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Topic: Balancing Equations using oxidation numbers  (Read 30101 times)

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777888

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Balancing Equations using oxidation numbers
« on: October 14, 2004, 01:26:23 PM »
I don't know how to balance the following redox equation using oxidation numbers:

H2SO4 + Al => Al2(SO4)3 + SO2 +H2O

I write out the half-reactions since the oxidation number of S and Al are changing between the reactants and products(only the S in SO2 is changing though)

Then I balance the atoms and charge within and between the 2 half-reactions! But I can't get the right coefficients!

The answer is 6,2,1,3,6
But I got 3,2,1,3,X
X is not yet given a coefficient!
Can someone teach me how to do it? Thank you!

Demotivator

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Re:Balancing Equations using oxidation numbers
« Reply #1 on: October 14, 2004, 02:00:12 PM »
This is more complicated than normal.
Start out with. Notice that the charges must be balanced.
2S6+  + 2e ->  S6+  +  S4+
Al         -> Al3+  +  3e

Now to balance the electrons of both equations so they cancel on addition:
3(2S6+  + 2e ->  S6+ + S4+)
2(Al         -> Al3+ + 3e)

adding the above (all the charges and atoms stay balanced):
6S6+ + 2Al -> (2Al3+ + 3S6+)  + 3S4+
The above unit in parenthesis forms Al2(SO4)3

Fill in the compounds and add the water:
6H2SO4 + 2Al => Al2(SO4)3 + 3SO2 +H2O
Now balance whatever's left to balance ie H2O
6H2SO4 + 2Al => Al2(SO4)3 + 3SO2 +6H2O

777888

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Re:Balancing Equations using oxidation numbers
« Reply #2 on: October 14, 2004, 05:17:57 PM »
This is more complicated than normal.
Start out with. Notice that the charges must be balanced.
2S6+  + 2e ->  S6+  +  S4+
Al         -> Al3+  +  3e

Now to balance the electrons of both equations so they cancel on addition:
3(2S6+  + 2e ->  S6+ + S4+)
2(Al         -> Al3+ + 3e)

adding the above (all the charges and atoms stay balanced):
6S6+ + 2Al -> (2Al3+ + 3S6+)  + 3S4+
The above unit in parenthesis forms Al2(SO4)3

Fill in the compounds and add the water:
6H2SO4 + 2Al => Al2(SO4)3 + 3SO2 +H2O
Now balance whatever's left to balance ie H2O
6H2SO4 + 2Al => Al2(SO4)3 + 3SO2 +6H2O


2S6+  + 2e ->  S6+  +  S4+
Why do you put S6+ both on the reactant and product side?


For this one (where there are 2 Cl on the product side), how can I write the half reactions first to balance the equation? (using oxidaiton numbers)
Cl2 + AgNO3 +H2O => AgCl + HClO3 +HNO3


Half reactions:
2e + Cl2 -> 2Cl-
Cl2 -> 2Cl5+ + 10e

Balance the electrons:
5(2e + Cl2 -> 2Cl-)
Cl2 -> 2Cl5+ + 10e

Is this right? Or should I do it this way?
Cl2 -> Cl- + Cl5+ + 4e
But this way, I only have 1 half reaction

Demotivator

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #3 on: October 14, 2004, 06:03:52 PM »
2S6+  + 2e ->  S6+  +  S4+
It does seem redundant, but since Al2SO4 appears on the right it is useful because the coefficients get balanced earlier. It is more convenient than starting with:
S6+  + 2e ->   S4+
In which case the S will be added later and a rebalancing on the left side would need to be done.


2) Yes, its this way
5(2e + Cl2 -> 2Cl-)
Cl2 -> 2Cl5+ + 10e
« Last Edit: October 14, 2004, 06:10:47 PM by Demotivator »

777888

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #4 on: October 14, 2004, 08:52:54 PM »
Thanks! How about this one?

H2S + O2  ->  SO2 + H2O

The O in the reactant are spilt apart in the product!
How can I write the half reaction for O?

Start like this?  O2 ->  3O2-

777888

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #5 on: October 14, 2004, 09:41:14 PM »
Also,

Ca3(PO4)2 + SiO2 +C -> P4+ CaSiO3 + CO

20e + 2(2P5+) -> P4
10(C -> C2+ + 2e)

Add the 2 half reactions

4P5+ + 10C -> P4 + 10C2+

4Ca3(PO4)2 + SiO3 + 10C -> P4 + CaSiO3 + 10CO

The answer is: 2,6,10,1,6,10
Mine: 4,X,10,1,X,10
Why did I get wrong for the coefficient for Ca3(PO4)2  ?

This stuff is quite confusing!

Demotivator

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #6 on: October 14, 2004, 09:59:37 PM »
H2S + O2  ->  SO2 + H2O
start like this
2O2  +   4e  ->  2O2-  +   2O2-
See why? I added an extra O on the right to achieve a material balance of oxygens. I then balanced the charges with 4e.

For this one:
4P5+ + 10C -> P4 + 10C2+
You simply misapplied The 4P5+.   4P5+ is 4 total phosphorouses. In  4Ca3(PO4)2  you have a total of 8.  Obviously, the coefficient should be 2 to give a total of 4.

Well, you almost got it. You must be getting the hang of it   :)

« Last Edit: October 14, 2004, 10:35:46 PM by Demotivator »

777888

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #7 on: October 14, 2004, 11:05:43 PM »
H2S + O2  ->  SO2 + H2O
start like this
2O2  +   4e  ->  2O2-  +   2O2-
See why? I added an extra O on the right to achieve a material balance of oxygens. I then balanced the charges with 4e.

For this one:
4P5+ + 10C -> P4 + 10C2+
You simply misapplied The 4P5+.   4P5+ is 4 total phosphorouses. In  4Ca3(PO4)2  you have a total of 8.  Obviously, the coefficient should be 2 to give a total of 4.

Well, you almost got it. You must be getting the hang of it   :)



For H2S + O2  ->  SO2 + H2O
Why can you add an extra O on the right?


I have another question!
Balance the following equation in acidic solution using oxidation numbers:
MnO4 + H2C2O4 -> Mn2+ + CO2 + H2O

Half reactions:
8H+ + 5e + MnO4 -  ->  Mn 2+ + 4H2O    GER
H2C2O4 -> 2CO2 + 2e + 2 H+     LEO

Balance the charge:
16H+ + 10e + 2MnO4 -  ->  2Mn 2+ + 8H2O    2GER
5H2C2O4 -> 10CO2 + 10e + 10 H+     5LEO

Add 2 GER and 5LEO:
6H+ +2MnO4 - + 5H2C2O4 -> 2Mn 2+ + 8H2O + 10CO2
But what do I do with the H2O on the reactant side? (I did not account for it in the half reactions!

Thanks for your *delete me*

Demotivator

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #8 on: October 15, 2004, 11:48:41 AM »
FYI, in the previous problem for Ca3(PO4)2, a trick that some use when encountering subscripts is to preserve them in the primitive equation. For example, instead of 4P5+,  2P25+.  The 2 subscript comes from the (PO4)2 in the formula.  That way the resulting coefficient usually matches the final answer.

Now, for this:
6H+ +2MnO4 - + 5H2C2O4 -> 2Mn 2+ + 8H2O + 10CO2
"But what do I do with the H2O on the reactant side? (I did not account for it in the half reactions!"

Nothing wrong with it. Water is just the solvent and does not need to appear on the left. It appears on the right because it is formed by the reaction from the H+ combining with O.

For this
H2S + O2  ->  SO2 + H2O
I miscounted the charge and put 4e, should be 6e.
2O2  +  6e  ->  O22-  +  2O2-
I can add an extra O on the right because the oxygens need to balance. As long as the material and charge is balanced, the equation is true. The extra O represents an extra H2O molecule that will appear in the final balanced equation.




777888

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #9 on: October 15, 2004, 12:48:39 PM »
FYI, in the previous problem for Ca3(PO4)2, a trick that some use when encountering subscripts is to preserve them in the primitive equation. For example, instead of 4P5+,  2P25+.  The 2 subscript comes from the (PO4)2 in the formula.  That way the resulting coefficient usually matches the final answer.

Now, for this:
6H+ +2MnO4 - + 5H2C2O4 -> 2Mn 2+ + 8H2O + 10CO2
"But what do I do with the H2O on the reactant side? (I did not account for it in the half reactions!"

Nothing wrong with it. Water is just the solvent and does not need to appear on the left. It appears on the right because it is formed by the reaction from the H+ combining with O.

For this
H2S + O2  ->  SO2 + H2O
I miscounted the charge and put 4e, should be 6e.
2O2  +  6e  ->  O22-  +  2O2-
I can add an extra O on the right because the oxygens need to balance. As long as the material and charge is balanced, the equation is true. The extra O represents an extra H2O molecule that will appear in the final balanced equation.




For this one:
MnO4 + H2C2O4 -> Mn2+ + CO2 + H2O  (acidic)

Half reactions:(Before adding H2O and H+)
5e + MnO4 -  ->  Mn 2+       GER
H2C2O4 -> 2CO2 + 2e         LEO

I did not include the H2O on the product side initally! (only MnO4, H2C2O4, Mn2+ , CO2 are in the half reaction) Will I still get a correct answer when I add the 2 half reactions?
Do I have to add back the H2O to the final equation?
« Last Edit: October 15, 2004, 01:16:54 PM by 777888 »

Demotivator

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #10 on: October 15, 2004, 01:00:46 PM »
I posted then changed my mind and deleted. I have to think about this later.

Demotivator

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #11 on: October 15, 2004, 03:40:27 PM »
MnO4 + H2C2O4 -> Mn2+ + CO2 + H2O  (acidic)
You did:
5e + MnO4 -  ->  Mn 2+      GER
H2C2O4 -> 2CO2 + 2e        LEO

You will get the correct answer by adding the H+ and water at the end. The above are not balanced half reactions, but in the end you have to force everything to balance anyway.
Strictly speaking, the oxid number method just looks at the primitive element first, like this:

2( 5e + Mn7+   ->  Mn2+ )
5(C23+           -> 2C4+  +  2e)

Add, Then fill in everything else.
I hope you don't have an infinite supply of these  :P
« Last Edit: October 15, 2004, 03:45:38 PM by Demotivator »

777888

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #12 on: October 16, 2004, 05:13:34 PM »
FYI, in the previous problem for Ca3(PO4)2, a trick that some use when encountering subscripts is to preserve them in the primitive equation. For example, instead of 4P5+,  2P25+.  The 2 subscript comes from the (PO4)2 in the formula.  That way the resulting coefficient usually matches the final answer.

Now, for this:
6H+ +2MnO4 - + 5H2C2O4 -> 2Mn 2+ + 8H2O + 10CO2
"But what do I do with the H2O on the reactant side? (I did not account for it in the half reactions!"

Nothing wrong with it. Water is just the solvent and does not need to appear on the left. It appears on the right because it is formed by the reaction from the H+ combining with O.

For this
H2S + O2  ->  SO2 + H2O
I miscounted the charge and put 4e, should be 6e.
2O2  +  6e  ->  O22-  +  2O2-
I can add an extra O on the right because the oxygens need to balance. As long as the material and charge is balanced, the equation is true. The extra O represents an extra H2O molecule that will appear in the final balanced equation.





2O2  +  6e  ->  O22-  +  2O2-
Should this be balanced with 8e instead of 6e? (because there are a total of 4O on each side)

By the way, the "trick" is quite useful!

I have a short question about the first one:
H2SO4 + Al => Al2(SO4)3 + SO2 +H2O
You did it this way:
2S6+  + 2e ->  S6+  +  S4+
My friend told me that I can write the H2SO4 2 times:
H2SO4* + H2SO4 + Al => Al2(SO4)3 + SO2 +H2O
And then balance the original H2SO4 with S4+ in the product side and balance H2SO4* with S6+ in the product side by inspection!
It also works! I am just wondering which method is better?


Thanks a lot! :)
« Last Edit: October 16, 2004, 05:30:09 PM by 777888 »

777888

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #13 on: October 16, 2004, 08:16:21 PM »
I try to work on the first one myself:
H2SO4 + Al => Al2(SO4)3 + SO2 +H2O

Half reactions:
2e + 4S 6+  ->  S3 6+ + S 4+
2Al  ->  Al2 3+ + 6e

Balance the charges between the half reactions:
3(e + 4S 6+  ->  S3 6+ + S 4+) ----1
2Al  ->  Al2 3+ + 6e ----2

Add 1 and 2:
12S 6+ +2Al  ->  3S3 6+ + Al2 3+ + 3S 4+

But these are not the right coefficients...WHY? ???

Demotivator

  • Guest
Re:Balancing Equations using oxidation numbers
« Reply #14 on: October 17, 2004, 02:23:27 PM »
Right, 8e. However I realized unnecessary:

3(2O2 + 8e -> O22- + 2O2-)
4(S2-  ->  S4+  +  6e)
6O2 + 4H2S -> 4SO2 + 4H2O  which factors down to:
3O2 + 2H2S -> 2SO2 + 2H2O

Preseving the subscript doesn't work well in multiple places(since O is split), unless the element is bonded together like diatomic: Cl2, H2O2, Hg2Cl2 (mercury in form of bonded Hg2, not the chloride) etc..
This is simpler:
3(O2 + 4e  -> 2O2-)
2(S2-  -> S4+ +  6e)

3O2 + 2S2-  -> 2S4+ + 6O2-
3O2 + 2H2S -> 2SO2 + 2H2O

In the case of
H2SO4 + Al => Al2(SO4)3 + SO2 +H2O
S appears in multiple places as SO2 and (SO4)3
What happens is redundancy builds up in the final equation which though balanced, would have to be reduced.
eg your:
12S6+ +2Al  ->  3S3 6+ + Al2 3+ + 3S 4+
subtracting 6S6+ from both sides yields the correct:
6S6+ +2Al  ->  S3 6+ + Al2 3+ + 3S4+
So obviously you don't want to do it this way.

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