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Topic: Tin(II) chloride  (Read 15370 times)

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Offline -SERS-

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Tin(II) chloride
« on: September 24, 2009, 09:07:51 AM »
I added some tin(II) chloride with aqueous potassium manganate and it turned to a cloudy yellow-brown - what is the product?

SnCl2 + KMnO4 :rarrow: ???

I also added the tin(II) chloride to potassium dichromate which turned orange

SnCl2 + K2Cr207  :rarrow: ?

Offline sjb

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Re: Tin(II) chloride
« Reply #1 on: September 24, 2009, 12:41:21 PM »
I added some tin(II) chloride with aqueous potassium manganate and it turned to a cloudy yellow-brown - what is the product?

SnCl2 + KMnO4 :rarrow: ???

Which? Just a friendly heads up, as KMnO4 is not potassium manganate.

Offline whatamidoing

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Re: Tin(II) chloride
« Reply #2 on: September 24, 2009, 02:45:55 PM »
I added some tin(II) chloride with aqueous potassium manganate and it turned to a cloudy yellow-brown - what is the product?

SnCl2 + KMnO4 :rarrow: ???

Which? Just a friendly heads up, as KMnO4 is not potassium manganate.

^It's Potassium Permanganate.

With that said, I'm pretty new to this chemistry business so I'm not sure if everything I say after this is totally correct. If someone could swoop down and help me out with my errors I think we could both benefit maybe. :)

Tin would react with the manganate to get Sn(MnO4)2 (Tin has a charge of 2+, need manganate to balance- hence the 2)
Chlorine would get with Potassium and you'd be left with 2KCl (Chlorine has a charge of 1-, Potassium has a charge of 1+, 2 Chlorine needs 2 Potassium)

Balanced and fuzzy equation might look something like: SnCl2 + 2KMnO4 -> 2KCl + Sn(MnO4)2

y/n?

Offline Borek

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Re: Tin(II) chloride
« Reply #3 on: September 24, 2009, 04:12:46 PM »
Tin would react with the manganate to get Sn(MnO4)2 (Tin has a charge of 2+, need manganate to balance- hence the 2)
Chlorine would get with Potassium and you'd be left with 2KCl (Chlorine has a charge of 1-, Potassium has a charge of 1+, 2 Chlorine needs 2 Potassium)

Balanced and fuzzy equation might look something like: SnCl2 + 2KMnO4 -> 2KCl + Sn(MnO4)2

Have you ever thought about trying a career as a fantasy writer? You are so wrong it is even difficult to classify it as a science fiction  :P

These are going to be a redox processes.
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Offline whatamidoing

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Re: Tin(II) chloride
« Reply #4 on: September 24, 2009, 05:36:51 PM »
Tin would react with the manganate to get Sn(MnO4)2 (Tin has a charge of 2+, need manganate to balance- hence the 2)
Chlorine would get with Potassium and you'd be left with 2KCl (Chlorine has a charge of 1-, Potassium has a charge of 1+, 2 Chlorine needs 2 Potassium)

Balanced and fuzzy equation might look something like: SnCl2 + 2KMnO4 -> 2KCl + Sn(MnO4)2

Have you ever thought about trying a career as a fantasy writer? You are so wrong it is even difficult to classify it as a science fiction  :P

These are going to be a redox processes.

lololol
Well, listen. This is how things go in my fantasy world.

Offline AWK

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Re: Tin(II) chloride
« Reply #5 on: September 25, 2009, 02:38:40 AM »
Probably you used a lab solution of SnCl2 which is additionally acidified with HCl.
Then reaction proceeds as follows:
SnCl2 + KMnO4 + HCl = MnCl2 + SnCl4 + KCl + Cl2 + H2O (unbalanced - this redox reaction has many solutions depending on molar ratio of SnCl2 to KMnO4 assuming always an excess of HCl)

Quote
I also added the tin(II) chloride to potassium dichromate which turned orange

SnCl2 + K2Cr207

Dichromate itself is orange.

Quote
potassium manganate
potassium manganate(VII) will be correct
AWK

Offline -SERS-

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Re: Tin(II) chloride
« Reply #6 on: September 25, 2009, 10:51:47 PM »
Thanks for the replies. I haven't done redox equations for about nearly 2 years so I'm a bit rusty and probably need to open my 7th form textbook and have a bit of a read. Anywho,

5Sn2+ :rarrow: 5Sn4+ + 10e-
10e- + 16H+ + 2MnO4- :rarrow: 2Mn2+ + 8H2O

Adding:
16H+ + 5Sn2+ + 2MnO4- :rarrow: 2Mn2+ + 8H2O + 5Sn4+


3Sn2+ :rarrow: Sn4+ + 6e-
6e- + 14H+ + Cr2O72- :rarrow: 2Cr3+ + 7H2O

Adding:
14H+ + 3Sn2+ + Cr2O72- :rarrow: 3Sn4+ + 2Cr3+ + 7H2O


Correct?

Offline Borek

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Re: Tin(II) chloride
« Reply #7 on: September 26, 2009, 05:01:05 AM »
Yes and no.

Seems to me at first glance that these are correctly balanced redox reactions.

Whether they precisely describe what you have seen I don't know, but somehow I doubt. Your pH was probably too high, besides, as AWK stated, chlorides could be oxidized to chlorine.

No idea about the second experiment, but there is a strong observational hint that in the first one reaction has not proceeded that far, but stopped earlier, with Mn(IV).
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