Fried Zygote Sandwich
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Redox Chemistry, a little help please?
I'm pretty tired right now so I'm not thinking very clearly so this is probably stupidly simple, but I've got a bit of a hangup on a
question I'm doing.
I'm suppposed to balance this equation assuming it's in an acid solution (in the presence of H+ ions)
2CuS(s) + HNO3 --> Cu(NO3)2 + H20 + NO (g) + S (s)
Ordinarily I'd balance the half reactions (excluding H and O), add H20 to compensate for O, H+ to compensate for H, and electrons to compensate
for charge. But this equation already seems half done. The H2O has been accounted for, etc.
I tried just straight up balancing the equation but it's sort of unruly. Trying to break it apart into the half reactions also gave me troubles.
Any ideas?
Thanks.
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woelen
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3CuS + 8HNO3 ---> 3Cu(NO3)2 + 4H2O + 2NO + 3S
How I did this?
Just write linear equations for all of the elements.
x CuS + y HNO3 ---> p Cu(NO3)2 + q H2O + r NO + s S
For Cu: x == p
For S: x == s
For H: y == 2*q
For N: y == 2*p + r
For O: 3*y == 2*3*p + q + r
Now you have 5 equations with six unknowns. Take one of the unknowns equal to 1 and then solve the set of equations. In the resulting solution,
multiply, such that all solutions become integers.
What I described above is the 'highschool-level' method. I have written a program, which can do all these things automatically. The program
works along the lines I scetched above, but it uses the more general concept of null-space of a matrix. That concept is not suitable for
hand-calculations, but very useful for writing a computer program.
Here follows a link to my program. Feel free to play around with it.
http://woelen.scheikunde.net/science/chem/chemeq/index.html
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vulture
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Solving redox equations simply mathematically is dangerous. Half reactions are the way to go.
One shouldn't accept or resort to the mutilation of science to appease the mentally impaired.
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Fried Zygote Sandwich
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How would I create half equations for this?
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guy
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3CuS --> 3Cu2+ + 3S + 6e-
8H+ + 2NO3- + 6e- --> 2NO + 4H2O
---------------------------------------
3CuS + 8H+ + 2NO3- ----> 3Cu2+ + 3S + 2NO + 4H2O
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woelen
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| Quote: | Originally posted by vulture
Solving redox equations simply mathematically is dangerous. Half reactions are the way to go. |
I only agree, if you do not know the reaction products very well. The chemistry part is determining what is formed, when certain reactants are mixed
under certain conditions.
Once, you KNOW your products, then the purely methematical way of solving the equation is a guaranteed sure fire, no misses at all, although for
handwork it may not be the easiest one. The computer program I've written, however, does not have a table of of halfreactions (even better, it
can generate half reactions  if you know the products).
Another advantage of using the mathematical method is that it imemdiately detects possible non-stoichiometry in a reaction.
The reaction we are talking about here is a good example.
In reality you will get NO and NO2, depending on conditions. More concentrated acid will give NO2.
If you attempt to solve the equation, with both NO and NO2 at the right hand side, then an infinity of solutions is possible (try it!). The
mathematical solution sees that the null-space of the coefficient matrix has dimension larger than 1 and hence gives a warning. My program also
detects that and gives a warning and provides N independent solutions, with N being the dimension of the solution-space.
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