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Topic: Electrolysis products problem  (Read 8487 times)

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rjdfan

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Electrolysis products problem
« on: April 22, 2004, 07:47:47 PM »


Hi, ive got a couple of questions for an assisgnment, some of which i missed in class and am having trouble finding any info on it.

 

1. describe n explain wot wud occur when the following substances are elctrolysed?

* molton KBr

* Nal solution

* molton Al 0

* dilute sulphuric acid

 

I fink that potassium has a valency of +1, that goes to the cathode and bromine to the anode. Aluminium has a valency 0 3+, that goes to the cathode and oxygen to the anode.
Sodium has a valency of +1, that goes to the cathode and chlorine to the anode.

Im not sure if im on the rite lines, but i need to go into more detailbut am stuck on how to explain it as it just about get it myself.

Any one knowing where a can find help would be very gratefull, fanks
« Last Edit: April 24, 2004, 06:48:13 PM by hmx9123 »

Offline Mitch

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Re:Electrolysis
« Reply #1 on: April 22, 2004, 07:56:43 PM »
I think you're right for KBr but, I think you will form Br2 at the anode.
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rjdfan

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Re:Electrolysis
« Reply #2 on: April 22, 2004, 07:58:52 PM »
Ok, fanks!!

chemicalLindsay

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Re:Electrolysis products problem
« Reply #3 on: May 30, 2004, 07:22:46 AM »
I couldn't be bothered to figure it out for you (sorry) but from my recollection when trying to figure out the products of electrolysis you need to find out what ions are present in the solution or molten substance. From that you should be able to tell what the products are .However if there are more then two of the same charged ion then the one that will deposit will be the ion easiliest discharged or deposited.This is dependant on a number of factors including concentration,reactivity and more.If You could find a good book on electrolysis or that had electrolysis in it then You would almost defineately find it easier (chemical calculation books are good).

Offline AWK

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Re:Electrolysis products problem
« Reply #4 on: May 31, 2004, 01:37:43 AM »
1. molten KBr
2K+ + 2e- = 2K(l)
2Br- - 2e- = Br2(l)

2. Nal solution
2H2O + 2e- = H2(g) + 2OH-(aq)
2I- - 2e- =I2 (Iodine may be solid or form NaI3 soluble in water. If electrode compartments are not separated iodine can react with NaOH to form NaOI+NaI)

3. molten Al2O3
4Al3+ + 12e- = 2Al(l)
6O2- - 12e- = 3O2(g) (oxygen usually  oxidize graphite anode to form carbon monooxide)

3. diluted sulphuric acid
4H3O+ (aq)+ 4e- = 2H2 (g)+ 4H2O
6H2O - 4e- = 4H3O+ (aq)+ O2(g)
« Last Edit: May 31, 2004, 02:03:15 AM by AWK »
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Re:Electrolysis products problem
« Reply #5 on: May 31, 2004, 12:42:07 PM »
At the temperatures that the electrolysis of molten KBr is going on, I can guarantee you that the bromine will form as a vapor and definitely not as a liquid.   ;D
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Offline Donaldson Tan

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Re:Electrolysis products problem
« Reply #6 on: May 31, 2004, 03:56:20 PM »
The more reactive cation will be preferentially discharged in the multi-cation aq medium, and the anion in greater concentration will be preferentially discharged in multi-anion aq. medium.
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Offline AWK

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Re:Electrolysis products problem
« Reply #7 on: June 01, 2004, 01:03:35 AM »
Concerning molten KBr - of course, this is my error. Should be:
2Br- - 2e- = Br2(g)
Thanks, Jdurg.
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