Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Maigowai on September 11, 2011, 02:58:02 PM
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NaOH(aq) + Cu(NO3)2(aq) -> Cu(OH)2(s) + NaNO3(aq)
Hi I am having great difficulty with this question figuring out which ion is being reduced and oxidized.
I know how to balance redox reactions but I can't figure out the oxidation numbers on each.
Please please please show all your work...including if you have to split up each molecule etc.
Thanks! I really need to know soon!!!
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can you write the half-reactions?
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can you write the half-reactions?
To write the half reactions i need to figure out which is being oxidized etc....which is what i am having trouble with :(
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Do you know rules used when assigning oxidation numbers? Try to apply them. Show what you get.
Could be your trouble is that you don't believe result you are getting. Could be the result you are getting is correct!
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Do you know rules used when assigning oxidation numbers? Try to apply them. Show what you get.
Could be your trouble is that you don't believe result you are getting. Could be the result you are getting is correct!
ok i will try...
NaOH(aq) + Cu(NO3)2(aq) -> Cu(OH)2(s) + NaNO3(aq)
1+2-1+ 1+5+ 6- 2+ 2-1+ 1+ 5+6-
you see i don't know what i am doing wrong...only Cu changes oxidation number (being oxidized) =[
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Sorry for the slow responses, i just clicked ''notify'' so i should respond almost right away now!
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Your equation is not balanced. You have to do that before you can figure out the oxidization numbers.
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Your equation is not balanced. You have to do that before you can figure out the oxidization numbers.
4 HNO3(aq) + 1 Cu(s) -> 2 H2O(l) + 1 Cu(NO3)2(aq) + 2 NO2(g)
I thought that coefficients are not supposed to be mingled with oxidation numbers?
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NaOH(aq) + Cu(NO3)2(aq) -> Cu(OH)2(s) + NaNO3(aq)
Hi I am having great difficulty with this question figuring out which ion is being reduced and oxidized.
I know how to balance redox reactions but I can't figure out the oxidation numbers on each.
Please please please show all your work...including if you have to split up each molecule etc.
Thanks! I really need to know soon!!!
Well, it's pretty easy to solve this, if you know the method.
First of all, you have to write down the Oxidation Number of every atom present in the formulae. To do this, you have to follow some simple rules, to be applied in this same order:
1. The sum of the O.N., each multiplied by its stoichiometric coefficient, is equal to the net charge of the molecule / ion;
2. Ions from 1st group atoms always have +1 as O.N.;
3. Hydrogen has +1 as O.N., unless he forms a hydride;
4. Oxygen has -2 as O.N., unless he's in peroxides (-1) or in F2O (+2).
As stated above, Na has +1 O.N. We haven't hydrides there, so hydrogen will have +1 as O.N., and oxygen -2.
From the first rule, then, it is immediate to get nitrogen's O.N. in nitrate ion:
NO3- : x + (-2) * 3 = -1 :rarrow: x = 5.
And, from that, you get the O.N. of Cu:
Cu(NO3)2: x + 2 * (5 + (-2) * 3) = 0 :rarrow: x = + 2.
Cu(OH)2: x + 2 * (+1 + (-2)) = 0 :rarrow: x = + 2.
So, working out this equation, we obtain:
NaOH(aq) + Cu(NO3)2(aq) :rarrow: Cu(OH)2(s) + NaNO3(aq)
+1-2+1 +2+5-2 +2-2+1 +1+5-2
Looking at the oxidation numbers written above, however, it is immediate that this reaction isn't a redox, because I can't see a change of the Oxidation Numbers in any atom.
This, in fact, is a simple ion exchange reaction:
2 NaOH(aq) + Cu(NO3)2(aq) :rarrow: Cu(OH)2(s) + 2 NaNO3(aq)
I advice you to try with some other reactions. This method requires some practice to be mastered, but it's really nice and fast.
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Your equation is not balanced. You have to do that before you can figure out the oxidization numbers.
Completely wrong.
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Thank you very much for explaining it! It's helped clear this up for my next assignment as well.