IB DP Chemistry Structure 2.1 The ionic model HL Paper 2- Exam Style Questions - New Syllabus
Question
The structures of sodium bromide and sodium metal are shown below.
(a) Suggest a technique that could be used to determine these structures. [1]
(b) State and describe the bonding present in the two solid structures. [2]
NaBr: ……………………………………………………..
Na: ………………………………………………………..
NaBr: ……………………………………………………..
Na: ………………………………………………………..
(c) Write the half-equations for the formation of the products at the positive electrode (anode) and the negative electrode (cathode) when molten sodium bromide is electrolysed. [2]
Positive electrode (anode): ………………………………….
Negative electrode (cathode): ………………………………
Positive electrode (anode): ………………………………….
Negative electrode (cathode): ………………………………
Data booklet (extract) for part (d): Standard electrode potentials at 298 K
| Half-reaction (reduction) | \(E^\circ/\text{V}\) |
|---|---|
| \(\tfrac{1}{2}\,\mathrm{Br_2(l)} + e^- \rightarrow \mathrm{Br^- (aq)}\) | \(+1.09\) |
| \(\mathrm{Na^+(aq)} + e^- \rightarrow \mathrm{Na(s)}\) | \(-2.71\) |
| \(\tfrac{1}{2}\,\mathrm{O_2(g)} + \mathrm{H_2O(l)} + 2e^- \rightarrow 2\,\mathrm{OH^- (aq)}\) | \(+0.40\) |
| \(\mathrm{H_2O(l)} + e^- \rightarrow \tfrac{1}{2}\,\mathrm{H_2(g)} + \mathrm{OH^- (aq)}\) | \(-0.83\) |
| \(\mathrm{Zn^{2+}(aq)} + 2e^- \rightarrow \mathrm{Zn(s)}\) | \(-0.76\) |
| \(\mathrm{Mg^{2+}(aq)} + 2e^- \rightarrow \mathrm{Mg(s)}\) | \(-2.37\) |
(d) Determine the products formed at each electrode during the electrolysis of an aqueous solution of sodium bromide. Use the data above. [2]
Positive electrode (anode): ………………………………….
Negative electrode (cathode): ………………………………
Positive electrode (anode): ………………………………….
Negative electrode (cathode): ………………………………
(e) A sodium bromide solution can be used for a salt bridge in a voltaic cell. Annotate the diagram of the magnesium (Mg) and zinc (Zn) voltaic cell (identify anode/cathode and the ions in the two half-cells). Use the data above as needed. [2]
Data (given) for parts (f)–(g) and additional data-booklet extracts
| Process / quantity | \(\Delta H^\circ / \mathrm{kJ\,mol^{-1}}\) |
|---|---|
| \(\mathrm{Na(s)} \rightarrow \mathrm{Na(g)}\) | \(+107\) |
| \(\mathrm{Br_2(l)} \rightarrow \mathrm{Br_2(g)}\) | \(+30\) |
| \(\mathrm{Br_2(g)} \rightarrow 2\,\mathrm{Br(g)}\) (bond dissociation, \(D_{\mathrm{Br-Br}}=193\)) | \(+\tfrac{1}{2}\times 193 = +96.5\) |
| Ionization of Na: \(\mathrm{Na(g)} \rightarrow \mathrm{Na^+(g)} + e^-\) | \(+496\) |
| Electron affinity of Br: \(\mathrm{Br(g)} + e^- \rightarrow \mathrm{Br^-(g)}\) | \(-325\) |
| Formation: \(\mathrm{Na(s)} + \tfrac{1}{2}\mathrm{Br_2(l)} \rightarrow \mathrm{NaBr(s)}\) | \(-361.5\) |
| Hydration: \(\Delta H^\circ_{\text{hyd}}(\mathrm{Na^+})\) | \(-424\) |
| Hydration: \(\Delta H^\circ_{\text{hyd}}(\mathrm{Br^-})\) | \(-328\) |
(f) Determine the lattice enthalpy of sodium bromide using the data given and sections 8 and 11 of the data booklet. [3]
(g) Calculate the enthalpy of solution of sodium bromide. Use your answer from (f) and the hydration data above. (If you did not obtain an answer to (f), use \(754\ \mathrm{kJ\,mol^{-1}}\), though this is not the correct answer.) [2]
(h) Predict the observations when separate aqueous \(\mathrm{I_2}\) and \(\mathrm{Cl_2}\) solutions are added to aqueous \(\mathrm{NaBr}\). Write an equation where a reaction occurs. [2]
▶️ Answer/Explanation
Markscheme (with detailed working)
(a)
Technique: X-ray crystallography. A1
[1 mark]
Technique: X-ray crystallography. A1
[1 mark]
(b)
NaBr: ionic lattice; electrostatic attraction between \(\mathrm{Na^+}\) cations and \(\mathrm{Br^-}\) anions. A1
Na (metal): metallic bonding; \(\mathrm{Na^+}\) cations in a lattice with a sea of delocalized electrons. A1
[2 marks]
NaBr: ionic lattice; electrostatic attraction between \(\mathrm{Na^+}\) cations and \(\mathrm{Br^-}\) anions. A1
Na (metal): metallic bonding; \(\mathrm{Na^+}\) cations in a lattice with a sea of delocalized electrons. A1
[2 marks]
(c)
Molten \(\mathrm{NaBr}\):
Positive electrode (anode, oxidation of bromide): \[\boxed{2\,\mathrm{Br^- (l)} \rightarrow \mathrm{Br_2(g)} + 2e^-}\] A1
Negative electrode (cathode, reduction of sodium): \[\boxed{\mathrm{Na^+ (l)} + e^- \rightarrow \mathrm{Na(l)}}\] A1
(Award full credit if correct equations are given but assigned to the opposite electrodes.)
[2 marks]
Molten \(\mathrm{NaBr}\):
Positive electrode (anode, oxidation of bromide): \[\boxed{2\,\mathrm{Br^- (l)} \rightarrow \mathrm{Br_2(g)} + 2e^-}\] A1
Negative electrode (cathode, reduction of sodium): \[\boxed{\mathrm{Na^+ (l)} + e^- \rightarrow \mathrm{Na(l)}}\] A1
(Award full credit if correct equations are given but assigned to the opposite electrodes.)
[2 marks]
(d)
Aqueous \(\mathrm{NaBr}\): choosing products using \(E^\circ\) values
• At the anode (oxidation): compare oxidation of \(\mathrm{Br^-}\) to \(\mathrm{Br_2}\) vs oxidation of water to \(\mathrm{O_2}\). From the reduction data, \(\mathrm{Br_2/Br^-}\) has \(E^\circ=+1.09\ \mathrm{V}\) while \(\tfrac{1}{2}\mathrm{O_2}/\mathrm{OH^-}\) has \(+0.40\ \mathrm{V}\). Oxidation is favored for the lower \(E^\circ\) reduction partner, so water is harder to reduce (easier to oxidize) than \(\mathrm{Br^-}\) is to reduce; therefore \(\mathrm{Br^-}\) is oxidized.
\(\Rightarrow\) Anode product: \(\boxed{\mathrm{Br_2}}\). A1
• At the cathode (reduction): compare reduction of \(\mathrm{Na^+}\) (\(-2.71\ \mathrm{V}\)) with reduction of water to \(\mathrm{H_2}\) (\(-0.83\ \mathrm{V}\)). Water has the higher \(E^\circ\) (easier reduction).
\(\Rightarrow\) Cathode product: \(\boxed{\mathrm{H_2(g)}}\). A1
[2 marks]
Aqueous \(\mathrm{NaBr}\): choosing products using \(E^\circ\) values
• At the anode (oxidation): compare oxidation of \(\mathrm{Br^-}\) to \(\mathrm{Br_2}\) vs oxidation of water to \(\mathrm{O_2}\). From the reduction data, \(\mathrm{Br_2/Br^-}\) has \(E^\circ=+1.09\ \mathrm{V}\) while \(\tfrac{1}{2}\mathrm{O_2}/\mathrm{OH^-}\) has \(+0.40\ \mathrm{V}\). Oxidation is favored for the lower \(E^\circ\) reduction partner, so water is harder to reduce (easier to oxidize) than \(\mathrm{Br^-}\) is to reduce; therefore \(\mathrm{Br^-}\) is oxidized.
\(\Rightarrow\) Anode product: \(\boxed{\mathrm{Br_2}}\). A1
• At the cathode (reduction): compare reduction of \(\mathrm{Na^+}\) (\(-2.71\ \mathrm{V}\)) with reduction of water to \(\mathrm{H_2}\) (\(-0.83\ \mathrm{V}\)). Water has the higher \(E^\circ\) (easier reduction).
\(\Rightarrow\) Cathode product: \(\boxed{\mathrm{H_2(g)}}\). A1
[2 marks]
(e)
Voltaic cell annotations (Mg | Mg\(^{2+}\) || NaBr(aq) || Zn\(^{2+}\) | Zn):
• Anode (oxidation): \(\boxed{\text{Mg electrode}}\), reaction \(\mathrm{Mg(s)} \rightarrow \mathrm{Mg^{2+}(aq)} + 2e^-\). Indicate \(\mathrm{Mg^{2+}(aq)}\) in the left half-cell. A1
• Cathode (reduction): \(\boxed{\text{Zn electrode}}\), reaction \(\mathrm{Zn^{2+}(aq)} + 2e^- \rightarrow \mathrm{Zn(s)}\). Indicate \(\mathrm{Zn^{2+}(aq)}\) in the right half-cell. A1
• Electrons flow externally from Mg (anode) → Zn (cathode). The NaBr salt bridge provides \( \mathrm{Na^+} \) (to cathode) and \( \mathrm{Br^-} \) (to anode) to maintain charge balance.
[2 marks]
Voltaic cell annotations (Mg | Mg\(^{2+}\) || NaBr(aq) || Zn\(^{2+}\) | Zn):
• Anode (oxidation): \(\boxed{\text{Mg electrode}}\), reaction \(\mathrm{Mg(s)} \rightarrow \mathrm{Mg^{2+}(aq)} + 2e^-\). Indicate \(\mathrm{Mg^{2+}(aq)}\) in the left half-cell. A1
• Cathode (reduction): \(\boxed{\text{Zn electrode}}\), reaction \(\mathrm{Zn^{2+}(aq)} + 2e^- \rightarrow \mathrm{Zn(s)}\). Indicate \(\mathrm{Zn^{2+}(aq)}\) in the right half-cell. A1
• Electrons flow externally from Mg (anode) → Zn (cathode). The NaBr salt bridge provides \( \mathrm{Na^+} \) (to cathode) and \( \mathrm{Br^-} \) (to anode) to maintain charge balance.
[2 marks]
(f)
Lattice enthalpy \(\Delta H^\circ_{\text{lattice}}(\mathrm{NaBr})\) by a Born–Haber cycle
Write the enthalpy of formation in terms of steps (taking \(+\) for endothermic, \(-\) for exothermic):
\[ \Delta H^\circ_f(\mathrm{NaBr}) = \underbrace{107}_{\mathrm{Na(s)\to Na(g)}}+ \underbrace{\tfrac{1}{2}(30)}_{\mathrm{Br_2(l)\to \tfrac{1}{2}Br_2(g)}=15}+ \underbrace{\tfrac{1}{2}(193)}_{\mathrm{Br_2(g)\to 2Br(g)}=96.5}+ \underbrace{496}_{\mathrm{IE_1(Na)}}+ \underbrace{(-325)}_{\mathrm{EA(Br)}}- \Delta H^\circ_{\text{lattice}} \] Rearranging for the lattice enthalpy: \[ \Delta H^\circ_{\text{lattice}}= 107+15+96.5+496-325-\Delta H^\circ_f(\mathrm{NaBr}). \] With \(\Delta H^\circ_f(\mathrm{NaBr})=-361.5\ \mathrm{kJ\,mol^{-1}}\): \[ \Delta H^\circ_{\text{lattice}}= 107+15+96.5+496-325+361.5 = \boxed{751\ \mathrm{kJ\,mol^{-1}}}. \] M1 M1 A1
[3 marks]
Lattice enthalpy \(\Delta H^\circ_{\text{lattice}}(\mathrm{NaBr})\) by a Born–Haber cycle
Write the enthalpy of formation in terms of steps (taking \(+\) for endothermic, \(-\) for exothermic):
\[ \Delta H^\circ_f(\mathrm{NaBr}) = \underbrace{107}_{\mathrm{Na(s)\to Na(g)}}+ \underbrace{\tfrac{1}{2}(30)}_{\mathrm{Br_2(l)\to \tfrac{1}{2}Br_2(g)}=15}+ \underbrace{\tfrac{1}{2}(193)}_{\mathrm{Br_2(g)\to 2Br(g)}=96.5}+ \underbrace{496}_{\mathrm{IE_1(Na)}}+ \underbrace{(-325)}_{\mathrm{EA(Br)}}- \Delta H^\circ_{\text{lattice}} \] Rearranging for the lattice enthalpy: \[ \Delta H^\circ_{\text{lattice}}= 107+15+96.5+496-325-\Delta H^\circ_f(\mathrm{NaBr}). \] With \(\Delta H^\circ_f(\mathrm{NaBr})=-361.5\ \mathrm{kJ\,mol^{-1}}\): \[ \Delta H^\circ_{\text{lattice}}= 107+15+96.5+496-325+361.5 = \boxed{751\ \mathrm{kJ\,mol^{-1}}}. \] M1 M1 A1
[3 marks]
(g)
Enthalpy of solution
\[ \Delta H^\circ_{\text{solution}}(\mathrm{NaBr}) = \Delta H^\circ_{\text{lattice}} + \Delta H^\circ_{\text{hyd}}(\mathrm{Na^+}) + \Delta H^\circ_{\text{hyd}}(\mathrm{Br^-}). \] Using the values above: \[ \Delta H^\circ_{\text{solution}} = 751 + (-424) + (-328) = \boxed{-1\ \mathrm{kJ\,mol^{-1}}}. \] M1 A1
[2 marks]
Enthalpy of solution
\[ \Delta H^\circ_{\text{solution}}(\mathrm{NaBr}) = \Delta H^\circ_{\text{lattice}} + \Delta H^\circ_{\text{hyd}}(\mathrm{Na^+}) + \Delta H^\circ_{\text{hyd}}(\mathrm{Br^-}). \] Using the values above: \[ \Delta H^\circ_{\text{solution}} = 751 + (-424) + (-328) = \boxed{-1\ \mathrm{kJ\,mol^{-1}}}. \] M1 A1
[2 marks]
(h)
• With \(\mathbf{I_2(aq)}\): No reaction; iodine is less oxidizing than bromine (cannot oxidize \(\mathrm{Br^-}\)). A1
• With \(\mathbf{Cl_2(aq)}\): Brown/orange \(\mathrm{Br_2}\) produced; chlorine displaces bromide: \[ \boxed{\mathrm{Cl_2(aq)} + 2\,\mathrm{Br^-(aq)} \rightarrow \mathrm{Br_2(aq)} + 2\,\mathrm{Cl^-(aq)}} \] A1
[2 marks]
• With \(\mathbf{I_2(aq)}\): No reaction; iodine is less oxidizing than bromine (cannot oxidize \(\mathrm{Br^-}\)). A1
• With \(\mathbf{Cl_2(aq)}\): Brown/orange \(\mathrm{Br_2}\) produced; chlorine displaces bromide: \[ \boxed{\mathrm{Cl_2(aq)} + 2\,\mathrm{Br^-(aq)} \rightarrow \mathrm{Br_2(aq)} + 2\,\mathrm{Cl^-(aq)}} \] A1
[2 marks]
Total Marks: 16