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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Evaldas on March 10, 2010, 05:54:54 AM

Title: How to balance this equation?
Post by: Evaldas on March 10, 2010, 05:54:54 AM
I don't know how to balance this equation, given in the workbook, the teacher said that you can't balance this equation, and that there must be a mistake. Is she right?
K+I- + Fe+3Cl-3  :rarrow: I02 + Fe+2Cl-2 + K+Cl-
2I- - 2e  :rarrow: I02        | 2 | 1 |
Fe+3 + 1e  :rarrow: Fe+2  | 1 | 2 |
So this means that I'd have to write:
2KI + 2FeCl3  :rarrow: I2 + 3FeCl2 + 2KCl
But then I can't balance iron...

By the way, do you know a webpage where this whole oxidation-reduction method with showing how many electrons are lost or gained is explained? Because I get confused if there's an element in one compound before the reaction and then after the reaction it's in two compounds, I don't know how to write that line then, and many more nuances are unclear to me...
Title: Re: How to balance this equation?
Post by: Borek on March 10, 2010, 06:05:54 AM
2I- - 2e  :rarrow: I02        | 2 | 1 |
Fe+3 + 1e  :rarrow: Fe+2  | 1 | 2 |

OK

Quote
So this means that I'd have to write:
2KI + 2FeCl3  :rarrow: I2 + 3FeCl2 + 2KCl
But then I can't balance iron...

Why 3FeCl2 on the right? You should multiply both sides of the second reaction by 2.

Quote
By the way, do you know a webpage where this whole oxidation-reduction method with showing how many electrons are lost or gained is explained? Because I get confused if there's an element in one compound before the reaction and then after the reaction it's in two compounds, I don't know how to write that line then, and many more nuances are unclear to me...

Please give an example of what you mean.
Title: Re: How to balance this equation?
Post by: Evaldas on March 10, 2010, 06:17:34 AM
Hm, the teacher wrote the coefficient 3 next to FeCl3. I guess she was balancing chlorine??

Here's an example of what I meant:
Cl02 + NaOH  :rarrow: NaCl- + NaCl+5O3 + H2O
Now see, there's only one chlorine before the reaction, and after reaction it goes to two compounds. So in the example this is how it's solved:
Cl02 +2  :rarrow: 2Cl-         | 10 | 5 |
Cl02  :rarrow: 2Cl+5 + 10 e  |  2 | 1 |
And I simply don't understand why it's solved that way, I'm not good at this whole redox thing, so I'm asking for some good reference to study from...
Title: Re: How to balance this equation?
Post by: vhpk on March 10, 2010, 07:32:48 AM
Hm, the teacher wrote the coefficient 3 next to FeCl3. I guess she was balancing chlorine??

Here's an example of what I meant:
Cl02 + NaOH  :rarrow: NaCl- + NaCl+5O3 + H2O
Now see, there's only one chlorine before the reaction, and after reaction it goes to two compounds. So in the example this is how it's solved:
Cl02 +2  :rarrow: 2Cl-         | 10 | 5 |
Cl02  :rarrow: 2Cl+5 + 10 e  |  2 | 1 |
And I simply don't understand why it's solved that way, I'm not good at this whole redox thing, so I'm asking for some good reference to study from...
It's disproportionation process of chlorine,i.e. chlorine is both oxidizing and reducing agent so you have to write 2 process: oxidizing and reducing with the same reactant which is chlorine :)
Title: Re: How to balance this equation?
Post by: Borek on March 10, 2010, 07:59:27 AM
Hm, the teacher wrote the coefficient 3 next to FeCl3. I guess she was balancing chlorine?

No idea what the teacher was doing, follow simple rules and you will get there. Add both equations multiplying them by coefficients that allow cancellation of electrons.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method
Title: Re: How to balance this equation?
Post by: Evaldas on March 10, 2010, 08:14:15 AM
2KI + 2FeCl3  :rarrow: I2 + 2FeCl2 + 2KCl
Title: Re: How to balance this equation?
Post by: Evaldas on March 10, 2010, 09:16:24 AM
Ok, what about this one?
K+Cl+5O-23 + Cr+3Cl-3 + K+O-2H+  :rarrow: K+2Cr+6O-24 + K+Cl- + H+2O-2
We have disproportionation here, and the only element that changed the oxidation state is chlorine, but there's chlorine in two compounds before the reaction and after the reaction only in one, what do I do here?
Title: Re: How to balance this equation?
Post by: Borek on March 10, 2010, 09:18:35 AM
No, two elements changed ON.
Title: Re: How to balance this equation?
Post by: Evaldas on March 10, 2010, 09:30:54 AM
Oh yes, Cr did.
Ok then:
Cr+3 - 3e  :rarrow: Cr+6

But what about Cl?
Title: Re: How to balance this equation?
Post by: Borek on March 10, 2010, 10:01:34 AM
+5 -> -1
Title: Re: How to balance this equation?
Post by: Evaldas on March 10, 2010, 10:07:57 AM
Cl+5 +6e  :rarrow: Cl-
And that's it?
Title: Re: How to balance this equation?
Post by: JimClark on March 10, 2010, 10:47:07 AM
I have produced a whole section on this on Chemguide.  The page you want about writing equations for redox reactions is http://www.chemguide.co.uk/inorganic/redox/equations.html , but you might also need to explore some other pages by following suggestions as you work through that page.
Title: Re: How to balance this equation?
Post by: Evaldas on March 10, 2010, 02:24:20 PM
I have produced a whole section on this on Chemguide.  The page you want about writing equations for redox reactions is http://www.chemguide.co.uk/inorganic/redox/equations.html , but you might also need to explore some other pages by following suggestions as you work through that page.
I don't see anything about disproportionation  ???

Anyways back to the KClO3 reaction...
So there's only one line for Cl?
Cl+5 + 6e  :rarrow: Cl-
And that's it??
Title: Re: How to balance this equation?
Post by: Borek on March 10, 2010, 02:40:38 PM
Cl+5 +6e  :rarrow: Cl-
And that's it?

Yes.
Title: Re: How to balance this equation?
Post by: Evaldas on March 10, 2010, 02:51:04 PM
Ok...
Then:
Cr+3 - 3e  :rarrow: Cr+6 | 3 | 6
Cl+5 + 6e  :rarrow: Cl-    | 6 | 3
So this would mean that I'd have to write 6 and 3 as coefficients, but the balanced equation SHOULD look like this:
KClO3 + 2CrCl3 + 10KOH  :rarrow: 2K2CrO4 + 7KCl + 5H2O

No 3s or 6s in sight... What's the problem?
Title: Re: How to balance this equation?
Post by: Borek on March 10, 2010, 05:29:37 PM
You don't have to use 6 & 3, to cancel electrons it is enough to multiply first reaction by 2.

However, let's assume I have not told you multiplying by 2 will do. Write reaction as you think you should - that is, multpiplying by 6 & 3.
Title: Re: How to balance this equation?
Post by: Evaldas on March 11, 2010, 04:32:22 AM
KClO3 + CrCl3 + KOH  :rarrow: 6K2CrO4 + 3KCl + H2O ??? Doesn't make any sense :(
Title: Re: How to balance this equation?
Post by: Borek on March 11, 2010, 05:09:46 AM
But you have not followed the rule. You should multiply equations on both sides. That means 6 CrCl3 on the left and 6 K2CrO4 on the right. Do the same to chlorate and chlorides.

Note: you are trying to balance full equation, that means you will need to spend substantial amount of time trying to balance spectators.
Title: Re: How to balance this equation?
Post by: Evaldas on March 11, 2010, 05:30:21 AM
3KClO3 + 6CrCl3 + KOH  :rarrow: 6K2CrO4 + 3KCl + H2O
Title: Re: How to balance this equation?
Post by: Borek on March 11, 2010, 06:43:13 AM
Now, you need to balance everything else - that is, oxygen, hydrogen and potassium. Do it with KOH and water, don't change coefficients in compounds that are getting reduced or oxidized, as they are already in correct ratio.

Unfortunately after that - as you are trying to balance full reaction equation - you also have to balance chlorine. However, chlorides from CrCl3 are just spectators, so they don't take part in the redox process. You may treat them as separate from chlorides that are being produced in chlorate reduction (imagine you have started not with chromium chloride, but with chromium nitrate - then you will have to balance NO3- as spectator and it will not confuse you by being identical with chlorides produced in the reduction process).
Title: Re: How to balance this equation?
Post by: Evaldas on March 11, 2010, 01:48:23 PM
Now, you need to balance everything else - that is, oxygen, hydrogen and potassium. Do it with KOH and water, don't change coefficients in compounds that are getting reduced or oxidized, as they are already in correct ratio.

Unfortunately after that - as you are trying to balance full reaction equation - you also have to balance chlorine. However, chlorides from CrCl3 are just spectators, so they don't take part in the redox process. You may treat them as separate from chlorides that are being produced in chlorate reduction (imagine you have started not with chromium chloride, but with chromium nitrate - then you will have to balance NO3- as spectator and it will not confuse you by being identical with chlorides produced in the reduction process).
I can't manage to do this... I'm lost.
Title: Re: How to balance this equation?
Post by: Borek on March 11, 2010, 02:16:42 PM
I suppose chlorides confuse you. So let's get rid of those that are not involved in the redox part of the reaction.

Try now:

3KClO3 + 6CrBr3 + KOH  =  6K2CrO4 + 3KCl + KBr + H2O

Just remember - don't touch 3KClO3, 6CrBr3, 6K2CrO4 and 3KCl - these are already balanced.
Title: Re: How to balance this equation?
Post by: Evaldas on March 11, 2010, 02:25:33 PM
Ok:
3KClO3 + 6CrBr3 + 30KOH = 6K2CrO4 + 3KCl + 18KBr + 15H2O ?
Title: Re: How to balance this equation?
Post by: Borek on March 11, 2010, 03:09:40 PM
1. Replace now all Br by Cl - just a moment ago we did the reverse to not confuse you, time to get back to the original equation.

2. There is something wrong with this equation - coefficients can be smaller. Do you see it?
Title: Re: How to balance this equation?
Post by: Evaldas on March 11, 2010, 03:18:57 PM
1. 3KClO3 + 6CrCl3 + 30KOH = 6K2CrO4 + 3KCl + 18KCl + 15H2O
2. Yes, I see that it can be divided by 3:
3KClO3 + 6CrCl3 + 30KOH = 6K2CrO4 + 3KCl + 18KCl + 15H2O /:3 =>
KClO3 + 2CrCl3 + 10KOH = 2K2CrO4 + KCl + 6KCl + 5H2O
Title: Re: How to balance this equation?
Post by: Evaldas on March 11, 2010, 03:34:27 PM
But see now, I don't get one thing: we have to write that number of electrons as a coefficient on both sides, but how do I act in this situation:
P0(s) + O02(g)  :rarrow: P+52O-25(g)
P0  :rarrow: P+5 + 5e       | 5 | 4 |
O02 + 2x2e  :rarrow: 2O-2 | 4 | 5 |
So I write 4 next to solid phosphorus and a 5 next to oxygen, but obviously I won't write a 4 neither a 5 next to phosphorus(V) oxide, right?
4P0(s) + 5O02(g)  :rarrow: 2P+52O-25(g)
How can this be explained, when does this rule apply?
Title: Re: How to balance this equation?
Post by: Borek on March 11, 2010, 04:15:07 PM
There are two P atoms in a product molecule, so it will be easier to write phosphorus part as

2P0 -> 2P+5 + 10e-

(you did the same for oxygen, just the other way around).
Title: Re: How to balance this equation?
Post by: Evaldas on March 11, 2010, 04:25:37 PM
But wouldn't it then be:
2P0 -> 2P+5 + 10e-  | 10 | 5 | 2 |
O02 + 2x2e- -> 2O-2 |  4 | 2 | 5 |
So accordingly I'd have to write:
2P0(s) + 5O02(g)  :rarrow: ?P+52O-25(g)
Or not? Or for some reason I'm supposed to know that the five will go to oxygen, the two to phosphorus(V) oxide, and then accordingly I'd have to count that I have to write 4 next to phosphorus?
Title: Re: How to balance this equation?
Post by: Borek on March 11, 2010, 06:01:19 PM
2P+5 is equivalent to P+52 which you have in the oxide molecule.
Title: Re: How to balance this equation?
Post by: Evaldas on March 12, 2010, 02:56:22 PM
Ok. Let's go through another reaction  :)
Cr2+3(SO4)3 + Br20 + NaOH  :rarrow: Na2Cr+6O4 + 3NaBr- + Na2SO4 + H2O
Br20 + 2x1e-  :rarrow: 2Br-     | 2 | 1 | 3
Cr2+3 - 2x3e-  :rarrow: 2Cr+6 | 6 | 3 | 1
So you said that I should follow the rules and write on both sides the numbers, so I'd have to write on both sides 3s, and I don't need to write 1s:
Cr2(SO4)3 + 3Br2 + NaOH  :rarrow: Na2CrO4 + 3NaBr + Na2SO4 + H2O
But now I don't have a choice but to write a six next to NaBr instead of 3. IS IT OK TO DO SO?
Title: Re: How to balance this equation?
Post by: Borek on March 12, 2010, 05:01:26 PM
1. You have two atoms of Cr on both sides of the oxidation equation - but you have lost 2 on the RHS of the full equation.

2. No idea why do you think it should be 6 INSTEAD of 3. 3 is wrong - your reduction reaction after being multiplied by 3 will give you 6 atoms of Br (more precisely - six Br- anions) on the RHS.

Seems like you did the same mistake twice.
Title: Re: How to balance this equation?
Post by: Evaldas on March 13, 2010, 03:25:58 PM
First, what is RHS?
Title: Re: How to balance this equation?
Post by: Borek on March 13, 2010, 05:25:36 PM
Right Hand Side, Left Hand Side (of the equation).
Title: Re: How to balance this equation?
Post by: Evaldas on March 17, 2010, 04:24:04 AM
There are two P atoms in a product molecule, so it will be easier to write phosphorus part as

2P0 -> 2P+5 + 10e-

(you did the same for oxygen, just the other way around).
First of all, my teacher says you're suppsed to write P0 -> P+5 + 5 e-, in the P + O2 -> P2O5 reaction.
Title: Re: How to balance this equation?
Post by: Evaldas on April 12, 2010, 02:35:20 PM
I still haven't cracked this topic and I still need help, does anyone know a good web page that explains in a detailed way how to balance the more difficult reactions?
Title: Re: How to balance this equation?
Post by: AWK on September 14, 2011, 02:22:54 AM
I still haven't cracked this topic and I still need help, does anyone know a good web page that explains in a detailed way how to balance the more difficult reactions?
Try these examples
Pb(N3)2 + Cr(MnO4)2 = Cr2O3 + MnO2 + Pb3O4 + NO
P2I4 + P4 + H2O = PH4I + H3PO4
K4[Fe(CN)6] + Ce(NO3)4 + KOH = K2CO3 + Fe(OH)3 + Ce(OH)3 + KNO3 + H2O