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PracticeChap17-1key

# PracticeChap17-1key - 65 The NaOH is a source of base and...

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Recommended Problems for Chapter 17 Problems from Oxtoby: Ch. 11: 27-38 Ch. 17: 1-7, 9, 17-26, 43-47, 49-50, 65-66, 73 Oxtoby is more calculation-focused than we are; these are intended to help you understand the various aspects of the material and not generally representative of exam questions. Questions similar to some of the more conceptual questions could be exam-like. Practicing the numerical computations is an important component of learning these concepts. You will need to solve questions covering the same equations and ideas to do well in the course, just at a lower level of mathmetical difficulty or that emphasize what the equations mean , which practice calculations can help you see. The answers to selected problems from the above will be posted on a later date. You should rely on your own work, your classmates, your GSI, and office hours to determine if you have the correct answers to the other problems. Additional problems will be posted separately on a later date.
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Unformatted text preview: 65: The NaOH is a source of base, and can initially be ignored. Balanced half-reactions for the redox chemistry: 2 Al → 2 Al 3+ + 6 e-6 H 2 O + 6 e-→ 3 H 2 + 6 OH-So the overall redox process is: 2 Al + 6 H 2 O → 2 Al 3+ + 3 H 2 + 6 OH-And the completely balanced process is: 2 Al(s) + 6 NaOH(s) + 6 H 2 O(l) → 2 Al 3+ (aq) + 3 H 2 (g) + 6 OH-(aq) 66: Assuming that the sulfur does not change oxidation state, balanced half-reactions are: 2 e-+ Ag 2 S → 2 Ag + S 2-S 2-+ Zn → ZnS + 2 e-So the overall reaction is: Ag 2 S(s) + Zn(s) → 2 Ag(s) + ZnS(s) 73: 1.5x10 10 kg of Al is ~ 55.5 x 10 10 mol of Al. Al in Al 2 O 3 has a +3 oxidation state, so ~166.5x10 10 mol e-are required. From Q = nF (using F ~ 10 5 C mol-1 ), that means Q ~1.67x10 17 C. ExQ = energy, so 5 V x 1.67x10 17 C ~8.35x10 17 J. Using 1 kWh = 3.6 x 10 6 J, we convert to ~2.3x10 11 kWh. At \$0.1 per kWh, then the cost is ~\$2.3x10 10 , or \$23 billion. 1...
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