Chem1BExam4SampleTestKey

Chem1BExam4SampleTestKey - Chem 1B Exam 4 Sample Questions...

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Unformatted text preview: Chem 1B Exam 4 Sample Questions Instructions: For all mathematical problems, show your work. Give the basic formula (if there is one), rearrange it if necessary, plug in data with units, round properly to the correct number of significant figures, and place a box around the answer. For essay questions, be complete, but concise. Complete sentences with proper spelling and punctuation are expected. kJ C J Useful information: R = 0.0083145 mol.K ; 1 F = 96,485 mol e- = 96,485 V.mol e- ; 0.0592 V c = 2.9979 × 108 m/s, 1 amu = 1.660540 × 10-27 kg 1. Given the following half-reactions: MnO41-(aq) + 8H1+(aq) + 5e1- → Mn2+(aq) + 4H2O(l) IO41-(aq) + 2H1+(aq) + 2e1- → IO31-(aq) + H2O(l) E° = E° = 1.52 V 1.60 V (a) (8 pts.) Write the balanced equation for the galvanic cell based on these half-reactions. - 2Mn2+(aq) + 8H2O(l ) → 2 MnO41-(aq) + 16H1+(aq) + 10e1- E° = -1.52 V 1.60 V 5IO41-(aq) + 10H1+(aq) + 10e1- → 5IO31-(aq) + 5H2O(l ) E° = 2Mn2+(aq) + 3H2O(l ) + 5IO41-(aq) → 2 MnO41-(aq) + 6H1+(aq) + 5IO31-(aq) (b) (3 pts.) What is E° for the overall reaction? E° = E°ox + E°red = -1.52 V + 1.60 V = 0.08 V 2. (6 pts.) From the half-reactions: Co3+(aq) + e1Fe3+(aq) + e1SO42-(aq) + 4H+(aq) + 2e1Sn2+(aq) + 2e1Ni2+(aq) + 2e1Fe2+(aq) + 2e1→ → → → → → Co2+(aq) Fe2+(aq) H2SO3(aq) + H2O(l) Sn(s) Ni(s) Fe(s) E° = E° = E° = E° = E° = E° = 1.95 V 0.77 V 0.20 V -0.14 V -0.23 V -0.44 V determine which of the following species (standard conditions): Fe3+, Co3+, Fe, Ni2+, Sn (a) is the strongest oxidizing agent? (b) is the strongest reducing agent? (c) can be oxidized by SO42-(aq) in acid? Co3+(aq) Fe(s) Sn(s) and Fe(s) 3. Consider the following cell: Al(s) + 3Cr3+(aq) (a) (6 pts.) Calculate ∆G°. ∆G° = -nFE° 96485 J ∆G° = -3mol e1- × V × mol e1- × 1.25 V ∆G° = -3.62 × 105 J → Al3+(aq) + 3Cr2+(aq) E° = 1.25 V (b) (8 pts.) Calculate the cell potential if [Cr3+] = 0.0250 M, [Cr2+] = 0.875 M and [Al3+] = 1.50 M. 0.0592 V [Al3+][Cr2+]3 E = E° × log [Cr3+]3 n 0.0592 V (1.50)(0.875)3 E = 1.25 V × log (0.0250)3 1.25 V 3 E = 1.25 V 0.0592 V × log(6.43 × 104) 3 E = 1.25 V - 0.0949 V = 1.16 V 4. (9 pts.) A platinum rod is placed in a beaker containing a solution of chromium(II) nitrate and chromium(III) nitrate. Another beaker has a nickel rod dipping into a solution of nickel(II) nitrate. The metal rods are connected to each other with a wire and the solutions in the two beakers are connected with a salt bridge. The resulting spontaneous reaction has the nickel rod as the cathode. Describe, using proper chemical terminology, all the changes that occur in this galvanic cell. Note: Your description must be with words; reaction equations and drawings will not be graded. Nickel(II) ion is reduced at the cathode to Ni(s), resulting in an increase in the mass of the nickel rod. Electrons travel through the external circuit from the platinum electrode to the nickel cathode. At the anode chromium(II) ions are oxidized to chromium(III). During the course of the redox reaction nickel(II) ions decrease in concentration in the nickel beaker and chromium(II) ions decrease while chromium(III) ion increase in concentration in the chromium beaker. To maintain electrical neutrality, nitrate ions will move through the salt bridge from the nickel beaker to the chromium beaker. 5. (8 pts.) An antique automobile bumper is to be chrome plated. The bumper is dipped into an acidic Cr2O72- solution and serves as one of the electrodes of an electrolytic cell. If the current is 10.00 amperes, how long will it take to deposit 100.0 g of Cr(s) on to the bumper? Cr6+ → Cr0 1 mol Cr 6 mol e1- 96485 C 1s 100.0 g Cr × 51.996 g Cr × mol Cr × mol e1- × 10.00 C = 1.113 × 105 s 6. (12 pts.) Write a balanced nuclear equation for: (a) spontaneous fission of 254Cf to 106Mo, four neutrons and another nuclide 98 42 254 Cf 98 → 106 Mo 42 + 41n + 0 144 Ba 56 (b) alpha particle capture followed by neutron emission for 239Pu 94 239 Pu 94 + 4He → 2 243 Cm 96 → 242 Cm 96 + 1n 0 (c) neutron capture followed by beta emission for 75As 33 75 As 33 + 1n → 0 76 As 33 → 76 Se 34 + 0 e -1 7. (8 pts.) Predict the type of nuclear decay 8B is expected to undergo, explain your reasoning, and 5 give the equation for the decay process. In this region of the periodic table, stable nuclei have an equal number of neutrons and protons. This proton rich nucleus can undergo either positron emission or electron capture, both generating a neutron at the expense of a proton. 8 B 5 8 B 5 → 8 Be 4 + 0e 1 + -0e → 8Be 1 4 8. (8 pts.) Compare positron emission with β-particle emission. Be as complete as possible. Positron emission occurs through decay of a proton to the positron and a neutron: 1 1 p → 0 n + 1e 1 0 This type of decay occurs with proton rich nuclides. The result is the product nuclide has the same mass number but the atomic number decreases by one compared to the starting nuclide. β -Particle emission occurs through decay of a neutron to the β -particle and a proton: 1 0 n → -1 e + 1 p 0 1 This occurs with neutron rich nuclides and gives a product nuclide with the same mass number but the atomic number increases by one compared to the starting nuclide. 9. (8 pts.) Fresh water contains enough tritium (3H) to show 5.5 decompositions per minute per 100 g of water. Tritium has a half-life of 12.3 years. A vintage wine shows 0.14 decompositions per minute per 100 g. How old is the wine? N -0.6931 lnNo = -kt = t1/2 × t N -t1/2 × lnNo 0.6931 = t 0.14 decompositions /min -12.3 y × ln 5.5 decompositions /min t= = 65 y 0.6931 The wine is 65 years old. 10. (8 pts.) Is the following reaction expected to be spontaneous? Explain your reasoning. 1 H + 18O → 19F 1 8 9 1 18 Nuclide masses are: 1H 1.00728 amu, 8O 17.99477 amu, and 19F 18.99346 amu. 9 ∆E = mass of products – mass of reactants ∆E = 18.99346 amu – (1.00728 amu + 17.99477 amu) = -0.00859 amu This nuclear process involves a loss of 0.00859 amu due to energy released by the reaction. This exothermic nuclear process is spontaneous. 11. (8 pts.) Compare fission with fusion and explain why one requires elevated temperatures while the other does not. Fission is the splitting of a heavy nucleus into two smaller nuclei. This is accomplished by bombardment of the heavy nucleus with neutrons. Since the neutron has no charge, there is no repulsion between it and the heavy nucleus. Fusion is the merging of two light nuclei to generate a heavier nucleus. Because both starting nuclei are positively charged, there is a strong electrostatic repulsion force which must be overcome. This high activation energy is accomplished only at extremely high temperatures. ...
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This note was uploaded on 12/21/2010 for the course CHEM 1B taught by Professor Garyj.decicco during the Fall '10 term at Santa Rosa.

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