Chapter 19: ElectrochemistryProblems# 1, 11, 19, 22, 25, 27, 31, 35, 37, 41, 45, 59, 63, 65, 75, 82, 90 & Ex 10 A, B in the chapter 1. From the observations listed, estimate the value of E⁰ for the half-reaction M2+ + 2e- → M(s).
11. Using Important Standard Reduction Potentials and assuming that reactants and products are in their standard states, dDecide if each reaction will be spontaneous in the forward direction and calculate Ecell for each.
19. Given the following cell diagrams write the cell reaction and calculate E⁰ cell for each reaction. For the reaction below, use data from The Activity Series Table to calculate E⁰ cell for each reaction.
22. Sketch a voltaic cell. Label the cathode and anode, indicate the direction of electron flow, write a balanced equation for the cell reaction and calculate E0cell
25. Determine the values of ΔG for the following reactions carried out in voltaic cells.
27. For the reaction below, use data from The Activity Series Table and Important Standard Reduction Potentials to determine (a)E०cell, (b) ΔG०, (c)K, (d) whether the reaction goes substantially to completion when the reactants and products are initialy in their standard states. Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
35.The following cell diagram has an Ecell=1.083v. What concentration of Cu2+ is required in the cell in order to get the same E⁰ cell value? Assuming room temperature and 1 atm. Use The Activity Series Table for required information. Zn(s)|Zn2+(2.M)||Cu2+(xM)|Cu(s) 37. Use the Nerst equation to calculate Ecell for each of the following cells. Fe(s) → Fe3++3e- E0 = -.04V a)Fe(s)lFe3+(.23M)llZn2+(.76M)lZn(s) 41. If [Fe2+] is maintained at 1.0 M,
45. A voltaic cell is constructed as follows: Cu(s)|Cu2+(satd CuCrO4)||Cu2+(0.125M)|Cu(s) What is the value of E०cell ? For CuCrO4, Ksp = 3.6 x 10-6 63. How many grams of metal are deposited at the cathode in the electrolysis reactions containing (a)Pb2+ (b)K+ (c) Al3+ (d)Ni2+ When 2.0A current passes through the reaction for 1.25 hours? 65. Which of these reactions occur spontaneously, and which can only happen through electrolysis, assuming that the reaction and products are in standard states? For those that do require electrolysis, what is the minimum voltage? a)Cu(s)+Pb2+(aq) → Cu2+(aq)+Pb(s) 75. Two voltaic cells are assembled in which the following reactions occur. NO2(g) + 2H+(aq) + Ni(s) → NO(g) + H2O(l) + Ni2+ Eocell = +1.287 NO3(aq) + 4H+(aq) +3Cr2+(aq) → NO(g) +2H2O(l) + 3Cr3+(aq) Eocell = +1.38 Use this data and other values from the standard electron reduction potential table to calculate Eo for the half-reaction Ni2++2e- → Ni(s) 82. Look over the following reaction: 4HI(g) + O2(g) → 2I2(g) + 2H2O(l) The theoretical E०cell of this fuel cell is 1.247V. Use this information and the data below to calculate a value of ΔG०f for [HI(g)]. 90. A 1.075g sample was dissolved in nitric acid to produce Pb2+ (aq) and Ag+ (aq). The solution was then diluted to 600.0mL with water. An Ag electrode was immersed in the solution, and the potential difference between this electrode and a SHE was found to be 0.670V. What was the percent Ag by mass in the lead metal? Example A: Will the following cell diagram be spontaneous? Solutions1. From the observations listed, estimate the value of E⁰ for the half-reaction M2+ + 2e- → M(s). For a review of electropotentials, visit the page "The Cell Potential". a) If the metal M reacts with HSO4(aq), then its reduction potential must be smaller than the reduction potential of HSO4(aq), which is +0.17V. If the metal M does not react with H2S(aq). then its reduction potential must be greater than the reduction potential of H2S(aq), which is 0V. (The reduction potential of 2H+(aq) to H2(g) is 0V.) If the metal M displaces Fe3+(aq) , then it is oxidized. This means it has a reduction potential that is less than the reduction potential of Fe3+(aq), which is +0.771V. If the metal M does not displace Sn4+(aq), then it is reduced meaning it has a reduction potential greater than the reduction potential of Sn4+(aq), which is +0.154V. b) If the metal M reacts with H2S(aq), then its reduction potential must be smaller than the reduction potential of H2S(aq), which is +0.14V. If the metal M does not displace Pb2+(aq) or Sn2+(aq), then it must have a reduction potential that is smaller (more positive) than the reduction potentials of Pb2+(aq) and Sn2+(aq), which are -0.125V and -0.137V, respectively. 11. For review on electrochemical cells, visit the page "Connection between Ecell, ∆G, and K". 19. For the reaction below, use data from The Activity Series Table a) 2Al(s)+3Cu2+(aq) b) Sn2+(aq)+ 2Ag+(aq) → Sn4+(aq) + 2Ag(s) c) 10Fe2+(aq) +2IO3-(aq)+ 12H+(aq)→ 10Fe3+(aq)+I2(s)+6H2O d) 5Pb(s)+ MnO4-(aq) + 8H+(aq) → 5Pb2+(aq) + Mn2+(aq) + 4H2O(l) 21. a) [--<-<--<--<--<---] [ [ ]…...................[ ] ] [ [ ] [ ] ] [Fe3+] [ Zn ] cathode anode reduction: Fe3+(aq)+e- → Fe2+(aq) oxidation: Zn(s) → 2e-+Zn2+(aq) 2Fe3+(aq)+Zn(s) → 2Fe2+(aq)+Zn2+(aq) ΔEocell = Ecathode - Eanode = 0.77V-(.-763V) = 1.533V b) [--<-<--<--<--<--] [ [ ]…......................[ ] ] [ [ ] [ ] ] [Fe2+] [ Al ] cathode anode reduction: Fe2+(aq) + 2e- → Fe(s) oxidation: Al(s) → 3e-+Al3+(aq) 3Fe2+(aq) + 2Al(s) → 3Fe(s)+2Al3+(aq) ΔEocell = Ecathode - Eanode = -0.44V-(-1.676V) = 1.236V
c) [--<-<--<--<--<--] [ [ ]….....................[ ] ] [ [ ] [ ] ] [ F2 ] [H2O ] cathode anode reduction: F2(g)+2e- → 2F-(aq) oxidation: H2O(l) → 0.5O2(g)+2H+(aq)+2e- F2(g)+H2O(l) → 2F-(aq)+.5O2(g) + 2H+(aq) ΔEocell = Ecathode - Eanode = 2.87V-(1.229V) = 1.641V
d) [--<-<--<--<--<--] [ [ ]…....................[ ] ] [ [ ] [ ] ] [ Pb ] [Cu ] cathode anode reduction: Pb2+(aq)+2e- → Pb(aq) oxidation: Cu(s) → Cu2+(aq)+2e- Pb2+(aq)+Cu(s) → Cu2+(aq)+Pb(aq) ΔEocell = Ecathode - Eanode = -0.13V-(0.340V) = -0.47V 25. (a) reduction: Fe3+(aq) + 3e- → Fe(s) Eo = +.771V oxidation: Mn2+(aq) + 2e- → Mn(s) Eo = -1.18V ΔEocell = Ecathode - Eanode = 0.771V - (-1.18V) = 1.951V ΔG = -nFEocell = -(6mol e-)(96,485C/mol e-)(1.951V) = -1,129.5kJ (b) reduction: MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H20 (l) Eo = +1.23V oxidation: Br2(l) + 2e- → 2Br(s) Eo = +1.065V ΔEocell = Ecathode - Eanode = 1.23V - (1.065) = .161V ΔG = -nFEocell = -(2mol e-)(96,485C/mol e-)(.161V) = -31.07kJ (c) reduction: ClO3-(aq) + 6H+(aq) + 6e- → Cl-(aq) + 3H2O(l) Eo = +1.450V oxidation: 6Cu+(aq) + 6e-(aq) → 6Cu(s) Eo = +.520V ΔEocell = Ecathode - Eanode = 1.450 - .520 = .93V ΔG = -nFEocell = -(6mole-)(96,485C/mol e-)(.93V) = -538.39kJ 27. For the reaction below, use data from The Activity Series Table and Important Standard Reduction Potentials to determine (a)E०cell, (b) ΔG०, (c)K, (d) whether the reaction goes substantially to completion when the reactants and products are initialy in their standard states. Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq) For review on electrochemical cells, visit the pages "Electrochemistry 3: Cell Potentials and Thermodynamics" and "Connection between Ecell, ∆G, and K". a) Please note that the E०cell equation above takes into account the negative E० from flipping the reduction equation of Fe3+ to make it into an oxidation equation so there is no need to use “-0.771V” in the E०cell equation; aka the equation should not be “1.33V - (-0.771V)”. b) ΔG० = -nFE०cell = -(6)(96485C/mol)(0.559V) = 323610.69J/mol = 323.61kJ/mol c) d) Since K is very small, this reaction will not go to completion substantially. 31. ΔGo=ΔGo(products)-ΔGo(reactants)= (-318.3kJ/mol)-(-11.20kJ/mol)=-307.1kJ/mol For review on how to combine electrochemistry and free gibbs energy, visit the pages "Electrochemistry 3: Cell Potentials and Thermodynamics" and "Connection between Ecell, ∆G, and K". 35.The following cell diagram has an Ecell=1.083V. What concentration of Cu2+ is required in the cell to get the same E⁰ cell value? Assuming room temperature and 1 atm. Use The Activity Series Table for required information. Oxidized: Zn2+(aq)+2e- → Zn(s) E०= -0.763v n= number of mols of electrons being transferred looking at the equation we can see that 2 electrons are being transferred so n=2 Plugging all the values we get 37. b)Eocell = Ecathode - Eanode=0.340V-(-0.137V) = 0.477V reduction: Pb2+(aq) + 2e- → Pb(s) Eo= -0.125V oxidation: Fe2+(aq) + 2e- → Fe(s) Eo = -0.440V Net ionic: Pb2+(aq) + Fe(s) → Fe2+(aq) + Pb(s) 41. (a) For reaction to be spontaneous, Ecell must be greater than 0. Solving for [Pb2+]... Ecell = Eocell - ((.0592)/n)logQ if Eccell = Ecathode - Eanode = -0.125 - (-0.440) = 0.315V n = 2 mol e- assuming Ecell = 0 then Eocell - Ecell = (.0592/2)log([Fe2+]/[Pb2+]) (0.315V - 0) = (0.0592/2)log(1.0M/x) 2(.315V)/(.0592) = log(1/x) 10^(2(0.315V)/(0.0592)) = 1/x 4.38x10-10 = 1/x x = 2.28x10-11M (b) Yes, the displacement of Pb2+(aq) by Fe(s) should go to completion. The reaction is spontaneous if Ecell is greater than 0, and this holds true if the [Pb2+] is greater than 2.28x10-11M. So if the [Pb2+] is greater than 2.28x10-11M, then the displacement should go to completion spontaneously.
45. A voltaic cell is constructed as follows: Cu(s)|Cu2+(satd CuCrO4)||Cu2+(0.125M)|Cu(s) What is the value of E०cell ? For CuCrO4, Ksp = 3.6 x 10-6 reduction: Cu2+ (0.125M) + 2e- → Cu(s) E०= +0.340V 59. "The Activity Series Table" Potassium and Magnesium because they are more active than Zinc on the activity series. Lead and Cobalt are less active so they cannot be used. Sacrificial anode – an active metal that protects the metal on the cathode from corrosion. 63. In order to find the mass of the metal we first need to find the number of mols of e- are passed through the current in 1.25 hours.
65. If E0cell>0 then ΔG<0 and is therefore spontaneous. E0cell=Ecathode-Eanode view Connection between Ecell, ∆G, and K for further information. a) Cu is oxidized (anode), Pb is reduced (cathode), so b) Zn is oxidized (anode), Sn is reduced (cathode), so c) Ag is oxidized (anode), Cl is reduced (cathode), so d) reduced (cathode) 2H+(aq) → 2e-+H2(g) 75. Write out half reactions involved in the reactions:: cathode: NO2(g) + 2H+(aq) + 2e- → NO(g) + H2O(l) Eo = +1.03V anode: Ni2+(aq) + 2e- → Ni(s) Eo= in question and cathode: NO3-(aq) + 4H+(aq) + 3e- → NO(g) + H2O(l) Eo= +.956V anode: Cr3+(aq) + e- → Cr2+ Eo= -.424V We can forget about the second cell (third and fourth half reactions) due to the lack of Ni2++2e- → Ni(s) reaction in this cell. Looking at the first cell, Eocell = Eocathode - Eoanode so Eoanode = Eocathode - Eocell = +1.03V - 1.287 = -.257 So: Eo = -.257 for half reaction Ni2++2e- → Ni(s) 82. Look over the following reaction: 4HI(g) + O2(g) → 2I2(g) + 2H2O(l) The theoretical E०cell of this fuel cell is 1.247V. Use this information and the data below to calculate a value of ΔG०f for [HI(g)]. To review Gibb's Free Energy, visit the page ""Connection between Ecell, ∆G, and K". ΔG० = -nFE०cell = -(4 mol e-)(96485C/mol e)(1.247V) = -481.2kJ 90. Oxidation: H2(g) → 2H+(aq) + 2e- E°cell = +0.00V To review Gibb's Free Energy, visit the page ""Connection between Ecell, ∆G, and K". Example A: Will the following cell diagram be spontaneous? We need to find the Ecell value to determine if the reaction is spontaneous since we know when Ecell>0; ΔG<0 This page viewed 6960 times
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