Balancing Organic Redox Reactions

[f***@hotmail.com]

Why is so little attention paid to balancing organic redox reactions in
chemistry courses? Is there a _systematic_ way of finding the balancing
coefficients? I use a rather artificial way for simple cases i.e
without assigning or identifying oxidation number of carbon undergoing
reduction or oxidation. The reaction of iron(III) with hydroquinone
HO-Ph-OH:

eg: Fe(3+) + 2HO-Ph-OH ----> Fe(2+) + 2O=Ph=O + 2H(+)

This is the artificial way:
Fe(3+) + e -------> Fe(2+)

HO-Ph-OH -------> O=Ph=O + 2H(+) Now in order to balance the charge
we add 2 electrons on the right without calculating oxidation state of
relevant carbon atom.

HO-Ph-OH -------> O=Ph=O + 2H(+) + 2e

So the final equation is

2Fe(3+) + HO-Ph-OH -------> O=Ph=O + 2H(+) + Fe(2+)


What to do with the following complex examples such as this one in
BASIC MEDIUM?
K3V5O14 + HOCH2CHOHCH2OH ------> VO(OH)2 + HCOO(-) + K(+)

the inorganic part is relatively easy:
K3V5O14 + 9H2O + 5e -----> 5VO(OH)2 + 8OH(-) + 3K(+)

How to deal with the organic part in BASIC medium, the above trick for
seems to fail here? Any suggestions for balanicng the organic part: The
answer is given in the book for the organic part is.
C3H8O3 + 11OH(-) ----> HCO2(-) + 8e + 8H2O
but how did the authors come up with 8 electrons?

Googled for organic redox reactions but the technique I understood is
very cubersome and requires that the carbon atom/s undergoing oxdiation
and reduction be identified first (that way does give the answer for 8
electrons) but are there any short-cuts or easier techniques?(Please no
algebraic methods).

Thanks.

[Bob]

Post by f***@hotmail.com
Why is so little attention paid to balancing organic redox reactions in
chemistry courses? Is there a _systematic_ way of finding the balancing
coefficients? I use a rather artificial way for simple cases i.e
without assigning or identifying oxidation number of carbon undergoing
reduction or oxidation. The reaction of iron(III) with hydroquinone
eg: Fe(3+) + 2HO-Ph-OH ----> Fe(2+) + 2O=Ph=O + 2H(+)
Fe(3+) + e -------> Fe(2+)
HO-Ph-OH -------> O=Ph=O + 2H(+) Now in order to balance the charge
we add 2 electrons on the right without calculating oxidation state of
relevant carbon atom.
HO-Ph-OH -------> O=Ph=O + 2H(+) + 2e
So the final equation is
2Fe(3+) + HO-Ph-OH -------> O=Ph=O + 2H(+) + Fe(2+)
What to do with the following complex examples such as this one in
BASIC MEDIUM?
K3V5O14 + HOCH2CHOHCH2OH ------> VO(OH)2 + HCOO(-) + K(+)

Given the massive oxidation of the organic part, why not go back and
use the usual "inorganic" method. It works fine for organic compounds;
it's just that what you did above with the phenol is simpler for
simple redoxes of complex molecules.

Remember that you will be calculating an average ox# for the C in the
reactant, and it may well come out as a fraction. No problem.

bob

Post by f***@hotmail.com
K3V5O14 + 9H2O + 5e -----> 5VO(OH)2 + 8OH(-) + 3K(+)
How to deal with the organic part in BASIC medium, the above trick for
seems to fail here? Any suggestions for balanicng the organic part: The
answer is given in the book for the organic part is.
C3H8O3 + 11OH(-) ----> HCO2(-) + 8e + 8H2O
but how did the authors come up with 8 electrons?
Googled for organic redox reactions but the technique I understood is
very cubersome and requires that the carbon atom/s undergoing oxdiation
and reduction be identified first (that way does give the answer for 8
electrons) but are there any short-cuts or easier techniques?(Please no
algebraic methods).
Thanks.

[f***@hotmail.com]

Post by Bob

Post by f***@hotmail.com
What to do with the following complex examples such as this one in
BASIC MEDIUM?
K3V5O14 + HOCH2CHOHCH2OH ------> VO(OH)2 + HCOO(-) + K(+)

Given the massive oxidation of the organic part, why not go back and
use the usual "inorganic" method. It works fine for organic

compounds;

Post by Bob
it's just that what you did above with the phenol is simpler for
simple redoxes of complex molecules.
Remember that you will be calculating an average ox# for the C in the
reactant, and it may well come out as a fraction. No problem.
bob

Post by f***@hotmail.com
answer is given in the book for the organic part is.
C3H8O3 + 11OH(-) ----> HCO2(-) + 8e + 8H2O
but how did the authors come up with 8 electrons?

C3H8O3 ----> 3HCO2(-)
Lets try your technique for C3H8O3:

3x + 8(+1) + 3(-2) = 0
So 3x = 6-8 = -2, (ignore individual ox. state of C. we would be
interested in overall change in oxidation numbers)

HCO2(-)
1 + x + 2(-2) = -1
x = -1 + 4 + -1= +2

There need to be 3 HCO2(-)
C3H8O3 [Ox# = -2] -------> 3HCO2(-) [3 X 2= +6]

-2 to +6 equivalent to a loss of 8 electrons.

Thats excellent...simple.

Now how about balancing in acidic medium first and then adding OH(-) on
both sides?

[number6]

Post by f***@hotmail.com
Thats excellent...simple.
Now how about balancing in acidic medium first and then adding OH(-) on
both sides?

I should have read further and saved myself some typing ...
You got it ... What gets tough is when there are multiple organics ...
use the same method ... but the algebra gets trickier ... Diophantine
Equations ...
One can always move protons, hydroxides and waters around for
convenience ...

[number6]

Post by f***@hotmail.com
What to do with the following complex examples such as this one in
BASIC MEDIUM?
K3V5O14 + HOCH2CHOHCH2OH ------> VO(OH)2 + HCOO(-) + K(+)
K3V5O14 + 9H2O + 5e -----> 5VO(OH)2 + 8OH(-) + 3K(+)
How to deal with the organic part in BASIC medium, the above trick for
seems to fail here? Any suggestions for balanicng the organic part: The
answer is given in the book for the organic part is.
C3H8O3 + 11OH(-) ----> HCO2(-) + 8e + 8H2O

I use the same half cell reactions for all ...

C3H8O3 = HCO2(-)

Balance carbons ...

C3H8O3 = 3HCO2 (-)

Balance O with water

3H2O + C3H8O3 = 3HCO2 (-)

Balance H with H(+)

3H2O + C3H8O3 = 3HCO2 (-) + 11 H(+)

Balance charge with electrons ...

3H2O + C3H8O3 = 3HCO2 (-) + 11 H(+) + 8e(-)
(there's the 8 electrons ...)

Make it alkaline ...
3H2O + C3H8O3 + 11 OH(-) = 3HCO2 (-) + 11 H(+) + 11OH(-) 8e(-)

Collect terms ...

C3H8O3 + 11 OH(-) = 3HCO2 (-) + H2O + 8 e (-)

As Kelly Bundy would say ... Viola'

[f***@hotmail.com]

Post by number6

Post by f***@hotmail.com
What to do with the following complex examples such as this one in
BASIC MEDIUM?
K3V5O14 + HOCH2CHOHCH2OH ------> VO(OH)2 + HCOO(-) + K(+)
K3V5O14 + 9H2O + 5e -----> 5VO(OH)2 + 8OH(-) + 3K(+)
How to deal with the organic part in BASIC medium, the above trick

for
The

Post by f***@hotmail.com
answer is given in the book for the organic part is.
C3H8O3 + 11OH(-) ----> HCO2(-) + 8e + 8H2O

I use the same half cell reactions for all ...
C3H8O3 = HCO2(-)
Balance carbons ...
C3H8O3 = 3HCO2 (-)
Balance O with water
3H2O + C3H8O3 = 3HCO2 (-)
Balance H with H(+)
3H2O + C3H8O3 = 3HCO2 (-) + 11 H(+)
Balance charge with electrons ...
3H2O + C3H8O3 = 3HCO2 (-) + 11 H(+) + 8e(-)
(there's the 8 electrons ...)
Make it alkaline ...
3H2O + C3H8O3 + 11 OH(-) = 3HCO2 (-) + 11 H(+) + 11OH(-) 8e(-)
Collect terms ...
C3H8O3 + 11 OH(-) = 3HCO2 (-) + H2O + 8 e (-)
As Kelly Bundy would say ... Viola'

Thanks, this is the way I wanted, altogether avoiding oxidation
numbers. Note that I have learnt your ellipses...rather involuntary in
my writings.
I can not find a J.Chem.Ed paper here in the mess, where a really
complex redox equation of naphthalene is balanced by a rather odd
method (perhaps breaking into hypothetical radicals-to further confuse
the poor), I would try to find it and post to get suggestions on it.

[Lloyd Parker]

Post by number6

Post by f***@hotmail.com
What to do with the following complex examples such as this one in
BASIC MEDIUM?
K3V5O14 + HOCH2CHOHCH2OH ------> VO(OH)2 + HCOO(-) + K(+)
K3V5O14 + 9H2O + 5e -----> 5VO(OH)2 + 8OH(-) + 3K(+)
How to deal with the organic part in BASIC medium, the above trick

for
The

Post by f***@hotmail.com
answer is given in the book for the organic part is.
C3H8O3 + 11OH(-) ----> HCO2(-) + 8e + 8H2O

I use the same half cell reactions for all ...
C3H8O3 = HCO2(-)
Balance carbons ...
C3H8O3 = 3HCO2 (-)
Balance O with water
3H2O + C3H8O3 = 3HCO2 (-)
Balance H with H(+)
3H2O + C3H8O3 = 3HCO2 (-) + 11 H(+)
Balance charge with electrons ...
3H2O + C3H8O3 = 3HCO2 (-) + 11 H(+) + 8e(-)
(there's the 8 electrons ...)
Make it alkaline ...
3H2O + C3H8O3 + 11 OH(-) = 3HCO2 (-) + 11 H(+) + 11OH(-) 8e(-)
Collect terms ...
C3H8O3 + 11 OH(-) = 3HCO2 (-) + H2O + 8 e (-)
As Kelly Bundy would say ... Viola'

Kelly would probably say, "What's French for 'Voila'?"

[number6]

Post by Lloyd Parker

Post by number6
As Kelly Bundy would say ... Viola'

Kelly would probably say, "What's French for 'Voila'?"

I'll always remember an episode with her frantically running through
the streets of London looking for help with her English-American
dictionary ... and the Lassie theme music playing in the background ...

[Lloyd Parker]

Post by number6

Post by Lloyd Parker

Post by number6
As Kelly Bundy would say ... Viola'

Kelly would probably say, "What's French for 'Voila'?"

I'll always remember an episode with her frantically running through
the streets of London looking for help with her English-American
dictionary ... and the Lassie theme music playing in the background ...

Bud: "Kelly, call 911."
Kelly: "What's the number?"

[Repeating Rifle]

Post by f***@hotmail.com
Why is so little attention paid to balancing organic redox reactions in
chemistry courses? Is there a _systematic_ way of finding the balancing
coefficients? I use a rather artificial way for simple cases i.e
without assigning or identifying oxidation number of carbon undergoing
reduction or oxidation. The reaction of iron(III) with hydroquinone

To me, the answer is obvious even though I am not a professionl chemist. If
you know the chemistry of the reactants, balancing equations is an excercise
in the solving of simultaneous linear algebraic equations. Period. Using the
term *organic* is a red herring. It can be done using Maple or Mathematica
or any similar program. It is possible that there are multiple ways of
balancing the equations.

All this stuff with electron transfer and half equations is done to avoid
the dog work involved with hand solution of large systems of equations.
Modern computers will handle a reaction involving 5 reactants and 15 kinds
of atoms easily.

In short, know your chemistry so that you know reactants and their reaction
products under the various conditions and let your Maple toss out integer
values for the number of molecules for each reactant and product.

Bill

[number6]

Post by Repeating Rifle
All this stuff with electron transfer and half equations is done to avoid
the dog work involved with hand solution of large systems of

equations.

Post by Repeating Rifle
Modern computers will handle a reaction involving 5 reactants and 15 kinds
of atoms easily.

Au contraire ... the half reactions and electron transfer show the
understanding of what is going on ...The purpose is not to get the
numbers right ... Afterall Deep Thought found the secret to the life
the universe and everything to be 42 ... it got the number right ...
but then had to take millions more years to determine the question ...
A reaction with 5 reactants and 15 kinds of atoms ... would indeed be
just an algebraic exercise ...without basis in reality ...

[Repeating Rifle]

Post by Repeating Rifle

Post by Repeating Rifle
All this stuff with electron transfer and half equations is done to

avoid

Post by Repeating Rifle
the dog work involved with hand solution of large systems of

equations.

Post by Repeating Rifle
Modern computers will handle a reaction involving 5 reactants and 15

kinds

Post by Repeating Rifle
of atoms easily.

Au contraire ... the half reactions and electron transfer show the
understanding of what is going on ...The purpose is not to get the
numbers right ... Afterall Deep Thought found the secret to the life
the universe and everything to be 42 ... it got the number right ...
but then had to take millions more years to determine the question ...
A reaction with 5 reactants and 15 kinds of atoms ... would indeed be
just an algebraic exercise ...without basis in reality ...

I thought that I was making the same point. Balancing equations is not the
same as knowing the chemistry. If you know the chemistry, what amounts to
valence changes and reaction products, the rest is practially rote. You have
to know wheter, in unbalance form, you get C + O2 -> CO2, or CO. That is
chemistry or experiment, not balancing.

Bill

[number6]

Post by Repeating Rifle
I thought that I was making the same point. Balancing equations is not the
same as knowing the chemistry.

I thought your point was to just use a program to balance equations ...
and mine was to write the half cell reactions whilst doing so ... to
understand the chemistry involved ...

[f***@hotmail.com]

Some complex organic redox from various J.Chem.Ed. articles.

Oxidation of naphthalene with concentrated sulfuric acid:
I wish it were possible to draw here, so the equation goes in words.

H2SO4 + Naphthalene ------> Phthalic acid + CO2 + H2O + SO2

C10H8 + H2SO4 -------> C6H4(COOH)2 + CO2 + SO2 + H2O

The authors balance the equations with a very cubersome method, i.e
balacing with hypothetical free radicals with complex rules of their
own.

Trying the conventional method:
(H2SO4 + 2e + 2H(+) ------> 2H2O + SO2) X 9

C10H8 + 8H2O -------> C6H4(COOH)2 + 2CO2 + 18H(+) + 18e

9H2SO4 + C10H8 -----> 10H2O + 9SO2 + C6H4(COOH)2 + 2CO2

which is correct; instead going into the complex free radical
procedure, our conventioanl method works fine here.


Now this one is a killer:
10 K4[Fe(CN)6] + 122 KMnO4 + 299 H2SO4 ---------->

162 KHSO4 + 5Fe2(SO4)3 + 122 MnSO4 + 60HNO3 + 60CO2 + 188H2O

Can this be balanced manually?


A rather simple looking: P2I4 + P4 + H2O ---> PH4I + H3PO4

is quite confusing. The authors balance it by algebraic method by
using,

aP2I4 + bP4 + cH2O ---> dPH4I + eH3PO4
and write 4 equations:

For P : 2a + 4b = d + e
For I : 4a = d
For H : 2c = 4d + 3e
For O : c = 4e

And solve it.

How to solve the above conventially when one does not know which P is
being converted into PH4I or H3PO4?

Thanks.

[number6]

Post by f***@hotmail.com
10 K4[Fe(CN)6] + 122 KMnO4 + 299 H2SO4 ---------->
162 KHSO4 + 5Fe2(SO4)3 + 122 MnSO4 + 60HNO3 + 60CO2 + 188H2O
Can this be balanced manually?

I'll try later ... It may be possible albeit complicated ...
the half(third .. ?? quarter??) cell reactions are ...

1) MnO4- = Mn(2+)
2) CN(-) = CO2
3) CN(-) = NO3 (-)
4) Fe (2+) = Fe(3+)
Sulfate is a spectator ion ...
2 and 3 may be combinable ...

Post by f***@hotmail.com
A rather simple looking: P2I4 + P4 + H2O ---> PH4I + H3PO4
is quite confusing. The authors balance it by algebraic method by
using,
aP2I4 + bP4 + cH2O ---> dPH4I + eH3PO4
For P : 2a + 4b = d + e
For I : 4a = d
For H : 2c = 4d + 3e
For O : c = 4e
And solve it.
How to solve the above conventially when one does not know which P is
being converted into PH4I or H3PO4?

I did it thusly

half cells
32e(-) + 32H(+)+ 3P4 + 2P2I4 = 8PH4I
16H2O + P4 = 4H3PO4 + 20H(+)+ 20e(-)

P4 being both oxidized and reduced ...

I hope there are no math errors ...

[f***@hotmail.com]

Post by number6
I did it thusly
half cells
32e(-) + 32H(+)+ 3P4 + 2P2I4 = 8PH4I
16H2O + P4 = 4H3PO4 + 20H(+)+ 20e(-)
P4 being both oxidized and reduced ...
I hope there are no math errors ...

Speaks volumes of your graduate school.
How & why did you think of the first part, did it involve some special
phosphorous chemistry?
32e(-) + 32H(+)+ 3P4 + 2P2I4 = 8PH4I

The answer given there is:
13P4 + 10P2I4 + 128 H2O -------> 40PH4I + 32H3PO4
and is readily obtained from your half reactions.

[number6]

Post by f***@hotmail.com

Post by number6
I did it thusly
half cells
32e(-) + 32H(+)+ 3P4 + 2P2I4 = 8PH4I
16H2O + P4 = 4H3PO4 + 20H(+)+ 20e(-)
P4 being both oxidized and reduced ...
I hope there are no math errors ...

Speaks volumes of your graduate school.
How & why did you think of the first part, did it involve some

special

Post by f***@hotmail.com
phosphorous chemistry?
32e(-) + 32H(+)+ 3P4 + 2P2I4 = 8PH4I
13P4 + 10P2I4 + 128 H2O -------> 40PH4I + 32H3PO4
and is readily obtained from your half reactions.

Actually ... there was a similar problem in Shaums outline of College
Chemistry from 1966 Freshman Year that I still remember ... :-) I
remember because the arrogant head of the Chemistry department in a
problem solving session stated unequivocally that it was unsolvable
using the usual method ... When I showed him up ... and how easy it
really was ... I earned his emnity for life ... which was not a problem
at all ... as other faculty members took a shine to me exactly for
doing so ... He was on a high horse for them also ... and loved seeing
him taken down a peg ...

[Bob]

Post by number6

Post by f***@hotmail.com
10 K4[Fe(CN)6] + 122 KMnO4 + 299 H2SO4 ---------->
162 KHSO4 + 5Fe2(SO4)3 + 122 MnSO4 + 60HNO3 + 60CO2 + 188H2O
Can this be balanced manually?

I'll try later ... It may be possible albeit complicated ...
the half(third .. ?? quarter??) cell reactions are ...
1) MnO4- = Mn(2+)
2) CN(-) = CO2
3) CN(-) = NO3 (-)
4) Fe (2+) = Fe(3+)
Sulfate is a spectator ion ...
2 and 3 may be combinable ...

Obviously the half cell method works most easily when the number of
halves is two, one gaining and one losing electrons. One can write
this one that way -- using a creative choice of ox #.

For example... I choose to assign ALL C an ox # of +4, and ALL N +5. I
chose these to agree with convention on the RHS. But it has a
consequence for the LHS. It now logically follows that the LHS Fe has
an ox# of -58. The useful consequence is that the entire eqn can now
be described by two half cells, simplified as:

Fe(-58) --> Fe(+3) + 61 e(-)

and the conventional Mn(+7) + 5e(-) -->Mn(+2)

Thus we see that Fe:Mn will be 5:61, which is reflected in the answer
shown above. I'll leave it to others to check that the numbers all
work out.


As has already been noted in the thread, ox# serve two purposes. At
one level, assigning realistic ox# helps us visualize the reaction.
But we also use ox# for bookkeeping while balancing. And for that
purpose, they need not be realistic. Choose ox# that make your life
easy -- given the purpose.


A few years ago some prof published an article in JCE about
difficult-to-balance redox eqns. One that he included he claimed that
he had never seen anyone balance it. I did it, in a few minutes using
the approach above. I think I ended up with Cr at +60 or so (it had
lost each of its electrons twice!). (As with the eqn above, it had too
many things going on -- and lots of big numbers.) I wrote the guy
privately, and received no reply. Others responded with letters to the
editor; the original author found the method quite objectionable, as
it used unrealistic ox#. Clearly, he was unable to separate the two
roles of ox#. (I wonder if this is the same guy number6 referred to.)

bob

[Wilco Oelen]

***@hotmail.com wrote:
<text snipped for brevity>

Post by f***@hotmail.com
A rather simple looking: P2I4 + P4 + H2O ---> PH4I + H3PO4
is quite confusing. The authors balance it by algebraic method by
using,
aP2I4 + bP4 + cH2O ---> dPH4I + eH3PO4
For P : 2a + 4b = d + e
For I : 4a = d
For H : 2c = 4d + 3e
For O : c = 4e
And solve it.
How to solve the above conventially when one does not know which P is
being converted into PH4I or H3PO4?

For balancing the equation it does not matter, which P goes into PH4I
and which goes into H3PO4. It is just a matter of linear algebra,
nothing more, nothing less.
Of course, from a mechanistic point of view, it does matter which P
goes where, but that question cannot be solved by math. That is a
question of chemistry.

Nothing special about the reaction equation itself:
10*P2I4 + 13*P4 + 128*H2O --> 40*PH4I + 32*H3PO4

At a more advanced level of chemisrty, I think that we should not worry
about how to balance equations. That is a fairly dumb, tedious and
error prone job, but nothing special.


-----------------------------

As others already stated, the real chemistry question is the
determination of what is formed. A nice example is the following:

Cu + HNO3 --> H2O + Cu(2+) + NO3(-) + NO + NO2

The experimental conditions determine the ratio of NO and NO2 formed.
The math just gives two linearly independent solutions:

1) 3*Cu + 8*HNO3 ---> 4*H2O + 3*Cu(2+) + 6*NO3(-) + 2*NO
2) Cu + 4*HNO3 ---> 2*H2O + Cu(2+) + 2*NO3(-) + 2*NO2

The real solution will be a linear combination of (1) and (2). The
factors for (1) and (2) are determined by the experimental conditions.
What affects these factors, that is a question of chemistry.

Wilco

[f***@hotmail.com]

Post by Wilco Oelen
<text snipped for brevity>

Post by f***@hotmail.com
A rather simple looking: P2I4 + P4 + H2O ---> PH4I + H3PO4
is quite confusing. The authors balance it by algebraic method by
using,
aP2I4 + bP4 + cH2O ---> dPH4I + eH3PO4
For P : 2a + 4b = d + e
For I : 4a = d
For H : 2c = 4d + 3e
For O : c = 4e
And solve it.
How to solve the above conventially when one does not know which P is
being converted into PH4I or H3PO4?

For balancing the equation it does not matter, which P goes into PH4I
and which goes into H3PO4. It is just a matter of linear algebra,
nothing more, nothing less.
Of course, from a mechanistic point of view, it does matter which P
goes where, but that question cannot be solved by math. That is a
question of chemistry.
10*P2I4 + 13*P4 + 128*H2O --> 40*PH4I + 32*H3PO4
At a more advanced level of chemisrty, I think that we should not worry
about how to balance equations. That is a fairly dumb, tedious and
error prone job, but nothing special.
-----------------------------
As others already stated, the real chemistry question is the
Cu + HNO3 --> H2O + Cu(2+) + NO3(-) + NO + NO2
The experimental conditions determine the ratio of NO and NO2 formed.
1) 3*Cu + 8*HNO3 ---> 4*H2O + 3*Cu(2+) + 6*NO3(-) + 2*NO
2) Cu + 4*HNO3 ---> 2*H2O + Cu(2+) + 2*NO3(-) + 2*NO2
The real solution will be a linear combination of (1) and (2). The
factors for (1) and (2) are determined by the experimental

conditions.

Post by Wilco Oelen
What affects these factors, that is a question of chemistry.
Wilco

Wilco,
I know that your program "chemeq" can balance all the examples
mentioned above. I have not tried it for organic compounds. But I was
doing it for fun...pure _chemical_ entertainment.

[Wilco Oelen]

But I was doing it for fun...pure _chemical_ entertainment.

I like to see that people still are doing scientific things for fun.
I would say "I was doing it for fun...pure _mathematical_ entertainment
with a _chemical_ flavor." :-)