Sample MT II Solutions W06

Sample MT II Solutions W06 - Solutions Chemistry 123...

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Solutions Chemistry 123 Midterm II Sample 1) E No oxidation number changes in the other choices 2) D E 0 = 0.154V – (-0.74V) = + 0.894 (A and C are not in fact redox reactions, limiting the choices). 3) C (1.0 amp)(2.0 Hr)(3600sec/Hr) = 7,200 amp-sec = 7,200 Coulomb (7,200 Coulomb)(1 / 96,500 Coulomb) = .075 F = .075 moles e- (3.93g)(1 mole/157 g) = 0.025 mole Gd. (.075 moles e-)/ 0.025 mole Gd = 3.0 4) D Cr 2 0 7 -2 : -14 from 7 oxygens. Cr must be +6 (total of + 12) 5) C E = E 0 - .0592/2 x log [dilute]/[concentrated] E 0 = 0 for concentration cell 6) A E 0 cell = E 0 redn - E 0 oxid 0.46 = E 0 redn (+1.36) and E 0 redn = 1.82V 7) C (1.0g)(1 mole / 107 g) = 0.0093 moles Ag (0.0093 moles Ag)(1.0 mole e- / mole Ag) (1 / mole e-) (96,500 Coulomb /1 F ) = 897.4 Coulombs = 897.4 amp-sec. 897.4 amp-sec / 30 amp = 29.9 sec 8) C Page 811 9) B 2e- + S 4 O 6 -2 2 S 2 O 3 -2 All that’s needed is to balance charge 10) A E 0 cell = E 0 redn - E 0 oxid E 0 cell = -0.44 – ( - 0.74) = 0.30 V 11) A Page 786 12)
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Sample MT II Solutions W06 - Solutions Chemistry 123...

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