I'm in desperate need of help with balancing a reaction.

Na₂S₂O₆ + -NO₂ + 6 H⁺ --> Na + -NH₂ + 2H₂O

I have no idea what the oxidation products are therefore I'm stuck.
Help  

Na2S2O6 + 2 RNO2 + 10H+ --> 2 NaHSO4 + 2 RNH2 + 2 H2O

As dithionites are oxidized to hydrogensulfites, I thought that dithionates might be oxidized to hydrogensulfates.

The reduction half reaction is:
R-NO₂ + 6 H⁺ + 6 e^{-  __}>  R-NH₂ + 2 H₂O

The oxidation half reaction is:
S₂O₆^2- + 2 H₂O  ^__>  2 SO₄^2- + 4H⁺ + 2 e^-

Hence, the total reaction is:
R-NO₂ + 3 S₂O₆^2- + 4 H₂O  ^__>  R-NH₂ + 6 SO₄^2- + 6 H⁺

EDIT: GC_MS, you got to it first. You're right:  NaHSO₄ will bee the product (screwed up the oxidation half reaction, lost track of the damned sodium ions LOL   )

Then the oxidation half reaction should be:


Na₂S₂O₆ + 2 H₂O  ^__>  2 NaHSO₄ + 2 H⁺ + 2 e^-

By multiplying the oxidation half reaction by three and adding it to the reduction half reaction, we obtain:
 R-NO₂ + 3 Na₂S₂O₆ + 4 H₂O  ^__> R-NH₂ + 6 NaHSO₄

Megatherium is a bit confused over here ... the mass balances of both equations seem to bee correct ...
GC_MS: Na2S2O6 + 2 RNO2 + 10H+ --> 2 NaHSO4 + 2 RNH2 + 2 H2O
Mega : R-NO2 + 3 Na2S2O6 + 4 H2O  --> R-NH2 + 6 NaHSO4

Is Megatherium doing too many drugs & is he losing his mind ?

HELP ...

GC_MS has 10 positive charges on the left, but none on the right!?
that isn't ok, is it?

Indeed, you 're right: it isn't.

3 x (Na2S2O6 + 2 H2O -->  2 NaHSO4 + 2 H+ + 2 e-)  (ox)
   R-NO2 + 6 H+ + 6 e-  -->  R-NH2 + 2 H2O  (red)

What is nice here is that the reaction seems to be auto-catalyzed: the protons that are created in the oxidation are used in the reduction.  However, I think the addition of some acid (1 eq. during the reactiton) would be required since: R-NH₂ + H⁺  -->  R-NH₃⁺


Oeff, what a relief: I can continue with my old habits  

I mistook dithionate S₂O₆^2- for dithionite S₂O₄^2-. Feeling really silly  
Dithionite is the reducing agent I'm curious about. But what is the oxidation product?

Bisulfate, HSO₄^- or
Sulfate, SO₄^2-

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Oeff, what a relief: I can continue with my old habits

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Yes, recalculated after hypo said my ion balance was way out. Have something like 2 S2O6^2- + 4 H2O + RNO2 --> 6 HSO4^- + RNH2
Shit, and that while wasn't on my drugs...  

when dithionite is oxidised by Fe³⁺ or CrO4^2-the oxidation product is SO3^2- (sulfite)

not sure what the oxidation product is when reducing nitro groups tho'.

Dithionite is the reducing agent I'm curious about. But what is the oxidation product?

Bisulfate, HSO4- or
Sulfate, SO42-

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When dithionite is oxidised by Fe3+ or CrO42-the oxidation product is SO32- (sulfite)

not sure what the oxidation product is when reducing nitro groups tho'.

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It seems indeed more reasonable to expect that the oxidation product here will be SO₃^2- instead of SO₄^2- or HSO₄^-.

I can't imagine that R-NO₂ would be a better oxidant than CrO4^2-, and has the power to oxidate the sulfur from Na₂S₂O₄ to an oxidation state +VI (like in SO₄^2-).  It is indeed pretty sure that the oxidation state from the sulfur in dithionite will change from +III to +IV in the oxidation half reaction.

Hence, we get for the oxidation half reaction:
S₂O₄^2- + 2 H₂0 --> 2 SO₃^2- + 4 H⁺ + 2 e^-

The reduction half reaction remains the same (cfr supra), so after multiplying the oxidation half reaction by three and adding it to the reduction half reaction, we obtain:
R-NO₂ + 3 S₂O₄^2- + 4 H₂0  -->  R-NH₂ + 6 SO ₃^2- + 6 H⁺, which can be rewritten as:
R-NO₂ + 3 Na₂S₂O₄^2- + 4 H₂0  -->  R-NH₂ + 6 NaHSO ₃

OK, enough redox chemistry for today already.  Now, its time for a good joint   .

This will hopefully be very beneficial to all of us. Just give me some time to get dirty in the lab  

No prob.  Are you going to reduce a aryl-2-nitropropane with the Na₂S₂O₄?  That would bee rather cool   .

Naaaaaah, no way he is going to try that cheap OTC product on nitroalkanes    Might be nice thing if it worked. It works for phenyl-1-propanone -> phenyl-1-propanol.