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Electron transfer in Redox equation |
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| Dec4-08, 10:22 AM | #1 |
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Electron transfer in Redox equation
1. The problem statement, all variables and given/known data
How many electrons are transferred in the following reaction? 6 Br^- (aq) + Cr2O7^2- (aq) + 14 H^+ --> 2Cr^3+ (aq) + 7H2O + 3Br2 (l) a) 1 b) 2 c) 3 d) 6 e) 14 2. Relevant equations 3. The attempt at a solution Redox reactions really confuse me, but I gave it a shot. I took a look at the number of electrons on each side of the equation and found the following: Left side: 6Br^- = -6 Cr2O7^2- = -2 14 H^+ = +14 Total: +6 Right side: 2Cr^3+ = +6 7H2O = 0 3 Br2 = 0 Total: 6 But this is as far as I can get. Somehow I don't think the answer is D) 6, but I really don't know. Someone please help me understand this problem? |
| Dec4-08, 10:52 AM | #2 |
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Recognitions:
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Easily concentrate your attention to just one half-reaction, assuming the reaction as written is balanced; that being for the bromide. How many electron difference is it from 2 bromides to one bromine(compound, not the separate atoms)? Now, how many electrons change is this for the same half-reaction as represented in the fully written reaction (in which you start with 6 bromides instead of just 2 bromides)?
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| Dec4-08, 10:54 AM | #3 |
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Recognitions:
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In other words, from post #2, your best choice would seem to be D. (6 electrons)
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| Dec4-08, 11:01 AM | #4 |
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Electron transfer in Redox equation
Okay, so D it is. If I understand correctly then in the half-reaction for bromide... there is 1 electron difference between 2 bromides to 1 bromine. As a whole reaction, there are 6 bromides... so 6 electrons.
What about the other half-reaction, though? Cr2O7^2- + 14H^+ --> 2Cr^3+ +7H2O I suppose it is the dichromate ion that confuses me. |
| Dec4-08, 11:22 AM | #5 |
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Recognitions:
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If you know your reaction as written is balanced, then you know that the chromate part of the reaction is also for 6 electrons.
Look at dicrhromate anion. Account for all the charges which give the -2 charge for this anion. Two chromiums, seven oxygens, the sum of the charges must be -2. You want to find the charge on the chromium. You know: 1 oxygen is -2, dichromate anion charge is -2. You do not yet know the charge on 1 chromium in the dichromate anion. This is what you want to find (using simple algebra). |
| Dec4-08, 11:26 AM | #6 |
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Ahhh... I see. Thank you for your help. I understand now. :) Redox reactions make my head spin.
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