June 19, 2019, 09:17:46 AM
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Topic: Hydrogen sulfide + nitric acid redox (why hydrogen sulfide in oxidation?)  (Read 365 times)

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Offline blurry_light

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Hey, everyone!

So, I have a question regarding the ion-electron method to balance the oxidation-reduction of gaseous hydrogen sulfide and nitric acid to produce sulfur, nitrogen monoxide and water.

Verbatim, the question states the following:

"HNO3 reacts with gaseous H2S to form sulfur (S) and NO.*

Write down the chemical reaction and balance it via the ion-electron method."
* Yes, I know it doesn't mention water; I was told it doesn't matter because H2O pops up by itself when summing the two half-reactions.

HNO3 + H2S :rarrow: S + NO

So, I'd first need to write down the two half-reactions, one for the oxidation and for the reduction, after identifying that the sulfur in H2S on the left-hand side goes from oxidation state of -2 to 0 as elemental sulfur, and that the nitrogen in HNO3 goes from +5 to +2 on the right-hand side's NO:

Oxidation: S2-  :rarrow:  S + 2e-
Reduction: NO3- + 4H+ + 3e- :rarrow: NO + 2H2O

So, after multiplying the first one by 3 and the second one by 2 and then adding, I end up with:

3S2- + 2NO3- + 8H+ :rarrow: 3S + 2NO + 4H2O

Bringing in the spectator ions to get the balanced molecular equation, we get

2HNO3 + 3H2S + 6H+ :rarrow: 3S + 2NO + 4H2O

So, the problem is that there's still 6 protons on the left-hand side that didn't combine fully with anything.

Now, I've tried searching the Internet to figure out what was going wrong here -- and apparently, in the oxidation reaction, one needs to consider H2S as a whole, and not just the sulfide ion. This way, the oxidation half-reaction would be

H2S :rarrow: S + 2H+ + 2e-

And with the previous reduction half-reaction, it all adds up perfectly.

But in other redox reactions that I've done it is the sulfide that is written in the oxidation half-reaction; for example, this unbalanced chemical reaction, also in acidic medium (like my example):

H2S + NaMnO4 + HBr  ::equil:: S + NaBr + MnBr3 + H2O

Here, the two half-reactions are

S2-  :rarrow: S + 2e-
MnO4- + 8H+ + 4e- → Mn3+ + 4H2O

What gives? How do I know when to consider just the ion or the whole compound other than "it doesn't give the right answer"? Is there a way to figure it out beforehand? ???

Offline Borek

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2HNO3 + 3H2S + 6H+ :rarrow: 3S + 2NO + 4H2O

So, the problem is that there's still 6 protons on the left-hand side that didn't combine fully with anything.


Is it really a problem? Why? Does it make the reaction unbalanced?

(but read on)

Quote
Now, I've tried searching the Internet to figure out what was going wrong here -- and apparently, in the oxidation reaction, one needs to consider H2S as a whole, and not just the sulfide ion.

Not exactly. It depends on the circumstances. S2- and HS- are weak acids. What forms do you expect to be present in low/high pH? What can you tell about pH of nitric acid solution?
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Offline blurry_light

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2HNO3 + 3H2S + 6H+ :rarrow: 3S + 2NO + 4H2O

So, the problem is that there's still 6 protons on the left-hand side that didn't combine fully with anything.


Is it really a problem? Why? Does it make the reaction unbalanced?

(but read on)

Well, there seems to be a charge imbalance, since there's 6 protons on the left-hand side that give that side 6 positive charges that aren't balanced on the right-hand side. My understanding of redox reactions is that an oxidation reaction always accompanies a reduction so that net charge ends up being conserved (which made sense later on when we covered galvanic cells and electrolysis.)

Additionally, the hydrogens don't seem to add up; on the left-hand side, there are 14 of them, and 8 on the right-hand side. Now, I know that if I don't consider the hydrogen ions (protons), the count is correct, but I understand that these must also be considered as hydrogens in the total count to check if a reaction is balanced, since in the half-reactions you have to balance the hydrogens of the water molecules that are introduced with as many H+ as are necessary.

Now, I've tried searching the Internet to figure out what was going wrong here -- and apparently, in the oxidation reaction, one needs to consider H2S as a whole, and not just the sulfide ion.

Not exactly. It depends on the circumstances. S2- and HS- are weak acids. What forms do you expect to be present in low/high pH? What can you tell about pH of nitric acid solution?

Well, I know nitric acid is considered a strong acid. So, basically, depending on concentration of each reagent, different ions can show up in the oxidation half-reaction? If that is true, how does one deal with the proton imbalance if one doesn't consider the whole neutral molecule as a reagent of the oxidation?

EDIT: So while I was doing more research to figure this out with your pointers, I came across this article:

May, P. M., Batka, D., Hefter, G., Königsberger, E., & Rowland, D. (2018). Goodbye to S2− in aqueous solution. Chemical Communications, 54(16), 1980–1983. doi:10.1039/c8cc00187a

(URL: https://pubs.rsc.org/en/content/articlelanding/2018/cc/c8cc00187a#!divAbstract )

In it, the authors claim the following in the abstract:

"New Raman spectra of Na2S dissolved in hyper-concentrated NaOH(aq) and CsOH(aq) cast serious doubt on the widely-assumed existence of S2−(aq). To avoid conceptual and practical problems with sulfide equilibria in numerous applications, S2−(aq) should be expunged from the chemical literature. Thermodynamic databases involving sulfide minerals also need careful revision."

This is even more confusing ... So, we can't actually assume S2−(aq) even exists? Is this why we need to consider the whole neutral compound? But that doesn't explain why that other redox reaction can be balanced properly by considering S2−(aq) in the oxidation half-reaction, and what you said about different species existing at different concentrations (if I understood you correctly) also makes sense, so now I am totally confused.
« Last Edit: May 31, 2019, 12:04:01 PM by blurry_light »

Offline AWK

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S2- ion practically does not exist in the presence of strong acid. Its concentration in 1 M strong acid is below 10-20 M.
Hence the correct reaction in both your redox reactions is
H2S = S + 2H+ + 2e-
AWK

Offline blurry_light

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S2- ion practically does not exist in the presence of strong acid. Its concentration in 1 M strong acid is below 10-20 M.
Hence the correct reaction in both your redox reactions is
H2S = S + 2H+ + 2e-

Interesting, thanks. I'll have to attempt the other redox reaction with H2S in the oxidation half reaction and see how it goes. I wonder why the reference solution uses S2-, though ...

Offline Borek

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Somehow I misread your initial post:

3S2- + 2NO3- + 8H+ :rarrow: 3S + 2NO + 4H2O

I checked that you were correct here and assumed what you did later was correct as well.

Well... it wasn't

Quote
2HNO3 + 3H2S + 6H+ :rarrow: 3S + 2NO + 4H2O

Check it again. You had 8H+ on the left, that nicely combines with 3S2- and 2NO3-, not leaving even a single H+. You have just randomly added 6H+ on the left, no wonder your second equation was wrong.

Actual presence of S2- is not that important here, stoichiometry will be the same no matter what the exact form is present. And as with every weak acid ratio of HA/A- depends on pH, nothing unusual here.
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Offline blurry_light

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Somehow I misread your initial post:

3S2- + 2NO3- + 8H+ :rarrow: 3S + 2NO + 4H2O

I checked that you were correct here and assumed what you did later was correct as well.

Well... it wasn't

Quote
2HNO3 + 3H2S + 6H+ :rarrow: 3S + 2NO + 4H2O

Check it again. You had 8H+ on the left, that nicely combines with 3S2- and 2NO3-, not leaving even a single H+. You have just randomly added 6H+ on the left, no wonder your second equation was wrong.

Actual presence of S2- is not that important here, stoichiometry will be the same no matter what the exact form is present. And as with every weak acid ratio of HA/A- depends on pH, nothing unusual here.

Ah, gotcha, I only absorbed H+ into the nitric acid, forgetting the hydrogen sulfide. That explains that; gotta be careful with that.

However, just to be chemically correct, I guess -- which is actually the correct oxidation reaction, that is, the actual oxidation reaction that takes place in real life? Because as per the article I found and what AWK said, one needs to consider H2S as a whole in the oxidation, since S2- doesn't actually exist in aqueous solution, or at the very least its concentration is ridiculously tiny.

Offline Borek

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However, just to be chemically correct, I guess -- which is actually the correct oxidation reaction, that is, the actual oxidation reaction that takes place in real life? Because as per the article I found and what AWK said, one needs to consider H2S as a whole in the oxidation, since S2- doesn't actually exist in aqueous solution, or at the very least its concentration is ridiculously tiny.

As I wrote earlier - it doesn't matter much. We don't know what is the exact mechanism and it can use any of the forms present, or even go through several paths using different forms. The only thing we can more or less reliably say is which form dominates the solution (as a function of pH) according to the simple speciation model. Assuming pKa1 of 7 (which is more or less a number on which most chemical literature agrees) in the neutral solution concentrations of H2S and HS- are identical - good luck on deciding which form is a "correct" one when balancing the equation.

Definitely at low pH it is H2S that dominates the solution, above pH 7 it is HS-, at really high pH simple model predicts presence of S2-.
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Offline blurry_light

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As I wrote earlier - it doesn't matter much. We don't know what is the exact mechanism and it can use any of the forms present, or even go through several paths using different forms. The only thing we can more or less reliably say is which form dominates the solution (as a function of pH) according to the simple speciation model. Assuming pKa1 of 7 (which is more or less a number on which most chemical literature agrees) in the neutral solution concentrations of H2S and HS- are identical - good luck on deciding which form is a "correct" one when balancing the equation.

Definitely at low pH it is H2S that dominates the solution, above pH 7 it is HS-, at really high pH simple model predicts presence of S2-.

All right, that sounds reasonable enough. Thanks!

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