| Question 1 |
Q = It = 0.005 C/s * 127s = 0.635 C (1 point) # of e- = 0.635C / 96485 C mol-1 = 6.58 x 10-6 mol mol Cr3+ = 6.58 x 10-6 mol / 3 = 2.19 x 10-6 mol (1 point) [Cr3+] = 2.19 x 10-6 mol / 0.010 L = 2.19 x 10-4 M (1 point) |
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| Question 2 |
using E = Eo - (0.0592/n)
* log (1/[Cr3+]), remember that n = 3, Eo
= -0.740 high end E = -0.760V low end E = -0.858V |
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| Question 3 |
pHunk = pHstandard - (Eunk
- Estandard)/0.0592 = 3.00 - (0.483-0.572)/0.0592 = 4.50 |
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| Question 4 |
(a) Choose any three steps:
You could also say that one of the steps is a pre-reaction. |
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| (b) In any case, rate limitation will give a voltage
decrease (less than Eo) for a galvanic cell. For
an electrolytic cell, more voltage would be required to force the reaction
(greater than Eo) |
Last updated, April 5, 2006
Page maintained by Craig MacKinnon