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Topic: balancing equations using O.N  (Read 9140 times)

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Offline sapta

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balancing equations using O.N
« on: August 06, 2005, 11:01:44 AM »
As u all know,in this method u have to make oxidiser:reducer=decrease in O.N:increase in O.N.
According to my book this is done to balance the changes in O.N. "since oxidation most exactly compensate reduction".
But does multiplying the reactants with integers mean multiplying the increase/decrease in O.N with the same numbers?how?

besides how does the changes in O.N. balance each other in the following equation-
4Mg + 10HNO3 =4Mg(NO3)2 + N2O + 5H2O.

-----------------------------------------------------------
this method shud be made obsolete.  


Offline sdekivit

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Re:balancing equations using O.N
« Reply #1 on: August 06, 2005, 01:50:49 PM »
the difference in oxidation numbers tells you how much electrons are transported during the reaction from reductor to oxidator. In you example:

write down the oxidation numbers of all elements before and after the arrow ( i usually write them above the original reaction) and you'll see Mg and N have changes in oxidation numbers. You can also see how many of the N and Mg react ( 2 Ns and 4 Mgs) Now how many electrons are thus transported during this reaction. Note that the electrons donated cancels out the electrons taken.

« Last Edit: August 06, 2005, 01:51:36 PM by sdekivit »

Offline xiankai

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Re:balancing equations using O.N
« Reply #2 on: August 06, 2005, 03:36:48 PM »
i thought the reaction betweens metals and nitric acids yield water, NO2, and a metal oxide/nitrate? N2O is in my opinion, highly unlikely to form.

an oxidation number of a specific element in a given substance is sort of definite. the integers are molar ratios, and hence the product of the integers and O.N. would be the combined O.N. of the given moles of substance.

to balance the changes in O.N. an ionic equation should be written instead.

yes this method is quite pointless.
one learns best by teaching

Offline sdekivit

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Re:balancing equations using O.N
« Reply #3 on: August 06, 2005, 04:13:43 PM »
ionic reaction is not needed. It's clearly that Mg goes from O.N 0 to +2 and, if the reaction is correct, N goes from +5 to +1 in N2O ( O = -2)

thus we have 2 Ns going from +5 to +1 and thus a total change of -8 --> electrontransfer is 8 electrons. Now if we know what will happen:

Mg + HNO3 --> Mg(NO3)2 + N2O + H2O it's easy to balance this, since we need 8 Mg to yield 1 N2O:

4Mg + 10HNO3 --> 4Mg(NO3)2 + N2O + 5H2O

But i would never do this. I woul make halfreactions.


Offline sapta

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Re:balancing equations using O.N
« Reply #4 on: August 07, 2005, 03:33:52 AM »
since we need 8 Mg to yield 1 N2O:

4Mg + 10HNO3 --> 4Mg(NO3)2 + N2O + 5H2O

But i would never do this. I woul make halfreactions.



why 8 Mg?With each Mg donating 2 electrons 4 Mg should be needed.

I know this is an idiotic process but it is a compulsory question in our exams.
« Last Edit: August 07, 2005, 03:58:44 AM by sapta »

Offline sdekivit

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Re:balancing equations using O.N
« Reply #5 on: August 07, 2005, 04:08:42 AM »
i'm sorry, it's a tpyo. It must be 4 Mg.

Offline sapta

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Re:balancing equations using O.N
« Reply #6 on: August 07, 2005, 05:04:42 AM »
O.K.,now i am trying to  balance the following equation:

KMnO4 + HCl ------>Cl2 + KCl + MnCl2 + H2O..(1)

Mn goes from O.N +7 to +2(decrease 5)
Cl ....O.N. -1 to 0(increase 1/atom)2 Cl atoms undergo change,so total charge of 2;2 electron transfer.So,

2KMnO4 + 10HCl---->Cl2 +KCl + MnCl2 + H2O..(2)

Balancing K & Mn,

2KMnO4 + 10HCl---->Cl2 + 2KCl +2MnCl2 +H2O...(3)

Now what?H,Cl & O are not balancing.




Offline sdekivit

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Re:balancing equations using O.N
« Reply #7 on: August 07, 2005, 06:09:35 AM »
O.K.,now i am trying to  balance the following equation:

KMnO4 + HCl ------>Cl2 + KCl + MnCl2 + H2O..(1)

Mn goes from O.N +7 to +2(decrease 5)
Cl ....O.N. -1 to 0(increase 1/atom)2 Cl atoms undergo change,so total charge of 2;2 electron transfer.So,

2KMnO4 + 10HCl---->Cl2 +KCl + MnCl2 + H2O..(2)

Balancing K & Mn,

2KMnO4 + 10HCl---->Cl2 + 2KCl +2MnCl2 +H2O...(3)

Now what?H,Cl & O are not balancing.





Correct you state that Mn takes up 5 electrons and Cl loses 1 electron. So we need per Mn 5 Cl But 5 Cl gives us 2,5 Cl2 and thus we multiply wit 2:

--> 2 KMnO4 needs 10 HCl to form 5 Cl2

thus: 2KMnO4 + 10 HCl --> 5 Cl2 + KCl + MnCl2 + H2O

we see that there will be formed 2 KCl and 2 MnCl2 ( to balance for K and Mn) thus 2 + 4 = 6 Cl stay unchanged, so the coefficient of HCl changes to 16:

2 KMnO4 + 16 HCl --> 5 Cl2  + 2 KCl + 2 MnCl2 + 8H2O

as the balanced reaction.
« Last Edit: August 07, 2005, 06:10:30 AM by sdekivit »

Offline sapta

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Re:balancing equations using O.N
« Reply #8 on: August 07, 2005, 07:21:19 AM »
it's clear to me now ;)

Offline sapta

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Re:balancing equations using O.N
« Reply #9 on: August 08, 2005, 02:18:50 AM »

we see that there will be formed 2 KCl and 2 MnCl2 ( to balance for K and Mn) thus 2 + 4 = 6 Cl stay unchanged,

i went through your post one more time and some problems have arisen.what do you mean by Cl stay unchanged?& why did you go for Cl first-
Quote
--> 2 KMnO4 needs 10 HCl to form 5 Cl2

you prefered it to MnCl2;i mean you could have said "1KMnO4 needs 5HCl to form 1MnCl2".right?can it be done that way?

Offline sdekivit

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Re:balancing equations using O.N
« Reply #10 on: August 08, 2005, 12:52:23 PM »
i went through your post one more time and some problems have arisen.what do you mean by Cl stay unchanged?& why did you go for Cl first-you prefered it to MnCl2;i mean you could have said "1KMnO4 needs 5HCl to form 1MnCl2".right?can it be done that way?

Oxidation number of Cl in KCl and MnCl2 is the same as in HCl: -1.

Your statement is also true, look at stoichiometry of the reaction (only 1 KMnO4 needs 2,5 HCl). It's not a matter of preferring, but looking at what happens when the electrons are transferred. Cl with ON of -1 goes to ON of 0. Mn goes from +7 to +2. Thus the electrontransfer goes from Cl(-) to MnO4(-). Thereby forming Cl2 and MnCl2.

--> Thus we need 5 Cl per MnO4(-) and 5Cl(-) will go to 2,5 Cl2. And MnCl2 is then balanced with MnO4(-)
« Last Edit: August 08, 2005, 12:56:24 PM by sdekivit »

Offline sapta

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Re:balancing equations using O.N
« Reply #11 on: August 08, 2005, 10:42:11 PM »

Your statement is also true, look at stoichiometry of the reaction (only 1 KMnO4 needs 2,5 HCl).


But I said 1KMnO4 needs 5HCl ???
Quote
--> Thus we need 5 Cl per MnO4(-) and 5Cl(-) will go to 2,5 Cl2. And MnCl2 is then balanced with MnO4(-)

So,I balance the electron donor first?

Offline sdekivit

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Re:balancing equations using O.N
« Reply #12 on: August 09, 2005, 11:10:53 AM »
But I said 1KMnO4 needs 5HCl ???
So,I balance the electron donor first?


yes first you need to look at the electrondonor and acceptor. Then balance the rest.

Offline sapta

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Re:balancing equations using O.N
« Reply #13 on: August 10, 2005, 02:03:12 AM »
thanks for all the help ;)i think i have got it.

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