Chemistry 12 Assignment 8 - Solutions

Chemistry 12 Assignment 8-Solutions Fall 2009 Chapter 11: #17, 18, 19, 26, 29, 32, 39, 46, 53, 55, 56, 91, 95 Galvanic Cells, Cell Potentials, and Standard Reduction Potentials 17. A typical galvanic cell diagram is: Anode (oxidation) Salt Bridge cations anions Cathode (reduction) The diagram for all cells will look like this. The contents of each half-cell will be identified for each reaction, with all concentrations at 1.0 M and partial pressures at 1.0 atm. Note that cations always flow into the cathode compartment and anions always flow into the anode compartment. This is required to keep each compartment electrically neutral. a. Reference Table 11.1 for standard reduction potentials. Remember that EllE°(cathode) - E°(anode); in the Solutions Guide, we will represent E°(cathode) as E and represent -E°(anode) as-E. Also remember that standard potentials are not multiplied by the integer used to obtain the overall balanced equation. (Cl₂+2e2 CI) × 3 7 H₂O+2 Cr Cr₂O, +14 H+66 E= 1.36 V -E--1.33 V 7 H₂O(1) +2 Cr(aq) +3 Cl₂(g) Cr₂O, (aq) +6 Cl(aq) +14 H*(aq) Ecell = 0.03 V The contents of each compartment is: Cathode: Pt electrode; Cl, bubbled into solution, Cr in solution Anode: Pt electrode; Cr, H', and Cr₂O, in solution We need a nonreactive metal to use as the electrode in each case, since all of the reactants and products are in solution. Pt is the most common choice. Another possibility is graphite. с. d. 18. a. 2 Ag+2e+2 Ag Cu Cu²+26 E = 0.80 V -E=-0.34 V 5e+6H+10, 1/2 I₂ +3 H₂O (FeFe+e) x 5 E = 1.20 V -E=-0.77 V 6 H+IO,+5 Fe2+5 Fe + 1/2 +3 H₂O Ecell = 0.43 V or 12 H'(aq) + 2 IO, (aq) +10 Fe²(aq) →10 Fe³ (aq)+I,(s) +6 H₂O(I) E0.43 V Cathode: Pt electrode; 10,, I, and H₂SO₄ (H source) in solution. Anode: Pt electrode; Fe and Fe* in solution Note: I₂(s) would make a poor electrode since it sublimes. ( Ag+e+Ag) x 2 E= 0.80 V ZnZn2+ +2e -E = 0.76 V Zn(s) +2 Ag+(aq) → 2 Ag(s) + Zn(aq) E-1.56 V cell Cathode: Ag electrode; Ag in solution Anode: Zn electrode; Zn in solution 2 Ag'(aq) + Cu(s) Cu²(aq) +2 Ag(s) Ecell 0.46 V Spontaneous at standard conditions (E>0). b. Zn2+ +2e Zn E= -0.76 V C. Ni Ni²+26 -E= 0.23 V Zn2+(aq) +Ni(s) → Zn(s) + Ni²*(aq) E-0.53 V Not spontaneous at standard conditions (E<0). E=1.51 V (5e+8 H+ MnO, Mn2+4 H₂O) × 2 (21+1+2) ×5 -E=-0.54 V 16 H'(aq) + 2 MnO, (aq) +10 I(aq)51(s) +2 Mn²(aq) +8 H₂O(I) E = 0.97 V Spontaneous since El > 0. d. (5 e+8 H + MnO, Mn2+4 H₂O) × 2 (2F+F+2e) × 5 E = 1.51 V -E--2.87 V 16 H'(aq) + 2MnO, (aq) +10 F(aq)→5F2(g) +2 Mn²(aq) +8 H₂O(1) E = -1.36 V Not spontaneous since E <0. coll 19. Reference Exercise 11.17 for a typical galvanic cell design. The contents of each half-cell compartment is identified below with all solute concentrations at 1.0 M and all gases at 1.0 atm. For each pair of half-reactions, the half-reaction with the largest standard reduction potential will be the cathode reaction and the half-reaction with the smallest reduction potential will be reversed to become the anode reaction. Only this combination gives a spontaneous overall reaction, ie., a reaction with a positive overall standard cell potential. д. Cl₂+2+2 Cl 2 BrBr₂ +26 E= 1.36 V -E=-1.09 V Cl₂(g) +2 Br(aq) → Br₂(aq) + 2 Cl(aq) E=0.27 V cell The contents of each compartment is: Cathode: Pt electrode; Cl₂(g) bubbled in, Cr in solution Anode: Pt electrode; Br, and Br in solution (2e+2H+IO,10,+H₂O)×5 (4 H2O+MnMnO,+8H+5e)×2 E= 1.60 V 10 H+5 IO, +8 H₂O+2 Mn2+510, +5 H₂O+ 2MnO, + 16 H -E--1.51 V Ecell = 0.09 V This simplifies to: 3 H₂O(1) +5 IO, (aq) +2 Mn²(aq) →5 IO, (aq) +2 MnO, (aq) +6H(aq) Een = 0.09 V Cathode: Pt electrode; 10, 10, and H,SO, (as a source of H') in solution Anode: Pt electrode; Mn2+, MnO, and H₂SO, in solution c. H₂O₂+2H+2+2H₂O H₂O, O2+2H+2 c 2 H₂O₂(aq) →2H₂O(I) + O2(g) E = 1.78 V - E° = -0.68 V E = 1.10 V cell Cathode: Pt electrode; H₂O, and H in solution Anode: Pt electrode; O₂(g) bubbled in, H₂O, and H in solution (Fe+3eFe) x 2 (Mn+Mn2+2 e) x 3 E = -0.036 V -E-1.18 V 2 Fe³(aq) +3 Mn(s) +2 Fe(s) +3 Mn²*(aq) Ecell-1.14 V Cathode: Fe electrode; Fe in solution; Anode: Mn electrode; Mn² in solution d. b. 26. Cl₂ +2e+2 Cl Pb+2ePb E = 1.36 V E-0.13 V Ag+e+Ag Zn+2eZn E-0.80 V E-0.76 V Nae Na E°=-2.71 V a. Oxidizing agents (species reduced) are on the left side of the above reduction half-reactions. Of the species available, Ag' would be the best oxidizing agent since it has the most positive E° value. b. Reducing agents (species oxidized) are on the right side of the reduction half-reactions. Of the species available, Zn would be the best reducing agent since it has the most positive-E° value. c. SO2+4H+2H₂SO₄ + H₂O E0.20 V; SO, can oxidize Pb and Zn at standard conditions. When SO, is coupled with these reagents, E is positive. d. AlAl+3E1.66 V; Al can oxidize Ag and Zn at standard conditions since Eell >0. 29. CIO+H₂O+2e+2 OH + Cl 2 NH₂ + 2OHN₂H₄ +2 H₂O+20 E° = 0.90 V -E-0.10 V CIO (aq) + 2 NH,(aq) → Cl(aq) + NH(aq) + H₂O(1) E = 1.00 V cell Since El is positive for this reaction, then at standard conditions CIO can spontaneously oxidize NH, to the somewhat toxic N₂H₄. 32. Consider the strongest oxidizing agent combined with the strongest reducing agent from Table 11.1: F₂+2e2F (Li+Li+e)×2 E = 2.87 V - E = 3.05 V F2(g) +2 Li(s)+2 Li (aq) +2 F(aq) Ell = 5.92 V Cell Potential, Free Coray, and Equilibrium 39. The claim is impossible. The strongest oxidizing agent and reducing agent when combined only give Ecell of about 6 V. "cell AG-FEAH-TAS, E- TAS AH nF nF If we graph E° vs. T we should get a straight line (y=mx+b). The slope of the line (m) is equal to AS°/nF and the y-intercept (b) is equal to -AH°/nF. 46. Ag+e+ Ag Ag+2 S₂O, Ag(SO₂)+ Ag(aq) +2 S₂O, (aq) → Ag(S₂O₃)₂ (aq) For this overall reaction, E eril 0.0591 n E = 0.80 V -E=-0.017 V Ecell = 0.78 V log K log K= (1)(0.78) 13.20, K=101320 = 1.6 × 1013 nE 0.0591 0.0591 Galvanic Cells: Concentration Dependence 53. a. n2 for this reaction (lead goes from Pb Pb in PbSO₄). Ecell E Ecell 0.0591 2 log 2.04 V - (-0.077 V) = 2.12 V 1 [H] [HSO] = 2.04 V- 0.0591 2 log 1 (4.5) (4.5)2 b. We can calculate AG° from AGAH-TAS and then E° from AG°=-nFE; or we can use the equation derived in Exercise 11.39. E -20 TAS - AH nF (253 K) (263.5 J/K) +315.9×10 J (2 mole) (96,485 C/mol e) 1.98 J/C1.98 V c. E-20 - E-20- RT nF RT In Q-1.98 V- In nF 1 [H][HSO₄ ] E.20 1.98 V- (8.3145 JK mol) (253 K) In 1 (2 mole) (96,485 C/mol e) c) (4.5)²(4.5)2 1.98 V-(-0.066 V) = 2.05 V d. As the temperature decreases, the cell potential decreases. Also, oil becomes more viscous at lower temperatures, which adds to the difficulty of starting an engine on a cold day. The combination of these two factors results in batteries failing more often on cold days than on warm days. 55. As is the case for all concentration cells, E=0, and the smaller ion concentration is always in the anode compartment. The general Nernst equation for the Ni | Ni (x M) || Ni²*(y M) | Ni concentration cell is: celi E cell 0.0591 log Q= -0.0591 2 log [Ni2+] mode [Ni2+] cathode a. Since both compartments are at standard conditions ([Ni**] 1.0 M), then Ecelt = Ecell = 0V. No electron flow occurs. b. Cathode 2.0 M Ni²; Anode 1.0 M Ni²; Electron flow is always from the anode to the cathode, so electrons flow to the right in the diagram. Ecell = -0.0591 [Ni 2+] 2 log [Ni2+] anode Icathode cathode -0.0591 1.0 2 log = 8.9 × 10 V 2.0 c. Cathode = 1.0 M Ni; Anode-0.10 M Ni; Electrons flow to the left in the diagram. Ecell = -0.0591 0.10 2 log -0.030 V 1.0 d. Cathode 1.0 M Ni²; Anode-4.0×105 M Ni²; Electrons flow to the left in the diagram. En= cell -0.0591 4.0 x 105 2 log 1.0 -0.13 V e. Since both concentrations are equal, log (2.5/2.5) = log 1.00 and Ecell = 0. No electron flow occurs. 56. Cathode: M²+2e+M(s) E = 0.80 V M(s) + M² + 2e -E--0.80 V Anode: M² (cathode) M² (anode) E0.00 V East 0.44 V-0.00 V. 0.0591 2 log [Mode, 0.44- [M2] cathode 0.0591 2 log [M2] 1.0 log [M] 2(0.44) Jatinde 0.0591 -14.89, [M2] made 1.3 x 101 M Janade Since we started with equal numbers of moles of SO, and M², then [M²] [SO] at equilibrium. K= [M²][SO]=(1.3×10-5)=1.7 × 10-30 Challenge Problems 91. a. Zn(s) + Cu(aq) → Zn(aq) + Cu(s) = 1.10 V; E = 1.10 - 0.0591 2 Ecull 1.10 V- 0.0591 0.10 2 log 2.50 1.10 V-(-0.041 V) = 1.14 V log [Zn²] [Cu] b. 10.0 h x 60 min 60 s 10.0 C 1 mol e X h min X S X 1 mol Cu 96,485 C 2 mol e The Cu² concentration decreases by 1.87 mol/L and the Zn mol/L. [Cu2"]2.50-1.87 0.63 M; [Zn2"]=0.10+1.87-1.97 M Ecell = 1.10 V- 0.0591 1.97 log 2 0.63 1.10 V-0.015 V = 1.09 V = 1.87 mol Cu produced concentration will increase by 1.87 c. 1.87 mol Zn consumed x 65.38 g Zn mol Zn 122 g Zn; Mass of electrode = 200.-122-78 g Zn 1.87 mol Cu formed x 63.55 g Cu = 119 g Cu; Mass of electrode = 200.+119=319 g Cu mol Cu d. Three things could possibly cause this battery to go dead: 1. All of the Zn is consumed. 2. All of the Cu² is consumed. 3. Equilibrium is reached (E=0). 2.50 mol Cu² 2 mol e 96,485 C 2+ X mol Cu mole X Is 1h X 10.0 C 3600 5 13.4 h [Zn²] K = 101.10900591 = 1.68 × 1037 [Cu²] This is such a large equilibrium constant that virtually all of the Cut must react to reach equilibrium. So, the battery will go dead in 13.4 hours. We began with 2.50 mol Cut and 200. g Zn x 1 mol Zn/65.38 g Zn = 3.06 mol Zn. Cu² is the limiting reagent and will run out first. To react all the Cu²" requires: For equilibrium to be reached: E-0-1.10 V- 0.0591 2 log [Zn²] [Cu²] 95. a. 3x (e+2H+NO,NO₂+H₂O) E= 0.775 V 2 H₂O + NO→ NO, +4 H*+3 e -E=-0.957 V 2 H(aq) + 2NO, (aq) +NO(g)+3 NO2(g) + H₂O(1) E-0.182 V K-? log K = nE 3(-0.182) 0.0591 0.0591 -9.239, K-10-92395.77 × 10-10 b. Let C concentration of HNO, = [H] = [NO,] 5.77 x 10-10 = PNO PNO PNOX [H] x [NO, 2 PNO XC4 If 0.20 mol% NO, and P 1.00 atm: 0.20 mol NO, PNO, 100. mol total x1.00 atm 2.0 × 10 atm; Po 1.00-0.00201.00 atm 5.77x10-10 (2.0×103) (1.00) C4 -, C=1.9 M HNO,