[Borek]

O₂ + 4e^- 2O₂^2-

2O₂^2- + 4H⁺ 4H₂O

You can add these two - some things will cancel out.

Although I still don't see why stable O₂ would take up the electrons to form O^2-.

Not to form O^2-, to form water.

[Bob Sacamano]

So initially the O^2- will react with the acetic acid to produce water and and acetate ion. But then the acetic anhydride will react with the water to produce more acetic acid. That isssss clever.

So would you have to use a strong acid in the synthesis of SnI₂ to get a high enough proton concentration or would any acid (such as acetic acid) do the trick to oxidize the solid tin?

[Borek]

I doubt there will be O^2- present ever in the reaction mixture, it is rather exotic ion, present only in some solids.

Without looking at standard potential tables I can be wrong, but I would risk a statement that the only difference between acetic acid and hydrochloric acid would be in reaction speed. But there is a slight chance that I am wrong.

[Bob Sacamano]

I see:

O₂ + 4e^- + 4H⁺ 4H₂O

Thanks for your help.

[Bob Sacamano]

UGHHHHH!

I made a really really dumb mistake. We used I₂, not KI!!!

Has my entire theory been ruined?

Now I have this reaction:

I₂ + 2e^- 2I^-

Obviously I₂ will now act as the oxidizing agent...

[Borek]

No doubt about it.

If so, whole acetic system is there just to make sure there are no traces of water present.

[Bob Sacamano]

But why would the system need to be water free?

And any why the system producing SnI₄ allowed to come into contact with oxygen? Whereas the system producing SnI₂ is not.

Why do we need HCl to produce SnI₂ but we do not to produce SnI₄?

I can't rationalize any of these now.

[Borek]

While making SnI₂ have you used iodine, or iodide?

Presence of oxygen in the case of stannous iodide synthesis will probably oxidize tin(II) to tin(IV). That's not the problem when making SnI₄.

[Bob Sacamano]

We used I₂ in both cases.

SnI_{2 case:}

Think I figured out why the HCl was used. The protons in solution will drive the concentration of OH^- down so this reaction does not take place:

Sn²⁺ + OH^- Sn(OH)₂

Why would oxygen oxidize Sn²⁺ to Sn⁴⁺ but the I₂ would not?

SnI_{4 case:}

Why is oxygen not a problem when making SnI₄? What is stopping oxygen from just oxidizing some of the Sn leaving the I₂ in solution?

Why does H₂O have to be kept out of this solution?

[Borek]

Iodine vs oxygen, and final oxidation state of tin - it is all in redox potentials of all systems involved. I haven't checked them, but could be iodine on its own is too weak oxidizing agent to get tin up to +4. If so, you may need oxygen as an oxidizer when you need Sn(IV) - but if you plan to stop at Sn(II) you have to make sure there is no oxygen in the mixture, otherwise reaction won't stop where you want it to stop.

Water - no idea, but I am guessing tin will prefer to react with water to create some kind of hydroxide, oxyhydroxide or something like that. That's not unusual, especially for heavier elements that are partially amphoteric.

Note: I haven't checked any facts, so can be I am wrong, as I am mostly guessing.

[Bob Sacamano]

The standard reduction potentials are:

SnI₄ case:

Sn Sn⁴⁺ + 4e^-     (-0.01 V)

2I₂ + 4e^- 4I^-    (0.54 V)

O₂ + 4e^- + H⁺ 2H₂O     (1.23 V)

SnI₂ case:

Sn Sn²⁺ + 2e^-     (0.14 V)

I₂ + 2e^- 2I^-    (0.54 V)

--------

 

[Bob Sacamano]

A few thing I dug up:

"STANNIC IODIDE [SnI4, CAS RN: 7790-47-8] a yellow to reddish crystals; decomposes in water, soluble in alcohol, ether, chloroform, carbon disulfide, and benzene; melt point 144 C; sublime at 180 C; also known as tin tetraiodide."

"Under normal circumstances tin is stable in water. When it comes in contact with hot water vapour a reaction results, forming tinoxide and hydrogen:

Sn + 2 H2O -> SnO2 + 2 H2

Some tin compounds hydrolyse in water. Examples include tin (IV) chloride, which forms tinoxide when heated."

"Inorganic-tin compounds are divided into two series: stannous, or tin(II), compounds and stannic, or tin(IV), compounds.Chemically, tin exhibits valencies of 2 and 4. It resists attack by water but is dissolved by strong acids and alkalis."