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Topic: reaction between (S2O3)2- and (I3)- : redox or not  (Read 17468 times)

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Offline ziyaad89

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reaction between (S2O3)2- and (I3)- : redox or not
« on: October 11, 2008, 10:32:37 PM »
When i am balancing the following equation, its very easy to put a 2 in front of thiosulphate. But if oxidation numbers are assigned to the triiodide and iodide, we can clearly see that it is a redox reaction; but no electrons are being transferred from the sulphur from thiosulphate!!

What's the reason behind this?
Is this a redox reaction?
If so please balance it using 1/2 reactions

thank you

(S2O3)2-   +    (I3)-    ---->    (I)-    +     (S4O6)2-

Offline Astrokel

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Re: reaction between (S2O3)2- and (I3)- : redox or not
« Reply #1 on: October 12, 2008, 01:22:18 AM »
If you put a 2 infront of thiosulphate, the charge is not balance ye.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline wpenrose

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Re: reaction between (S2O3)2- and (I3)- : redox or not
« Reply #2 on: October 13, 2008, 01:46:19 PM »
Quote
no electrons are being transferred from the sulphur from thiosulphate!!

The two sulfurs in thiosulfate actually have different oxidation numbers, but you don't need to know that to solve this one.

Of course, electrons are being transferred from sulfur. The resulting dithionite ion has two charges, meaning two electrons are missing. Otherwise the charge would be -4 on the dithionite ion.

Now you can solve this using the algorithm that should be in your textbook.
1. Break out the two half reactions.
2. Balance the central atoms (the ones with variable oxidation numbers.
3. Balance the oxygens by adding water to either side of each half reaction.
4. Balance the hydrogens.
5. Balance the electrons.
6. Inspect the two equations and multiply until the electrons cancel out between the two half reactions.
7. Add them together and check your arithmetic.

Dangerous Bill
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