I was studying reductive behaviour of $\ce{H2S}$. So, there is one reaction in my textbook.

$$\ce{3H2S + K2Cr2O7 + 4H2SO4 -> 3S + Cr2(SO4)3 + K2SO4 + 7H2O}$$

This reaction oxidizes $\ce{S^2-}$ to $\ce{S}$. But I thought why not $\ce{SO2}$?

$$\ce{H2S + K2Cr2O7 + 4H2SO4 -> SO2 + Cr2(SO4)3 + K2SO4 + 5H2O}$$

The reaction seems stoichiometrically plausible, but this reaction must need highly concentrated oxidants. I know the 1st reaction happens at low concentration of $\ce{H2SO4}$. Maybe then using high concentrated $\ce{H2SO4}$ would do the work for 2nd one.

I am not sure, though I am confident that the 2nd reaction will happen. What do you think?

  • 1
    $\begingroup$ All reactions are possible. H2S can be oxidized to sulfur, sulfite, or sulfate. $\endgroup$ Apr 28 '17 at 8:32
  • $\begingroup$ Well, permeakra's answer suggests me it's quite hard to produce SO<sub>2</sub> in this case! $\endgroup$ Apr 28 '17 at 8:58
  • $\begingroup$ I never said it's easy. In fact, I think, permeakra is right. SO2 will be produced as a minor product, if at all. Still, it will be produced at least in some cases, hence the reaction is possible, like I said. $\endgroup$ Apr 28 '17 at 9:03

But I thought why not $\ce{SO2}$

While the reaction is stechiometrically correct, it is hard to perform.

  • In excess of $\ce{H2S}$ sulfur will be produced.
  • In excess of oxidant sulfate will be produced.

And exact balance is impossible to achieve.

That said, $\ce{SO2}$ reacts with $\ce{H2S}$ finally producing sulfur. This reaction, just like many inorganic reactions, is an unholy mess of many different processes with many products obtained simulationaly and the only way to produce understandable and reproduceable results is to simplify the situation. One of the way to simplify the situation is to use a vast excess of one of the reagents.

  • $\begingroup$ Have you seen the second answer? It says H2S will also be produced in this reaction. will it? $\endgroup$ Apr 28 '17 at 22:46

Actually the first reaction is possible at normal conditions.

$$\ce{3H2S + K2Cr2O7 + 4H2SO4 → 3S + Cr2(SO4)3 + K2SO4 + 7H2O}$$

Hydrogen sulfide react with potassium dichromate and sulfuric acid to produce sulfur, chromium(III) sulfate, potassium sulfate and water. Hydrogen sulfide - saturated solution. Sulfuric acid - diluted solution.(chemiday 1)

Now, if you observe the two reactions, you see that there is 7 moles of water molecule in 1st reaction and 5 moles of water molecule in 2nd reaction. So, the 2 moles of water must somehow reacts with sulfur to produce sulfur dioxide.

$$\ce{3S + 2H2O ->[400 C] 2H2S + SO2}$$

Sulfur react with water to produce hydrogen sulfide and sulfur dioxide. This reaction takes place at a temperature of over 400°C.(chemiday 2)

You observe that 2 moles of hydrogen sulfide gas is produced which get cancelled from each side and give the 2nd reaction.

$$\ce{3H2S + K2Cr2O7 + 4H2SO4 -> (S + 2H2O) + Cr2(SO4)3 + K2SO4 + 5H2O~~~~~~(1)}$$

$$\ce{3H2S + K2Cr2O7 + 4H2SO4 -> (SO2 + 2H2S) + Cr2(SO4)3 + K2SO4 + 5H2O}$$

$$\ce{H2S + K2Cr2O7 + 4H2SO4 -> (SO2) + Cr2(SO4)3 + K2SO4 + 5H2O~~~~~~~(2)}$$

  • $\begingroup$ My question is about the second one. $\endgroup$ Apr 28 '17 at 8:56
  • $\begingroup$ O.. Should I include H2S in this reaction's products... $\endgroup$ Apr 28 '17 at 22:44
  • $\begingroup$ @Mockingbird, yes but H2S gets cancelled out on both sides leaving H2S only on left hand side. Hope this edited answer helps. $\endgroup$ Apr 29 '17 at 3:43
  • $\begingroup$ Oops... Missed it. $\endgroup$ Apr 29 '17 at 5:21

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