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what type of rxn is the folowing: iodine ion (I^-1) + (S208^-2) --> I3^-1 + S04^-2 (ie redox, acid-base ect.)

2007-02-10 14:28:40 · 1 answers · asked by Kay 1 in Science & Mathematics ➔ Chemistry

1 answers

It is a redox reaction. The oxidation number of the iodine changes from 1- to 5+ (oxidation) and the oxidation number of sulfur changes from 7+ to 6+ (reduction).

2007-02-10 15:17:09 · answer #1 · answered by physandchemteach 7 · 0⤊ 0⤋

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