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Hydrogen sulfide + nitric acid redox (why hydrogen sulfide in oxidation?)

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blurry_light:
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? ???

Borek:

--- Quote from: blurry_light on May 30, 2019, 07:50:25 PM ---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.
--- End quote ---


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.
--- End quote ---

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?

blurry_light:

--- Quote from: Borek on May 31, 2019, 03:12:06 AM ---
--- Quote from: blurry_light on May 30, 2019, 07:50:25 PM ---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.
--- End quote ---


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

(but read on)
--- End quote ---

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.


--- Quote from: Borek on May 31, 2019, 03:12:06 AM ---
--- Quote from: blurry_light on May 30, 2019, 07:50:25 PM ---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.
--- End quote ---

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?

--- End quote ---

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.

AWK:
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-

blurry_light:

--- Quote from: AWK on May 31, 2019, 01:16:18 PM ---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-

--- End quote ---

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 ...

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