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Topic: Oxidation of a compound  (Read 2163 times)

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Offline confusedstud

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Oxidation of a compound
« on: August 28, 2012, 09:49:57 PM »
When a reaction of a metal ion M3+ and S2O3 2-, 2 electrons are being given out by th thiosulfate ion. So in the product's side of the reaction, if the new compound formed has only one sulfur eg H2SO3, then the new oxidation state of sulfur increases from +2 to +4 as there is only one sulfur at the products right? But if there is a compound XS2, then will the change be from +2 to +3 instead? Thanks for the *delete me*

Offline Hunter2

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Re: Oxidation of a compound
« Reply #1 on: August 29, 2012, 01:21:32 AM »
To much confusion.

Thiosulfate will be oxidized to Sulfate or Tetrathionate

S2O32- + 5 H2O => 2 SO42- + 10 H+ + 8 e-

Oxidation number goes from +2 to +6

2 S2O32- => S4O62- + 2 e-

Oxidation number goes from +2 to +2.5

Offline confusedstud

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Re: Oxidation of a compound
« Reply #2 on: August 31, 2012, 10:30:05 AM »
Oh so if I know how many electrons are given out, the original oxidation state and the number of that particle in a compound then you can find the new oxidation state? For example, a thiosulfate ion reacts and gives out 2 electrons. What are the possible products from these ions: H2SO3, S2C where C has a oxidation state of -6. So in this case, H2SO3's sulphur has a oxidation state of +4 which is correct since 2 electrons are given out to form a product with 1 sulphur in it. And S2C's sulphur has a oxidation state of +3 which is also correct as 2 electrons are given out to form a product with 2 sulphur in it. Is this correct? Thanks for the *delete me*

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