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Topic: Neutral redox equation!  (Read 18138 times)

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777888

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Neutral redox equation!
« on: October 19, 2004, 04:42:30 PM »
NaHSO3 is added to KMnO4 (neutral, ie non-basic and non-acidic condition). Write the balanced equation. (One of the product is a brown precipitate MnO2)

How would you know the other product(s)???

Please help me!

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Re:Neutral redox equation!
« Reply #1 on: October 19, 2004, 04:45:33 PM »
Well, you know that the manganese has lost two oxygens.  Where do you think that they could end up?

777888

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Re:Neutral redox equation!
« Reply #2 on: October 19, 2004, 05:15:03 PM »
Well, you know that the manganese has lost two oxygens.  Where do you think that they could end up?

Yes, the manganese is reduced! And I guess S will be oxidized, but is that a way I can know the other products formed in order to balance the final equation?

Thanks!

777888

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Re:Neutral redox equation!
« Reply #3 on: October 19, 2004, 07:15:16 PM »
Please help me! I have to hand in the homework tomorrow! Thank you!

Demotivator

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Re:Neutral redox equation!
« Reply #4 on: October 19, 2004, 07:58:25 PM »
neutral, ie non-basic and non-acidic condition means that in the half reactions, you don't introduce H+ or OH- to help balance the equations. The reactions can however use H2O if necessary, and can produce OH- or H+ naturally.
for example
H2O + MnO4-  +  ne  ->  MnO2 +  xOH-    unbalanced

Well, you know MnO2 is one product.
You Know SO3 2- is oxidized to SO4 2-.
You know Na+  is there as an ion.
You know H is in there too. Now where can H go?  It can become a H+  ion  or it can combine with OH- if it finds it to form H2O.  It depends on how the balanced equation adds up.

Note: although they are ions in solution like Na+,  in equations they're often paired with ion counterparts, like Na+   and Cl-  forms NaCl. ( a minor detail)
« Last Edit: October 19, 2004, 08:45:17 PM by Demotivator »

777888

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Re:Neutral redox equation!
« Reply #5 on: October 19, 2004, 08:57:29 PM »
H2O + MnO4-  +  ne  ->  MnO2 +  xOH-    unbalanced

But how can a reaction that is non-basic has OH- in the product side? How about the S, should I put it in both sides?

THANK

Demotivator

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Re:Neutral redox equation!
« Reply #6 on: October 19, 2004, 09:05:15 PM »
Remember what I said. H+ can also be produced from the other half reaction. What happens when H+ sees OH-?  I think you can figure that.

BTW, are you doing this neutral rxn by oxid number method or "ion electron method".  The half rxn I posted is by ion electron method.

777888

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Re:Neutral redox equation!
« Reply #7 on: October 19, 2004, 09:14:12 PM »
Remember what I said. H+ can also be produced from the other half reaction. What happens when H+ sees OH-?  I think you can figure that.

BTW, are you doing this neutral rxn by oxid number method or "ion electron method".  The half rxn I posted is by ion electron method.
We use oxidation numbers to write half reactions for neutral rxn!

Besides, would a balanced equation of
MnO4(-) + HSO3(-) -> MnO4(2-) + SO4(2-) (BASIC)

be

2 MnO4- + 2 OH- + SO32- --> 2 MnO42- + SO42- + H2O ??

OR

3OH(-) + 2MnO4(-) + HSO3(-) -> 2 MnO4(2-) + SO4(2-) + 2H2O??

777888

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Re:Neutral redox equation!
« Reply #8 on: October 19, 2004, 09:17:51 PM »
So will the equation for the netural one(where no base or acid is added) be 2 MnO4- + 3 HSO3- --> 3 SO42- + H2O + 2 MnO2 + H+ ?? But there is H+ ions on the product side..

Demotivator

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Re:Neutral redox equation!
« Reply #9 on: October 19, 2004, 09:30:34 PM »
Hey, you're close
2KMnO4 + 3NaHSO3 -> 2MnO2 +  3SO42-  + H2O +  H+  + 3Na+  +  2K+

reduces to NaHSO4 + 2KNaSO4
It is weakly acidic  (NaHSO4 is a weak acid) but that is the byproduct of the reaction.  The reaction started neutral unaided by acid or base. How it winds up is not under your control.
« Last Edit: October 19, 2004, 09:55:22 PM by Demotivator »

Demotivator

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Re:Neutral redox equation!
« Reply #10 on: October 19, 2004, 09:41:21 PM »
And oh,  it started with NaHSO3 wich is also a weak acid (contains H+), so in effect there was no change in condition.

777888

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Re:Neutral redox equation!
« Reply #11 on: October 19, 2004, 10:02:37 PM »
So would a balanced equation of
MnO4(-) + HSO3(-) -> MnO4(2-) + SO4(2-) (BASIC)

be

i)  2 MnO4- + 2 OH- + SO32- --> 2 MnO42- + SO42- + H2O ??

OR

ii)  3OH(-) + 2MnO4(-) + HSO3(-) -> 2 MnO4(2-) + SO4(2-) + 2H2O??

Will the H in HSO3(-) sneak out and form H2O by combining H+ to 1 OH(-)?

Is the second one acceptable?

Thanks!
« Last Edit: October 19, 2004, 10:03:09 PM by 777888 »

Offline AWK

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Re:Neutral redox equation!
« Reply #12 on: October 20, 2004, 09:03:34 AM »
Neutral redox is write down without H(+) or OH(-) on the left side of equation.
K2 of H2SO3 is so small that can be neglected. NaHSO3 in solution is slightly basic because of hydrolysis.
The reaction posted by Demotivator is correct.
AWK

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