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UT CH 302 - Advanced Electrochemistry Problems
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CH 302 Spring 2008Worksheet 10b: More Advanced Electrochemistry Problems 1. (a) Calculate the mass of copper metal produced at the cathode during the passage of 2.50 amps of current through a solution of copper (II) sulfate for 50.0 minutes. Cu2+ + 2e- → Cu (reduction/cathode) 1 mol 2(6.02 x 1023)e- 1 mol 63.5 g 2(9.65 x 104C) 63.5g 50.0 min x (60s/1 min) x (2.50C/s) = 7.50 x 103C 7.50 x 103C x (1 mol e-/9.65 x 104C) x (63.5 g Cu/2 mol e-) = 2.47 g Cu (b) What volume of oxygen gas (measured at STP) is produced by the oxidation of water at the anode in the electrolysis of copper(II) sulfate in part (a)? 2H2O → O2 + 4H+ + 4e- (oxidation/anode) 1 mol 4((6.02 x 1023)e- 22.4 L 4(9.65 x 104C) 7.50 x 103C x (1 mol e-/9.65 x 104C) x (22.4 L O2/4 mol e-) = 0.435 L O2 2. What is the E° for the following electrochemical cell where Zn is the cathode? Zn | Zn2+ (1.0 M) || Fe2+ (1.0 M) | Fe E° (Zn) = -0.76 E° (Fe) = -0.44 Ecell = E°cathode - E°anode = -0.76 – (-0.44) = -0.32 3. For the electrolysis of molten sodium bromide, write the two half-reactions and show write which electrode at which each occurs (cathode or anode). Two half-reactions: Na+ + 1e- → Na and 2Br- → Br2 + 2e- Oxidation occurs at the anode, so the 2Br- → Br2 + 2 e- reaction is at the anode, and Reduction occurs at the cathode, so Na+ + 1e- → Na is at the cathode. 4. Calulate the potential, E, for the Fe3+/ Fe2+ electrode when the concentration of Fe2+ is exactly five times that of Fe3+. Fe3+ + e- → Fe2+ E° = +0.771 V Q = [Red]y/[[Ox]x = [Fe2+]/[ Fe3+] = 5[Fe3+]/[ Fe3+] = 5 E = E° - 0.0592/n*logQ = +0.771 – 0.0592/1*log5 = (0.771-0.041)V = 0.730 V 5. At standard conditions, will chromium (III) ions, Cr3+, oxidize metallic copper to copper (II) ions, Cu2+, or will Cu2+ ozidize metallic chromium to Cr3+ ions? Write the cell reaction and calculate E°cell for the spontaneous reaction. Cu2+ + 2e- → Cu E° = 0.337 Cr3+ + 3e- → Cr E° = -0.74 E° 3(Cu2+ + 2e- → Cu) +0.337 V 2(Cr → Cr3+ + 3e-) + 0.74 V 2Cr + 3 Cu2+ → 2Cr3+ + 3Cu E°cell = 1.08 V Cu2+ ions spontaneously oxidize metallic Cr to Cr3+ ions and are reduced to metallic Cu.6. In an acidic solution at standard conditions, will tin(IV) ions, Sn4+, oxidize gaseous nitrogen oxide, NO, to nitrate ions, NO3-, or will NO3- ozidize Sn2+ to Sn4+ ions? Write the cell reaction and calculate E°cell for the spontaneous reaction. Sn4+ + 2e- → Sn2+ E° = +0.15 NO3- + 4H+ + 3e- → NO + 2H2O E° = +0.96 E° 2(NO3- + 4H+ + 3e- → NO + 2H2O) +0.96V 3(Sn2+ → Sn4+ + 2e-) -0.15V 2NO3- + 8H+ + 3Sn2+ → 3Sn4+ + 4H2O + 2NO E°cell = +0.81V Nitrate ions spontaneously oxidize tin(II) ions to tin(IV) ions and are reduced to nitrogen oxide in acidic solution. 7. Calculate the Gibbs free energy change, ∆G°, in J/mol at 25°C for the following reaction: 3 Sn4+ + 2Cr → 3Sn2+ + 2Cr3+ Sn4+ + 2e- → Sn2+ E° = +0.15 Cr3+ + 3e- → Cr E° = -0.74 E° 3(Sn4+ + 2e- → Sn2+) +0.15V 2(Cr → Cr3+ + 3e-) -(-0.74V) 3Sn4+ + 2Cr → 3Sn2+ + 2 Cr3+ E°cell = 0.89V ∆G° = -nF E°cell = -(6 mol e-/mol rxn)(9.65x 104 J/V.mol e-)(+0.89V) = -5.2 x 105 J/mol rxn 8. Use the standard cell potential to calculate the value of the equilibrium constant, K, at 25°C for the following reaction. 2Cu + PtCl62- → 2Cu+ + PtCl42- + Cl- Cu+ + e- → Cu; E° = 0.521V and PtCl62- + 2e- → PtCl42- + 2Cl-; E° = +0.68V E° 2(Cu → Cu+ + e-) -(+0.521V) PtCl62- + 2e- → PtCl42- + 2Cl- +0.68V PtCl62- + 2Cu → PtCl42- + 2Cl- + 2Cu+ E°cell =+0.16V lnK = -nF E°cell/RT = (2)(9.65x 104 J/V.mol e-)(0.16V)/(8.314J/mol.K)(298K) = 12.5 K = e12.5 = 2.7 x 105 9. The following cell is maintained at 25°C. One half-cell consists of a chlorine/chloride, Cl2/Cl-, electrode with the partial pressure of Cl2= 0.100 atm and [Cl-] = 0.100 M. The other half-cell involves the MnO4-/Mn2+ couple in acidic solution with [MnO4-] = 0.100 M, [Mn2+] = 0.100 M, and [H+] = 0.100 M. Apply the Nernst equation to the overall cell reaction to determine the cell potential for this cell. MnO4- + 8H+ + 5e- → Mn2+ + 4H2O E° = 1.507 V Cl2 + 2e- → 2Cl- E° = 1.360 V E° 2(MnO4- + 8H+ + 5e- → Mn2+ + 4H2O) +1.507V 5(2Cl- → Cl2 + 2e-) -1.360V 2MnO4- + 16H+ + 10Cl- → 2Mn2+ + 8H2O + 5Cl2 E°cell =0.147V Ecell = E°cell – 0.0592/n*log {[Mn2+]2(PCl2)5/[MnO4-]2[H+]16[Cl-]10} = 0.147V – (0.0592/10) * log [(0.100)^2(0.100)^5/(0.100)^2(0.100)^16(0.100)^10] = 0.147 V - (0.0592/10) * log (1.00 x 10^21) =


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UT CH 302 - Advanced Electrochemistry Problems

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