CHAPTER 12 Practice exercises 12.1 2Al(s) + 3Cl 2 (g) 2AlCl 3 (aq) Aluminium is oxidised and is therefore the reducing agent. Chlorine is reduced and is therefore the oxidising agent. 12.3 First the oxidation numbers of all atoms must be determined. Cl 2 + 2NaClO 2 2ClO 2 + 2NaCl Reactants: Products: Cl 0 Cl +4 O 2 Na +1 Cl +3 Na +1 O 2 Cl 1 The oxidation numbers for O and Na do not change. However, the oxidation numbers for chlorine atoms can change. There is no simple way of working out which chlorines are reduced and which are oxidised in this reaction. One possibility is that the Cl in Cl 2 ends up as the Cl in NaCl and the Cl in NaClO 2 ends up as the Cl in ClO 2. In this case Cl 2 is reduced and is the oxidising agent and NaClO 2 is oxidised and is the reducing agent. 12.5 3Sn 2+ + 16H + + 2TcO 4 2Tc 4+ + 8H 2 O + 3Sn 4+ 12.7 anode: Mg(s) Mg 2+ (aq) + 2e cathode: Fe 2+ (aq) + 2e Fe(s) cell notation: Mg(s) Mg 2+ (aq) Fe 2+ (aq) Fe(s)
12.9 2Cl + Br 2 Cl 2 + 2 Br 12.11 G o = 115.8 kj 12.13 Cell potential = 1.04 V 12.15 Sn(s) Review questions 12.1 (a) Oxidation is the loss of one or more electrons. Reduction is the gain of one or more electrons. (b) The oxidation number decreases in a reduction and increases in an oxidation. 12.3 The number of electrons involved in both the reduction and the oxidation must be the same; only those electrons that come from the reductant and go to the oxidant are involved. No electrons from external or uninvolved sources are involved and there cannot be any electrons left unaccounted for at the end. An oxidising agent is the species that is reduced or gains electrons in an oxidation/reduction reaction. A reducing agent is the species that is oxidised or loses electrons in an oxidation/ reduction reaction. 12.5 Nitrogen is reduced. This change in oxidation number requires that nitrogen gains five electrons. 12.7 These equations are not balanced since the charge is different on each side of the arrow. Balancing gives: (a) 2Ag + Fe 2+ 2Ag + + Fe (b) 2Cr 3+ + 3Zn 2Cr + 3Zn 2+ 12.9 (a) 2Ag 2Ag + + 2e is oxidation Fe 2+ + 2e Fe is reduction (b) 3Zn 3Zn 2+ + 3e is oxidation 2Cr 3+ + 3e 2Cr is reduction 12.11 A galvanic cell is one in which a spontaneous redox reaction occurs producing electricity. A half-cell consists of either the cathode or the anode, together with the accompanying electrolyte.
12.13 These must be kept separate otherwise Ag + ions are reduced directly by Cu metal and no external current is produced. 12.15 Electrolytic conduction is the flow of positive and negative ions. Metallic conduction is the flow of electrons. 12.17 Magnesium metal is oxidised at the anode and copper ions are reduced at the cathode to copper metal: Mg(s) + Cu 2+ (aq) Mg 2+ (aq) + Cu(s). The anode half-cell is a magnesium wire dipping into a solution of Mg 2+ ions and the cathode half-cell is a copper wire dipping into a solution of Cu 2+ ions. Additionally a salt bridge connects the two half-cell compartments. 12.19 Silver is reduced because it has a more positive E o than iron. The standard cell notation is Ag(s) Ag + (aq) Fe 3+ (aq) Fe(s). 12.21 A potential is the measure of the electron flow through an external circuit. The unit of potential is the volt. 12.23 The cell potential for the anode half-cell reaction is subtracted from the cell potential for the cathode half-reaction: E o cell = E o substance reduced E o substance oxidised E o cell = E o reduction E o oxidation 12.25 No, emf measurements require a complete circuit for current flow and there must be two half-cells connected with a salt bridge.
12.27 A positive reduction potential indicates that the system is more easily reduced than the hydrogen ion. Conversely, a negative reduction potential indicates that the system comprising the half-cell is less easily reduced than the hydrogen ion. 12.29 The negative terminal of the voltmeter must be connected to the anode in order to obtain correct readings of the voltage that is generated by the cell. 12.31 The strongest oxidising agent is H + (aq). 12.33 The half-reactions at the top of table 12.1 have the greatest tendency to occur as a reduction, the half-reactions at the bottom of the table have the greatest tendency to occur as an oxidation. 12.35 G o = nf E o cell 12.37 The Nernst equation: RT E cell = E o cell nf R = 8.314 J mol 1 K 1 T = 25 C = 298 K F = 96 485 C mol 1 RT F ln Q = 0.025 67 J C 1 In x = 2.303 log x 0.02567 J E cell = E o cell 1 1n Q n C 0.0592 J E cell = E o cell 1 log Q n C If the system is at equilibrium, Q = K c and E cell = 0 0.0592 J 0 = E o cell log K 1 n C c 0.0592 J E o cell = log K 1 n C c
12.39 We begin by separating the reaction into its two half-reactions in order to obtain the value of n: Pb(s) + SO 4 2 (aq) PbSO 4 (s) + 2e PbO 2 (s) + 4H 4 (aq) + SO 4 2 (aq) + 2e PbSO 4 (s) + 2H 2 O( l ) Thus, n = 2: E cell = E o cell 0.0592 n log Q E cell = 2.05 0.0592 1 log 4 2 2 + 2 H SO4 12.41 Electrons are negatively charged and move towards the electrode where reduction occurs. This is the cathode. 12.43 oxidation: 2H 2 O(l) 4H + (aq) + 4e + O 2 (g) reduction: 2H 2 O(l) + 2e H 2 (g) + 2OH (aq) 12.45 One Faraday (F) is equivalent to one mole of electrons. Also, one Faraday is equal to 96 500 Coulombs and a Coulomb is equivalent to an Ampere second: 1 F = 96 485 C and 1 C = 1 A s 12.47 Copper has a larger atomic mass than iron and therefore, the copper deposits a greater mass of metal. Both metals are in the same +2 oxidation state. 12.49 anode: PbSO 4 (s) + 2H 2 O(l) 2e + PbO 2 (s) + 4H + (aq) + SO 2 4 (aq) cathode: PbSO 4 (s) + 2e Pb(s) + SO 2 4 (aq) 12.51 anode: Zn(s) + 2OH (aq) ZnO(s) + H 2 O(l) + 2e cathode: 2MnO 2 (s) + H 2 O(l) + 2e Mn 2 O 3 (s) + 2OH (aq) 12.53 The hydrogen is absorbed by the alloy, Mg 2 Ni. The electrolyte is KOH. 12.55 The electrode materials in a typical lithium cell are graphite and cobalt oxide. When the cell is charged, Li + ions migrate to the graphite. When the cell discharges, the Li + ions migrate back to the cobalt oxide and the electrons move through the external circuit to maintain charge balance.
12.57 Rust is iron oxide hydroxide, FeO(OH), that is formed by the reaction of iron with oxygen in the presence of water. Rusting of iron is a three-step process. Firstly, iron hydroxide, Fe(OH) 2, is formed by the reaction: 2Fe + O 2 + 2H 2 O 2Fe(OH) 2 Fe(OH) 2 is then oxidised to Fe(OH) 3 : 4Fe(OH) 2 + O 2 + 2H 2 O 4Fe(OH) 3 And finally, FeO(OH) is formed by dehydration of Fe(OH) 3 : Fe(OH) 3 FeO(OH) + H 2 O 12.59 Passivation is a phenomenon shown by some elements, for example, aluminium and zinc, which are stable to air, although their reduction potentials are strongly negative. These metals form an impermeable layer of metal oxide that is stable over a wide ph range, which covers a corrosive metal and protects it from oxidation. Review problems 12.61 (a) substance reduced (and oxidising agent): HNO 3 substance oxidised (and reducing agent): H 3 AsO 3 (b) substance reduced (and oxidising agent): HOCl substance oxidised (and reducing agent): NaI (c) substance reduced (and oxidising agent): KMnO 4 substance oxidised (and reducing agent): H 2 C 2 O 4 (d) substance reduced (and oxidising agent): H 2 SO 4 substance oxidised (and reducing agent): Al (e) substance reduced (and oxidising agent): H 2 SO 4 substance oxidised (and reducing agent): Cu (f) substance reduced (and oxidising agent): HNO 3 substance oxidised (and reducing agent): SO 2 (g) substance reduced (and oxidising agent): H 2 SO 4 substance oxidised (and reducing agent): Zn (h) substance reduced (and oxidising agent): HNO 3 substance oxidised (and reducing agent): I 2 (i) substance reduced (and oxidising agent): NaClO 4 Substance oxidised (and reducing agent): HCl 12.63 (a) Na +1, H +1, P +5, O 2 (b) MnO 4 does not exist, however, for empirical MnO 4, Mn +8, O 2 (c) Na +1, S = + 2.5, O 2 (d) Cl + 3, F 1 (e) O 2, Cl +1 (f) Cl +3, O 2 (g) Cl + 5, O 2 (h) Cl +7, O 2 (i) Ca +2, V +5, O 2 (j) Sn +4, Cl 1 (k) Mn +6, O 2
(l) Mn +4, O 2 (m) Pb +2, S 2 (n) Ti +4, Cl 1 (o) Sr +2, I +5, O 2 (p) Cr +3, S 2 (q) F 1, O +0.5 (r) H +1, O 0, F 1 (s) Cl +4, O 2 (t) F 1, O +1 12.65 In the forward reaction, Cl 2 undergoes disproportion. One chlorine atom is reduced to Cl and the other is oxidised to Cl + (in HOCl). In the reverse reaction, HOCl is the oxidising agent and Cl is the reducing agent. 12.67 (a) BiO 3 + 6H + + 2e Bi 3+ + 3H 2 O 2 Reduction of BiO 3 (b) Pb 2+ + 2H 2 O PbO 2 + 4H + + 2e Oxidation of Pb 2+ (c) NO 3 + 10H + + 8e NH + 4 + 3H 2 O 2 Reduction of BiO 3 (d) 6H 2 O + Cl 2 2ClO 3 + 12H + + 10e Oxidation of Cl 2 12.69 (a) 2S 2 O 2 3 S 4 O 2 6 + 2e OCl + 2H + + 2e Cl + H 2 O Net: OCl + 2S 2 O 2 3 + 2H + S 4 O 2 6 + Cl + H 2 O (b) (NO 3 + 2H + + e NO 2 + H 2 O) 2 Cu Cu 2+ + 2e Net: 2NO 3 + Cu + 4H + 2NO 2 + Cu 2+ + 2H 2 O (c) (d) (e) IO 3 + 6H + + 6e I + 3H 2 O (H 2 O + AsO 3 3 AsO 4 3 + 2H + + 2e ) 3 Net: IO 3 + 3AsO 3 3 + 6H + + 3H 2 O I + 3AsO 4 3 + 3H 2 O + 6H + 3AsO 3 3 + IO 3 I + 3AsO 4 3 SO 4 2 + 4H + + 2e SO 2 + 2H 2 O Zn Zn 2+ + 2e Net: Zn + SO 4 2 + 4H + Zn 2+ + SO 2 + 2H 2 O NO 3 + 10H + + 8e NH 4 + + 3H 2 O (Zn Zn 2+ + 2e ) 4 Net: NO 3 + 4Zn + 10H + 4Zn 2+ + NH 4 + + 3H 2 O (f) 2Cr 3+ + 7H 2 O Cr 2 O 7 2 + 14H + + 6e (BiO 3 + 6H + + 2e Bi 3+ + 3H 2 O) 3 Net: 2Cr 3+ + 3BiO 3 + 18H + + 7H 2 O Cr 2 O 7 2 + 14H + + 3Bi 3+ + 9H 2 O 2Cr 3+ + 3BiO 3 + 4H + Cr 2 O 7 2 + 3Bi 3+ + 2H 2 O
(g) I 2 + 6H 2 O 2IO 3 + 12H + + 10e (OCl + 2H + + 2e Cl + H 2 O) 5 Net: I 2 + 5OCl + H 2 O 2IO 3 + 5Cl + 2H + (h) (Mn 2+ + 4H 2 O MnO 4 + 8H + + 5e ) 2 (BiO 3 + 6H + + 2e Bi 3+ + 3H 2 O) x 5 2Mn 2+ + 5BiO 3 + 30H + + 8H 2 O 2MnO 4 + 5Bi 3+ + 16H + + 15H 2 O Net: 2Mn 2+ + 5BiO 3 + 14H + 2MnO 4 + 5Bi 3+ + 7H 2 O (i) (H 3 AsO 3 + H 2 O H 3 AsO 4 + 2H + + 2e ) 3 Cr 2 O 7 2 + 14H + + 6e 2Cr 3+ + 7H 2 O Net: 3H 3 AsO 3 + Cr 2 O 7 2 + 3H 2 O + 14H + 3H 3 AsO 4 + 2Cr 3+ + 6H + + 7H 2 O 3H 3 AsO 3 + Cr 2 O 7 2 + 8H + 3H 3 AsO 4 + 2Cr 3+ + 4H 2 O (j) 2I I 2 + 2e HSO 4 + 3H + + 2e SO 2 + 2H 2 O Net: 2I + HSO 4 + 3H + I 2 + SO 2 + 2H 2 O (k) (Sn + 2H 2 O SnO 2 + 4H + + 4e ) 3 (ΝΟ 3 + 4H + + 3e NO + 2H 2 O) 4 Net: 3Sn + 4NO 3 + 16H + + 6H 2 O 3SnO 2 + 12H + + 4NO + 8H 2 O 3Sn + 4NO 3 + 4H + 3SnO 2 + 4NO + 2H 2 O (l) PbO 2 + 2Cl + 4H + + 2e PbCl 2 + 2H 2 O 2Cl Cl 2 + 2e Net: PbO 2 + 4Cl + 4H + PbCl 2 + Cl 2 + 2H 2 O (m) Ag Ag + + e NO 3 + 2H + + e ΝΟ 2 + Η 2 Ο Net: Ag + 2H + + NO 3 Ag + + NO 2 + H 2 O (n) (Fe 3+ + e Fe 2+ ) 4 2NH 3 OH + N 2 O + H 2 O + 6H + + 4e Net: 4Fe 3+ + 2NH 3 OH + 4Fe 2+ + N 2 O +6H + +H 2 O (o) 2I I 2 + 2e (HNO 2 + H + + e NO + H 2 O) 2 Net: 2I + 2HNO 2 + 2H + I 2 + 2NO + 2H 2 O (p) C 2 O 4 2 2CO 2 + 2e (HNO 2 + H + + e NO + H 2 O) 2 Net: C 2 O 4 2 + 2HNO 2 + 2H + 2CO 2 + 2NO + 2H 2 O (q) (HNO 2 + H 2 O NO 3 + 3H + + 2e ) 5
(MnO 4 + 8H + + 5e Mn 2+ + 4H 2 O) 2 Net: 5HNO 2 + 2MnO 4 + 16H + + 5H 2 O 5NO 3 + 2Mn 2+ + 15H + + 8H 2 O 5HNO 2 + 2MnO 4 + H + 5NO 3 + 2Mn 2+ + 3H 2 O (r) (H 3 PO 2 + 2H 2 O H 3 PO 4 + 4H + + 4e ) 3 (Cr 2 O 7 2 + 14H + + 6e 2Cr 3 + 7H 2 O) 2 Net: 3H 3 PO 2 + Cr 2 O 7 2 + 28H + + 6H 2 O 3H 3 PO 4 + 4Cr 3+ + 14H 2 O + 12H + 3H 3 PO 2 + 2Cr 2 O 7 2 + 16H + 3H 3 PO 4 + 4Cr 3+ + 8H 2 O (s) (VO 2 + + 2H + + e VO 2+ + H 2 O) 2 Sn 2+ Sn 4+ + 2e Net: 2VO 2 + + Sn 2+ + 4H + 2VO 2+ + Sn 4+ + 2H 2 O (t) XeF 2 + 2e Xe + 2F 2Cl Cl 2 + 2e Net: XeF 2 + 2Cl Xe + Cl 2 + 2F (u) S 2 O 3 2 + 5H 2 O 2SO 4 2 + 10H + + 8e (OCl + 2H + + 2e Cl + H 2 O) 4 Net: S 2 O 3 2 + H 2 O + 4OCl 2SO 4 2 + 2H + + 4Cl 12.71 O 3 + Br BrO 3 12.73 Cu(s) Cu 2+ + 2e H 2 SO 4 + 2H + + 2e SO 2 + 2H 2 O Cu + H 2 SO 4 + 2H + Cu 2+ + SO 2 + 2H 2 O 12.75 Since Ru reduces Pt 2+, it is more reactive than Pt (Ru > Pt) Since Tl reduces Ru 2+, it is more reactive than Ru (Tl > Ru) Since Pu reduces Tl +, it is more reactive than Tl (Pu > Tl) Therefore the increasing ease of oxidation is: Pu > Tl> Ru > Pt 12.77 Cd(s) + Pt 2+ (aq) Cd 2+ (aq) + Pt(s) 12.79 (a) Fe(s) Fe 2+ (aq) Cd 2+ (aq) Cd(s) (b) Pt(s) Br 2 (aq), Br (aq) Cl 2 (g), Cl - (aq), Pt(s) (c) Ag(s) Ag + (aq) Au 3+ (aq) Au(s) (d) Pt(s) Fe 2+ (aq), Fe 3+ (aq) NO - 3 (aq), NO(g) Pt(s) (e) Ag(s) Ag + (aq) NiO 2 (s), Ni 2+ (aq) Pt(s) (f) Mg(s) Mg 2+ (aq) Cd 2+ (aq) Cd(s)
12.81 (a) E o cell = 0.88 V spontaneous (b) E o cell = 1.40 V not spontaneous (c) E o cell = 2.02 V spontaneous (d) E o cell = 0.29 V not spontaneous (e) E o cell = +0.19 V spontaneous (f) E o cell = +0.90 V spontaneous (g) E o cell = 0.37 V not spontaneous 12.83 (a) BrO 3 (aq) + 6I (aq) + 6H + (aq) 3I 2 (s) + Br (aq) + 3H 2 O net reaction E o cell = 0.90 V (b) MnO 2 (s) + 4H + (aq) + Pb(s) + 2Cl (aq) Mn 2+ (aq) + 2H 2 O(l) + PbCl 2 (aq) net reaction E o cell = 1.5 V 12.85 Br 2 (aq) + 2e 2Br (aq) E o cell = 1.07 V I 2 (s) + 2e 2I (aq) E o cell = 0.54 V Since the first of these has the larger reduction half-cell potential, it is reduction and the second is reversed as oxidation: Br 2 (aq) + 2I (aq) I 2 (s) + 2Br (aq) E o cell = E o reduction E o oxidation = 1.07 V (0.54 V) = 0.53 V 12.87 (a) E o cell = 0.54 V (b) G o = 521.0 kj (c) K c = anti-ln(210.3) = 2.15 10 91 12.89 K c = anti-log ( 0.51) = 0.31 12.91 E cell = +0.135 V 8.314 J K 1 mol 1 298 K / 2 96 485 C mol 1 ln 6 10 [ 0.0001] [ 0.001] 5 34 [ 0.01] [ 0.1] = + 0.135 V 8.56 10 4 V ln (1 10 10 ) = + 0.135 V 8.56 10 4 V ( 23.03) = + 0.135 V ( 0.02 V) = 0.137 V 12.93 0.75 M 12.95 (a) 2400 C, equivalent to 0.025 moles of electrons (b) 12 000 C, equivalent to 0.124 moles of electrons (c) 32 400 C, equivalent to 0.336 moles of electrons
12.97 0.0371 mol Cr 3+ would be reduced. Electrolysis would take 45 minutes. 12.99 3.28g Cl 2 12.101 6.04 hr 12.103 0.0402 L HCl or 40.2 ml HCl 12.105 V = 0.0784 L or 78.4 ml 12.107 (a) 2H 2 (O)(l) + 2e H 2 (g) + 2OH (aq) (b) Cu 2+ (aq) + 2e Cu(s) (c) Cu 2+ (aq) + 2e Cu(s) Additional exercises 12.109 Increasing ease of oxidation: Cu Fe Pb Al. The second and fifth experiments are not required to establish this order. 12.111 (a) aluminium (b) zinc (c) magnesium 12.113 50.23 g PbO 2 is required. 12.115 Standard reduction potential = 1.118 V 12.117 E cell = 0.30 V 12.119 0.964 s