Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: paul.fr on 10/08/2007 09:46:06
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Also, Are there any toxic fumes released from burning it?
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Try this....
http://ptcl.chem.ox.ac.uk/~hmc/hsci/chemicals/ammonium_dichromate.html
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Would it be for one of these......?
http://www.chem.uiuc.edu/clcwebsite/ammvol.html
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Cheers, i must do my own research...one day. Yes, i was thinking of one of those. But not the one with green chromic acid Cr2O3.
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Also, Are there any toxic fumes released from burning it?
No:
(NH4)2Cr2O7 --> Heat --> N2 + 4H2O + Cr2O3
The only problem with this reaction is that you'll find a lot of Cr2O3 powder around the little volcano, at the end.
The toxicity of (NH4)2Cr2O7 is due to Cr2O7-- ion. The same in chromates: the toxicity is due to CrO4-- ion; the problem is caused by Cr(VI).
Cr2O3 is not chromic acid, it's chromium (III) oxide.
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(NH4)2Cr2O7 --> Heat --> N2 + 4H2O + Cr2O3
Is there a good way to describe the formula here? Could you run through both sides and show how it "balances"?
I never understood it at school, perhaps I'll get it here!
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Cr2O3 is not chromic acid, it's chromium (III) oxide.
Thank you for that correction...one day, just one day i want to get it right..
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(NH4)2Cr2O7 --> Heat --> N2 + 4H2O + Cr2O3
Is there a good way to describe the formula here? Could you run through both sides and show how it "balances"?
I never understood it at school, perhaps I'll get it here!
1. Find the oxidation numbers (O.N.) of all the atoms and take notice of those atoms which change its O.N. In this case we have: N(-III) in (NH4)2Cr2O7 and N(0) in N2 so every N atom has lost 3 electrons --> N has been oxydized. Cr(+VI) in (NH4)2Cr2O7 and Cr(+III) in Cr2O3 so Cr has taken 3 electrons --> Cr has been reduced.
2. Write the 2 semi-reactions of oxidation and reduction:
2NH4 - 6e- --> N2; here it's 6e- because there are 2 atoms of N.
Cr2O7 + 6e- --> Cr2O3; same as up.
3. Balance those semi-reactions. In this case thy are already balanced: the number of electrons involved in the first semi-reaction is = number in the other. If it was, e.g. 3e- in the first and 5e- in the second, you multiply all the first semi-reaction by 5 and all the second by 3, so you have 15 e- in both (minimum common multiple of 3 and 5).
4. Write the reaction with the coefficients you have already computed. In this case they are: 2 for (NH4), 1 for N2, 1 for Cr2O7 and 1 for Cr2O3 (in the other hypotetical case, you should have multiplied the first two coefficients by 5 and the other two by 3).
So you write the still unbalanced reaction but with these coefficients:
(NH4)2Cr2O7 --> N2 + Cr2O3.
5. Balance charges, if there are. Not in this case, so it's already balanced.
6. Balance atoms. You have: 2 atoms of N at left and 2 at right. Ok. 8 H atoms at left and 0 at right, you have to balance adding H2O molecules, so you add 4 H2O molecules at right. 2 Cr atoms at left and 2 at right. Ok. 7 O atoms at left and 4 + 3 = 7 at right. Ok.
Everything balanced, you can write the reaction:
(NH4)2Cr2O7 --> N2 + 4H2O + Cr2O3.
This was a very simple case. In other cases could be less simple to find O.N. and the semi-reactions. Ask if you want to know more.