[chronicfuture]

Hi all, I am wondering if you can help me out with balancing this long equations:
K₄Fe(CN)₆ + KMnO₄ + H₂SO₄ ------->  Fe₂(SO₄)₃ + K₂(SO₄) + CO₂ + HNO₃ + MnSO₄ + H₂O

Thanks for *delete me*

[FeLiXe]

first take out the spectator ions (K, SO4)

then you have a fairly complex redox reaction

the standard way would be to write the half reactions and combine them

after that, add the spectator ions

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you can always do them with equation systems

[Borek]

IMHO you should forget about going through the redox path - algebraic method mentioned by FeLiXe is the way to go

http://www.chembuddy.com/?left=balancing-stoichiometry&right=algebraic-method

[AWK]

Hi all, I am wondering if you can help me out with balancing this long equations:
K₄Fe(CN)₆ + KMnO₄ + H₂SO₄ ------->  Fe₂(SO₄)₃ + K₂(SO₄) + CO₂ + HNO₃ + MnSO₄ + H₂O

Thanks for *delete me*

Use artifitial oxidation numbers for C and N on the left side (+4 and +5 respectively). You will get a fantastic oxidation number -58 for Fe, but this redox reaction will behave absolutely "prettily", and in a few minutes you will solve the problem without algebraic method or "button-push method" (the highest coefficient 218)

[FeLiXe]

I have never heard about artificial oxidation numbers

how do you know what the artificial oxidation numbers on the right side are?

[AWK]

You shoud use +4 for C and +5 for N on both sides of equation (that of CO2 and HNO3 on the right side). Then you will have +7 for Mn and -58 for Fe on the left side and +2 for Mn and +3 for Fe on the right side. But electron balance will work gently.

[billnotgatez]

i think    -58   is a typo

[AWK]

-58 (minus 58) is the right number for these assumptions, and this number will work 

[billnotgatez]

ok - thanks

[Borek]

It is one of the reasons to not treat ON too seriously

This approach works in this case because it assumes that the Fe(CN)₆^4- is oxidised as a whole molecule - and we are artficially assuming that it is only Fe that happens to accept/give electrons, while C and N are only spectators.

[FeLiXe]

cool, now I get what you guys are talking about

[AWK]

cool, now I get what you guys are talking about

If you understand this method, then try balancing this reaction
C₂H₅OH + Hg + HNO₃ = Hg(ONC)₂ + NO₂ + CO₂ + H₂O

[FeLiXe]

the problem is that you have two different molecules that are reduced. so it does not quite work like above. not even an equation system works.

[FeLiXe]

I did it the straight forward with an equation system without any oxidation numbers. I don't think the reaction is uniquely determined

you could leave out Hg and get

2 C₂H₅OH + 0 Hg + 24 HNO₃ = 0 Hg(ONC)₂ + 24 NO₂ + 4 CO₂ + 18 H₂O

you could leave out NO₂:

14 C₂H₅OH + 12 Hg +  24 HNO₃ = 12 Hg(ONC)₂ + 0 NO₂ + 4 CO₂ + 54 H₂O

any linear combination of these two equations would also work

[AWK]

the problem is that you have two different molecules that are reduced. so it does not quite work like above. not even an equation system works.

But this problem can be easily overcome. Join fulminate and carbon dioxide together in one artificial molecule in any ratio except 1:0 or 0:1 (from a suggested mechanism of fulminate synthesis this ratio is Hg(ONC)₂+2CO₂ (more CO2 can be treated as an oxidation of an ethanol in a concentrate nitric acid). In this artificial molecule HgC₄N₂O₆ force ONs: 0(null) for Hg and +5 for N.
Then ON for C will be +1/2, and only the same two elements (C and N)  change ONs on both sides of equation.