Chemistry 31X, Autumn 2008 Problem Set 2, Answer Key: Due: 2:30 PM Monday, Oct. 6, 2008 You should read Chapters 4, 12, and 13. You should review carefully: Ch.4: 4.4 Ch.12: 12.1-12.3 Ch.13: 13.2, 13.3, 13.6 and 13.7 1. Oxtoby 4.39: Write a balanced equation to represent the reaction between each of the following pairs of substances in the presence of water: (a) hydrogen bromide and calcium hydroxide 2HBr(aq) + Ca(OH)2 (s) (b) ammonia and sulfuric acid 2NH3(aq) + H2SO4(aq) 2NH41+(aq) + SO42- (aq) Ca2+(aq) + 2Br1-(aq) + 2H2O(l) (c) lithium hydroxide and nitric acid LiOH(aq) + HNO3(aq) Li+(aq) + NO3-(aq) + H2O(l) 2. Tell which of the following are oxidation-reduction reactions. For the redox reactions, identify the reducing agent, oxidizing agent, the substance being reduced, and the substance being oxidized. (a) Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) Is oxidation-reduction reaction; Reducing agent: Zn, reducing H+ to H0 (in the form of H2) Oxidizing agent: H+, oxidizing Zn0 to Zn2+ (b) SiCl4(l) + 2H2O(l) 4HCl(aq) + SiO2(s) Not an oxidation-reduction reaction; (c) SiCl4(l) + 2Mg 2MgCl2(s) + Si(s) Is oxidation-reduction reaction; Reducing agent: Mg, reducing Si4+ to Si0(s) Oxidizing agent: Si4+, oxidizing Mg0 to Mg2+ 3. Rust stains can be removed by washing a surface with a dilute solution of oxalic acid (H2C2O4). The reaction is: Fe2O3(s) + 6H2C2O4(aq) 2Fe(C2O4)33-(aq) + 3H2O(l) + 6H+(aq) (a) Is this an oxidation-reduction reaction? No (b) What mass of rust can be removed by 1.0 L of a 0.14 M solution of oxalic acid? (0.14 mol/1 L)(1 L) = 0.14 mol oxalic acid (6 mol oxalic acid)/(1 mol rust) = (0.14 mol oxalic acid)/(X mol rust) => X mol rust = 0.023 (0.023 mol Fe2O3)(160 g Fe2O3/1 mol Fe2O3) = 3.68 g Fe2O3 4. Oxtoby 12.19: Acidified water can be electrolyzed to produced hydrogen and oxygen. Write equations for the half-reactions that occur at the anode and at the cathode, and verify that they combine correctly to give the overall decomposition of water to its elements. Cathode: 2 x [2H+ + 2eH2 ] O2 + 4H+ + 4eAnode: 2H2O ________________________________ 2H2O 5. 2H2 + O2 Oxtoby 12.34: A galvanic cell uses a cadmium cathode immersed in a CdSO4 solution and a zinc anode immersed in a ZnSO4 solution. (a) Write a balanced equation for the cell reaction. Cathode: Cd2+(aq)+ 2eCd(s) Anode: Zn(s) Zn2+(aq) + 2e_________________________________ Cd2+(aq) + Zn(s) Cd(s) + Zn2+(aq) (b) A steady current of 1.45 A is observed to flow for a period of 2.60 h. How much charge passes through the circuit during this time? How many moles of electrons is this charge equivalent to? 1.45 A = 1.45 (C/sec) (1.45 C/sec)*(2.6 h)*(60 min/1 h)*(60 sec/1 min) = 13,572 C (13572 C)*(1 mol electrons/(9.6485342*10^4 C)) = 0.141 moles electrons (c) Calculate the change in mass of the zinc electrode. (0.141 mol electrons) (1 mol Zn / 2 mol electrons) (65.409 g/1 mol Zn) = 4.60 g lost from the Zn anode (d) Calculate the change in mass of the cadmium electrode. (0.141 mol e-) (1 mol Cd/2 mol e-) (112.411 g/1 mol Cd) = 7.91 g increase in Cd 6. Oxtoby 12.37: An acidic solution containing copper ions is electrolyzed, producing gaseous oxygen (from water) at the anode and metallic copper at the cathode. For every 16.0 g of oxygen that is generated, 63.5 g of copper plates out. What is the oxidation state of the copper in solution? (16.0g O)*(1 mol O/ 16g O) = 1 mol O (63.5g Cu)*(1 mol Cu/63.5g Cu) = 1 mol Cu 1:1 molar ratio between the oxygen and copper generated on an atom to atom basis implies the same amount of electrons required to convert both ions. Since oxygen in solution has oxidation number of -2, the oxidation number of Cu ions in solution must be +2. 7. Oxtoby 12.78: Tarnish forms on silver spoons when airborne sulfur compounds react with the silver to produce a dark coating of silver sulfide (Ag2S). Tarnish can be removed by putting the corroded spoons in an aluminum container (a pie plate works) containing a warm solution of baking soda (sodium hydrogen carbonate, NaHCO3). The NaHCO3(aq) does not react but serves only as an electrolyte in an electrochemical cell in which the aluminum container is the anode. Write a balanced chemical equation for the tarnishremoving reaction. Since the aluminum container is the anode, aluminum is oxidized and silver is reduced. 6Ag(s) + 2Al3+(aq) + 3S2-(aq) 3Ag2S(s) + 2Al(s) 8. Oxtoby 13.10: A galvanic cell is constructed by connecting a standard Pt|Fe2+, Fe3+ halfcell to a standard Cd2+|Cd cathode half-cell. (a) Write balanced chemical equations for the half-reactions at the anode and the cathode and for the overall cell reaction. Cathode: Fe3+ + eFe2+ Anode: Cd Cd2+ + 2e_______________________________ 2Fe3+ + Cd 2Fe2+ + Cd2+ (b) Calculate the cell voltage at 25°C (use information from Appendix E). ∆E = E(Cathode) – E(Anode) = 0.771 – (-0.403) = 1.174 9. Oxtoby 13.20: Many bleaches, including chlorine and its oxides, oxidize dyes in cloth, destroying their color. Predict which of the following is the strongest bleach at a given concentration and pH 0: NaClO3(aq), NaClO(aq), Cl2(aq). How does the strongest chlorine-containing bleach compare in strength with ozone (O3(g))? Considering the standard reduction potenetials for the anions, NaClO is strongest bleach, followed by NaClO3 and Cl2. NaClO, however, is not as strong as O3 because its reduction potential is lower. ClO + H+ + eO3 + 2H+ + 2e1/2Cl2 + H2O E° = 1.611 V O2 + H2O E° = 2.076 V 10. Oxtoby 13.24: Extending Figure 13-5, the following additional reduction potentials are measured at pH 14: S(s) + H2O(l) + 2eHS-(aq) + OH-(aq) E° = -0.51 V S2O32-(aq) + 3H2O(l) + 4e2S(s) + 6OH-(aq) E° = -0.74 V (a) Does sulfur disproportionate spontaneously under standard basic conditions at 25°C? We are looking at the disproportionation of S. Reduction of S, E° = -0.51 V Oxidation of S, E° = 0.74 V Sum = ∆E° = -0.51 + 0.74 = 0.23 V. Since ∆E° > 0, S does disproportionate spontaneously. (b) Which is the stronger reducing agent: S(s) or HS-(aq)? The stronger reducing agent is S(s). 11. Oxtoby 13.54: What quantity of charge (in coulombs) is a fully charged 1.34 V zincmercuric oxide watch battery capable of furnishing through an external circuit if the mass of HgO in the battery is 0.50 g. (0.50 g HgO)(mol/216.6g) = 0.0023 mol HgO Since two moles of electrons are release1d per one mole of HgO that is reduced, 0.0046 mol electrons are released. (0.0046 mol e-)((9.6485342*10^4C)/(1mol e-) = 450 C 12. Oxtoby 13.73: Estimate the cost of the electrical energy needed to produce 2.1 x 1010 kg (a year’s supply for the world) of aluminum from Al2O3(s) if electrical energy costs 10 cents per kilowatt-hour (1kW-h = 3.6 MJ = 3.6 x 106 J) and if the cell voltage is 5 V. First, let’s calculate the number of aluminum atoms in a year’s supply of Al. 1000 g 1 mol Al (2.1× 1010 kg) × ( )×( ) = 7.78 × 1011 mol Al 1 kg 26.982 g The production of aluminum from alumina is a three electron process: Al3+ + 3e − → Al 0 With this information, we can calculate the number of amount of charge transfer required to produce the Al. 3 e96485.342 C (7.78 × 1011 mol Al) × ( )×( ) = 2.25 × 1017 C − mol Al mol e We know that the electrical work performed is equal to the product of the charge transferred and the cell voltage. Welec = Q ⋅ ∆E (2.25 × 1017 C) × (5 V) = 1.13 × 1018 J The electric company charges for this work. Let’s calculate our total bill. $ 0 .1 1 kW - h )×( ) = $3.13 × 1010 (1.13 × 1018 J) × ( 6 kW − h 3.6 × 10 J