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### Topic: How to balance this equation?  (Read 22590 times)

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#### Evaldas

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##### How to balance this equation?
« 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  I02 + Fe+2Cl-2 + K+Cl-
2I- - 2e  I02        | 2 | 1 |
Fe+3 + 1e  Fe+2  | 1 | 2 |
So this means that I'd have to write:
2KI + 2FeCl3  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...

#### Borek

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##### Re: How to balance this equation?
« Reply #1 on: March 10, 2010, 06:05:54 AM »
2I- - 2e  I02        | 2 | 1 |
Fe+3 + 1e  Fe+2  | 1 | 2 |

OK

Quote
So this means that I'd have to write:
2KI + 2FeCl3  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.
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #2 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  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  2Cl-         | 10 | 5 |
Cl02  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...

#### vhpk

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##### Re: How to balance this equation?
« Reply #3 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  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  2Cl-         | 10 | 5 |
Cl02  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
Genius is a long patience

#### Borek

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##### Re: How to balance this equation?
« Reply #4 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
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #5 on: March 10, 2010, 08:14:15 AM »
2KI + 2FeCl3  I2 + 2FeCl2 + 2KCl

#### Evaldas

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##### Re: How to balance this equation?
« Reply #6 on: March 10, 2010, 09:16:24 AM »
K+Cl+5O-23 + Cr+3Cl-3 + K+O-2H+   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?
« Last Edit: March 10, 2010, 09:29:30 AM by Evaldas »

#### Borek

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##### Re: How to balance this equation?
« Reply #7 on: March 10, 2010, 09:18:35 AM »
No, two elements changed ON.
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #8 on: March 10, 2010, 09:30:54 AM »
Oh yes, Cr did.
Ok then:
Cr+3 - 3e  Cr+6

#### Borek

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##### Re: How to balance this equation?
« Reply #9 on: March 10, 2010, 10:01:34 AM »
+5 -> -1
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #10 on: March 10, 2010, 10:07:57 AM »
Cl+5 +6e  Cl-
And that's it?

#### JimClark

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##### Re: How to balance this equation?
« Reply #11 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.

#### Evaldas

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##### Re: How to balance this equation?
« Reply #12 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   Cl-
And that's it??

#### Borek

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##### Re: How to balance this equation?
« Reply #13 on: March 10, 2010, 02:40:38 PM »
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#### Evaldas

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##### Re: How to balance this equation?
« Reply #14 on: March 10, 2010, 02:51:04 PM »
Ok...
Then:
Cr+3 - 3e  Cr+6 | 3 | 6
Cl+5 + 6e  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  2K2CrO4 + 7KCl + 5H2O

No 3s or 6s in sight... What's the problem?