HW_6 - CHEM 150 HOMEWORK #6 Due: Tuesday, 7/22/08 1. A...

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CHEM 150 HOMEWORK #6 Due: Tuesday, 7/22/08 1. A 0.1002 g sample containing only CCl 4 and CHCl 3 was dissolved in methanol and electrolyzed at the surface of a mercury electrode at –1.0 V, which required 50.20 s at a constant current of 0.200 A. The potential of the cathode was then adjusted and held constant at –1.80 V. Completion of the titration at this potential required 382.35 s at a constant current of 0.200 A. Calculate the respective percentages of CCl 4 and CHCl 3 in the mixture. The electrolysis reactions are given below. 2CCl 4 + 2H + + 2e - + 2Hg(l) 2CHCl 3 + Hg 2 Cl 2 (s) [occurs at –1.0V] 2CHCl 3 + 6H + + 6e - + 6Hg(l) 2CH 4 + 3Hg 2 Cl 2 (s) [occurs at –1.80V] 2. Phenol (94.11 g/mol) undergoes a quantitative reaction with bromine. 3Br 2 + 3H 2 O + phenol phenol·Br 3 + 3Br - + 3H 3 O + The bromine needed to react with the phenol in a 50.0 mL sample of disinfectant was generated electrically. The complete reaction required 10.40 min at a constant current of 80.0 mA (F =
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This note was uploaded on 08/06/2008 for the course CHEM 150 taught by Professor Buratto during the Summer '08 term at UCSB.

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