Answer to Question #153957 in Inorganic Chemistry for df

Question #153957

In equation 4 on the lab information sheet, what element is oxidized? What element is reduced? Justify using oxidation numbers

I3- (aq) + 2S2O32- (aq) → 3I- (aq) + S4O62- (aq)                   Equation 4

Expert's answer

Assuming that your equation is correct (It actually has issues because I3- does not exist)

S2O32- has an oxidation state of -2, when it changed to S4O62- the oxidation state remains the same, hence neither oxidation nor reduction has taken place. However, it is implied that Iodine initially had an oxidation state of -3 (assuming this (I3-) is the implied ion in the equation). Thus the oxidation state changed to -1 (I-) by loss of 2 electrons, hence it has undergone oxidation.

However, the equation is wrong all through

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