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NaCReS
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[*] posted on 2-5-2005 at 03:55
a simple equation...


hi everbody,

I have tree simple equations but i couldn't complete them... can you help me?

KMnO4 + H2SO4 + FeSO4 -----> ???

KMnO4 + H2SO4 + H2O2 -----> ???

KMnO4 + H2SO4 + Na2SO3 -----> ???

thank you...

[Edited on 2-5-2005 by NaCReS]
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Saerynide
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[*] posted on 2-5-2005 at 04:40


KNO4 and Fe2SO4 do not exist.

Do you mean iron (II) sulfate FeSO4?




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[*] posted on 2-5-2005 at 04:41


Have you tried them? Double check that you have written them correctly, I doubt the existence of 'KNO4'. If they all involve KMnO4 they are practically all redox rxns where KMnO4 is reduced and the other compound is oxidized, with the acid acting as an H+ 'donor'.

Edit: And yes, Saerynide also noted the other thingy... Fe2SO4

[Edited on 2-5-2005 by Esplosivo]




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NaCReS
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[*] posted on 2-5-2005 at 04:59


Quote:
Originally posted by Saerynide
KNO4 and Fe2SO4 do not exist.

Do you mean iron (II) sulfate FeSO4?
yes you are right... i changed them! ;)



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Saerynide
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[*] posted on 2-5-2005 at 07:24


first one is
8 H2SO4 + 2 KMnO4 + 10 FeSO4 --> 2 MnSO4 + K2SO4 + 5 Fe2(SO4)3 + 8 H2O

The rest work the same way. As Esplosivo pointed out, they are only redox reactions.

Work out the oxidation half reactions first, balance charge with H+, balance H+s with H2O. If you did it all right, the number of Os on each side will be equal. After you figure that out, replace the every 2 H+'s with 1 H2SO4. Balance out the SO4s on the other side. Youre done.

[Edit]: Since you didnt write them as ionic equations, I didnt put the answer in ions. But you really should (its easier for one, and it more accurately represents what happens, as those compounds dissociate in water) unless your teacher asked you not to :o

[Edited on 2-5-2005 by Saerynide]




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tom haggen
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[*] posted on 2-5-2005 at 07:31


How did the oxidation number on Iron from 2 to 3? I just got done with a thread on ferric chloride, and everyone was saying that oxidation numbers don't change very easilly.



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[*] posted on 2-5-2005 at 07:38


If you look at the standard reduction potential of the MnO4- in acid to Mn2+, it is 1.51 V the potential of Fe2+ --> Fe3+ + e- is -0.77. In the end, the reaction still happens.

I did this titration in school last semester acutally.




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[*] posted on 2-5-2005 at 08:58


thanx for your interest... ;)



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[*] posted on 2-5-2005 at 13:07


Tom: Oxidation states don’t change in precipitation reactions (like in your thread), but they do change in redox reactions (as we have here).

About #2, would that one give you oxygen? I know it will without the acid, but I can think of various side reactions that may or may not do make other things happen.
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[*] posted on 2-5-2005 at 18:34


Ya I don't know what my deal is. I've had my head up my ass lately. I could balance redox equations with my eyes closed a couple weeks ago, but now I just can't seem to do them as smoothly. In acid, and in bases you break every thing into half reactions, then you balance atoms, then I get stuck, are you supposed to balance electrons next, or oxygens and hydrogens?



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[*] posted on 2-5-2005 at 21:43


I've always followed this order:

1) Atoms
2) Oxygens (with H2O's on the side that needs the oxygens)
3) Add H+ to balance off the hydrogens
4) If in basic solution, add the same number of OH- to each side
5) Combine the H+ and OH- if they are present on one side to form H2O
6) Cancel out anything that can be cancelled
7) Balance charges (with electrons...but you knew that!)
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