Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Bob Sacamano on September 25, 2010, 05:00:23 PM
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3M hydrochloric acid was added to Sn(s) and KI(s) and heated under reflux in an inert atmosphere. Why must the reaction be carried out under an inert atmosphere?
Initially I had thought it was to prevent the moisture in the air from entering the system. But there is already plenty of water in the solution because aq HCl was added. Then I thought that the the following reaction would take place:
Sn (s) + 2 HCl (aq) → SnCl2 (aq) + H2 (g)
Then I have the Sn(II) required to form the stannous iodide but its bonded to Cl... And besides that still doesn't explain why the reaction must be carried out under an inert atmosphere.
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What is atmosphere composition?
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N2 - inert
H2O - already present in the system
Ar - inert
Leaving oxygen as the most likely suspect.
So assuming the inert atmosphere is to prevent the oxidation of the SnCl2, the SnCl2 must be somehow converted to SnI2 making SnCl2 an intermediate.
But I don't get why a chlorinated intermediate is formed first? why isn't the SnI2 formed immediately?
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Perhaps because HI is much more expensive and difficult to obtain?
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Yes, but the potassium iodide is also in solution. Why do we need the HCl to form stannous chloride before it is converted to stannous iodide?
Previously we synthesized tin(iv) iodide but did not use HCl as a solvent.
Is the stannous chloride complex somehow required to form as an intermediate in the process of synthesizing stannous iodide?
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You need to oxidize tin - what will be oxidizing it in just a KI solution?
How did you made SnCl4?
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We didn't make SnCl4 to get SnI4. We just heated Sn and KI in a glacial acetic acid / acetic anhydride solution at very high temperature and then crystallized the SnI4 out of the solution.
I don't see why the Cl was needed for the one reaction but not the other.
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Was the solution in contact with the air (oxygen)?
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Ahhhhh. I see. In that reaction the Sn is being oxidized by atmospheric oxygen to yield tin dioxide. And in the other reaction the tin is being oxidized by the Cl.
But why does the Cl induce the 2+ oxidation state and oxygen induce the +4 oxidation state?
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It is not Cl- that does the oxidation.
And atmospheric oxygen was not to yield tin dioxide. I think it is a clever combination of reagents - including acetic anhydride - that removes any traces of water that can be produced in the oxidation process, thus leaving just Sn(IV) that precipitates with iodide.
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Hmmm.
SnI2 case:
Sn(s) + 2H+ :rarrow: Sn2+ + H2
Now, we still have the Cl- and I- in solution and have oxidized the Sn to a 2+ state. Why does stannous iodide now crystalize instead of stannous chloride? Since chloride is a better nucleophile won't it bond preferentially?
SnI4 case:
The acetic anhydride will react with any water in the system to produce more acetic acid. Is H+ again acting as an oxidizing agent in this reaction? If so, why is this Sn complex oxidized to a 4+ state? The higher heat? and why is this system allowed to interact with atmospheric oxygen?
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I guess the solubility of the iodides is much smaller.
You are right about H+ being an oxidizing agent.
Oxygen is much stronger oxidizing agent than H+.
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These are my proposed mechanisms:
SnI4 case:
Sn(s) + O2 :rarrow: SnO2
SnO2 + 4I- + 4H+ :rarrow: SnI4 + 2H2O
SnI2 case:
Sn(s) + 2H+ :rarrow: Sn2+ + H2
Sn2+ + 2I- :rarrow: SnI2
Am I on the right track without these?
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I told you - I doubt SnO2 was present in the meantime.
Seems to me like you learned organic chemistry before inorganic, people that go that way try to find some strange mechanisms for inorganic reactions. That's not the way inorganic chemistry is done ;)
You have to write probable half reaction for oxidizing agent - how it gets reduced. How the tin is oxidized is quite obvious, it just loses 4 electrons.
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Ok I'll take another stab at it. Although I must admit that my redox chemistry is lacking.
Sn(s) :rarrow: Sn4+ + 4e-
O2 + 4e- :rarrow: 2O22-
2O22- + 4H+ :rarrow: 4H2O
Sn4+ + 4I- :rarrow: SnI4
That's my best guess. Although I still don't see why stable O2 would take up the electrons to form O2-.
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O2 + 4e- :rarrow: 2O22-
2O22- + 4H+ :rarrow: 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 O2-.
Not to form O2-, to form water.
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So initially the O2- 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?
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I doubt there will be O2- 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.
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I see:
O2 + 4e- + 4H+ :rarrow: 4H2O
Thanks for your help.
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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- :rarrow: 2I-
Obviously I2 will now act as the oxidizing agent...
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No doubt about it.
If so, whole acetic system is there just to make sure there are no traces of water present.
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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.
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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.
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We used I2 in both cases.
SnI2 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- :rarrow: Sn(OH)2
Why would oxygen oxidize Sn2+ to Sn4+ but the I2 would not?
SnI4 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?
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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.
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The standard reduction potentials are:
SnI4 case:
Sn :rarrow: Sn4+ + 4e- (-0.01 V)
2I2 + 4e- :rarrow: 4I- (0.54 V)
O2 + 4e- + H+ :rarrow: 2H2O (1.23 V)
SnI2 case:
Sn :rarrow: Sn2+ + 2e- (0.14 V)
I2 + 2e- :rarrow: 2I- (0.54 V)
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???
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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."