Dithionite is the reducing agent I'm curious about. But what is the oxidation product?
Bisulfate, HSO4- or
Sulfate, SO42-
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'.
It seems indeed more reasonable to expect that the oxidation product here will be SO
32- instead of SO
42- or HSO
4-.
I can't imagine that R-NO
2 would be a better oxidant than CrO4
2-, and has the power to oxidate the sulfur from Na
2S
2O
4 to an oxidation state +VI (like in SO
42-). 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:
S
2O
42- + 2 H
20 --> 2 SO
32- + 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-NO
2 + 3 S
2O
42- + 4 H
20 --> R-NH
2 + 6 SO
32- + 6 H
+, which can be rewritten as:
R-NO
2 + 3 Na
2S
2O
42- + 4 H
20 --> R-NH
2 + 6 NaHSO
3OK, enough redox chemistry for today already. Now, its time for a good joint
.