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Topic: Expected products from the electrolysis of 1.0M tin (IV) chloride  (Read 2918 times)

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Offline ajax0604

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I am doing a multiple choice question which asks which equation best describes the overall redox reaction occurring in the electrolysis of 1.0M tin(IV) chloride.
I have narrowed it down to these two:
Sn4+(aq) + 2H2O(l) --> Sn(s) +O2(g) + 4H+(aq)
Sn4+ (aq)+ 2Cl-(aq) --> Sn(s) +Cl2(g)
I first thought it was the first equation but since the concentration of Cl- ions will be 4.0M, would they preferentially oxidise instead of water?

Offline Borek

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Re: Expected products from the electrolysis of 1.0M tin (IV) chloride
« Reply #1 on: October 13, 2015, 02:55:58 AM »
Have you tried to consult the standard reduction potential table?
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Offline ajax0604

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Re: Expected products from the electrolysis of 1.0M tin (IV) chloride
« Reply #2 on: October 13, 2015, 05:18:35 AM »
Yes. The table shows that water is a better reductant but since the concentration of chloride will be considerably higher than the 1.0M  as per standard conditions, I think it may preferentially react.

Offline Borek

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Re: Expected products from the electrolysis of 1.0M tin (IV) chloride
« Reply #3 on: October 13, 2015, 06:31:12 AM »
This is tricky. You are not told what the pH is, plus oxygen evolution is heavily influenced by the kinetics (it is pretty slow thus requires huge overpotentials).

I have no idea what answer they expect. From what I remember during simple electrolysis of NaCl both gases - O2 and Cl2 - are produced.
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Offline ajax0604

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Re: Expected products from the electrolysis of 1.0M tin (IV) chloride
« Reply #4 on: October 13, 2015, 06:43:27 AM »
The question mentions that 1.50A is applied for 15 minutes but nothing about voltage and pH

Offline Enthalpy

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Re: Expected products from the electrolysis of 1.0M tin (IV) chloride
« Reply #5 on: October 13, 2015, 02:18:44 PM »
Are the charges balanced in every equation? This eliminates some answers.

Offline ajax0604

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Re: Expected products from the electrolysis of 1.0M tin (IV) chloride
« Reply #6 on: October 13, 2015, 06:45:17 PM »
Thanks! I didn't see that

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