are connected in series. All electrodes are inert.
(i) State the half equation for the reaction occurring at each electrode in Cell 2. [2]
Use these Data Booklet values: E°(K+/K) = −2.93 V; E°(2H2O + 2e− → H2 + 2OH−) = −0.83 V; E°(Cl2 + 2e− → 2Cl−) = +1.36 V; E°(O2 + 4H+ + 4e− → 2H2O) = +1.23 V; E°(O2 + 2H2O + 4e− → 4OH−) = +0.40 V.
(ii) Identify the product at the anode in Cell 1. [1]
(iii) Determine the mole ratio of non-ionic products formed at the cathode (negative electrode) in Cell 1 and Cell 2. [1]
(c) State the number of each type of subatomic particle in the potassium ion, \( {}^{41}_{19}\mathrm{K}^+ \). [1]
(d) Outline the evidence that the electrons in a potassium atom occupy four main energy levels. [2]
(e) Potassium reacts with water to produce potassium hydroxide.
\[ \mathrm{2K\,(s) + 2H_2O\,(l) \rightarrow 2KOH\,(aq) + H_2\,(g)} \] (i) Calculate the enthalpy of reaction, in kJ mol\(^{-1}\), when 1 mol of potassium reacts with water. \(\Delta H^\circ_\mathrm{f}\)(KOH, aq) \(=\,-481.8\ \mathrm{kJ\,mol^{-1}}\). [3]
\[ \mathrm{2K\,(s) + 2H_2O\,(l) \rightarrow 2KOH\,(aq) + H_2\,(g)} \] (i) Calculate the enthalpy of reaction, in kJ mol\(^{-1}\), when 1 mol of potassium reacts with water. \(\Delta H^\circ_\mathrm{f}\)(KOH, aq) \(=\,-481.8\ \mathrm{kJ\,mol^{-1}}\). [3]
Data booklet value used : \(\Delta H^\circ_\mathrm{f}\big(\mathrm{H_2O(l)}\big) = -285.8\ \mathrm{kJ\,mol^{-1}}\).
(ii) Describe the difference between the reactions of sodium and potassium with water. [1]
(iii) Demonstrate, with an equation, the acid–base nature of K2O (s). [1]
▶️ Answer/Explanation
Markscheme (with explanations)
(a)
K(s): conductor — metallic lattice with delocalized electrons free to move. A1
KCl(s): non-conductor — ions fixed in the ionic lattice; no mobile charge carriers. A1
[2 marks]
K(s): conductor — metallic lattice with delocalized electrons free to move. A1
KCl(s): non-conductor — ions fixed in the ionic lattice; no mobile charge carriers. A1
[2 marks]
(b)(i) (Cell 2 is dilute KCl(aq); inert electrodes.)
Using the given E° values: reduction at the cathode favors water over K+ (−0.83 V > −2.93 V), and oxidation at the anode favors water over Cl− in dilute solution (+1.23 or +0.40 V vs +1.36 V). Thus:
Anode (oxidation): \[ \mathrm{2\,OH^-(aq) \rightarrow \tfrac{1}{2}\,O_2(g) + H_2O(l) + 2e^-} \quad\text{(from E°(O}_2\!/\mathrm{OH^-})=+0.40\ \mathrm{V)} \] (or \( \mathrm{H_2O(l) \rightarrow \tfrac{1}{2}\,O_2(g) + 2H^+(aq) + 2e^-} \) based on E°(O2/H+) = +1.23 V).
Cathode (reduction): \[ \mathrm{H_2O(l) + e^- \rightarrow \tfrac{1}{2}\,H_2(g) + OH^-(aq)} \quad\text{(E° = −0.83 V)} \] (K+ + e− → K(s) is not favored; E° = −2.93 V). [2 marks]
Using the given E° values: reduction at the cathode favors water over K+ (−0.83 V > −2.93 V), and oxidation at the anode favors water over Cl− in dilute solution (+1.23 or +0.40 V vs +1.36 V). Thus:
Anode (oxidation): \[ \mathrm{2\,OH^-(aq) \rightarrow \tfrac{1}{2}\,O_2(g) + H_2O(l) + 2e^-} \quad\text{(from E°(O}_2\!/\mathrm{OH^-})=+0.40\ \mathrm{V)} \] (or \( \mathrm{H_2O(l) \rightarrow \tfrac{1}{2}\,O_2(g) + 2H^+(aq) + 2e^-} \) based on E°(O2/H+) = +1.23 V).
Cathode (reduction): \[ \mathrm{H_2O(l) + e^- \rightarrow \tfrac{1}{2}\,H_2(g) + OH^-(aq)} \quad\text{(E° = −0.83 V)} \] (K+ + e− → K(s) is not favored; E° = −2.93 V). [2 marks]
(b)(ii)
Product at anode in Cell 1 (CuSO4(aq), inert anode): \( \mathrm{O_2} \). [1 mark]
Product at anode in Cell 1 (CuSO4(aq), inert anode): \( \mathrm{O_2} \). [1 mark]
(b)(iii)
Non-ionic products at the cathode in both cells are \( \mathrm{H_2} \); mole ratio \(=\ \mathbf{1:1} \). [1 mark]
Non-ionic products at the cathode in both cells are \( \mathrm{H_2} \); mole ratio \(=\ \mathbf{1:1} \). [1 mark]
(c)
\( {}^{41}_{19}\mathrm{K}^+ \): Protons \(=19\); Electrons \(=18\); Neutrons \(=22\). [1 mark]
\( {}^{41}_{19}\mathrm{K}^+ \): Protons \(=19\); Electrons \(=18\); Neutrons \(=22\). [1 mark]
(d)
Successive ionization energies show three large jumps after removing the 1st, 9th, and 17th electrons, evidencing four main energy levels (shells) in K. [2 marks]
Successive ionization energies show three large jumps after removing the 1st, 9th, and 17th electrons, evidencing four main energy levels (shells) in K. [2 marks]
(e)(i) Using enthalpies of formation (\(\Delta H^\circ_\mathrm{f}\)):
Reaction: \(\mathrm{2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2(g)}\).
\[ \Delta H_\text{rxn}^\circ = \big[2\,\Delta H^\circ_\mathrm{f}(\mathrm{KOH(aq)}) + 0\big] – \big[0 + 2\,\Delta H^\circ_\mathrm{f}(\mathrm{H_2O(l)})\big] = 2(-481.8) – 2(-285.8) = -392.0\ \mathrm{kJ} \] For 1 mol of K (half the stoichiometric amount of K): \[ \boxed{\Delta H = -196.0\ \mathrm{kJ\,mol^{-1}}}. \] [3 marks]
Reaction: \(\mathrm{2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2(g)}\).
\[ \Delta H_\text{rxn}^\circ = \big[2\,\Delta H^\circ_\mathrm{f}(\mathrm{KOH(aq)}) + 0\big] – \big[0 + 2\,\Delta H^\circ_\mathrm{f}(\mathrm{H_2O(l)})\big] = 2(-481.8) – 2(-285.8) = -392.0\ \mathrm{kJ} \] For 1 mol of K (half the stoichiometric amount of K): \[ \boxed{\Delta H = -196.0\ \mathrm{kJ\,mol^{-1}}}. \] [3 marks]
(e)(ii)
Sodium reacts less vigorously/slower; potassium reacts more vigorously and often ignites (lilac flame). [1 mark]
Sodium reacts less vigorously/slower; potassium reacts more vigorously and often ignites (lilac flame). [1 mark]
(e)(iii)
\[ \mathrm{K_2O(s) + H_2O(l) \rightarrow 2K^+(aq) + 2OH^-(aq)} \quad(\text{or } \mathrm{K_2O + H_2O \rightarrow 2KOH(aq)}) \] [1 mark]
\[ \mathrm{K_2O(s) + H_2O(l) \rightarrow 2K^+(aq) + 2OH^-(aq)} \quad(\text{or } \mathrm{K_2O + H_2O \rightarrow 2KOH(aq)}) \] [1 mark]
Total Marks: 13