[777888]

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!

[movies]

Well, you know that the manganese has lost two oxygens.  Where do you think that they could end up?

[777888]

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]

Please help me! I have to hand in the homework tomorrow! Thank you!

[Demotivator]

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)

[777888]

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]

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]

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]

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]

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.

[Demotivator]

And oh,  it started with NaHSO3 wich is also a weak acid (contains H+), so in effect there was no change in condition.

[777888]

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!

[AWK]

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.