When SO2 is passed in a solution of potassium iodate the oxidation state of iodine changes from
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KIO3+SO2--->K2SO4+I2+H2SO4
On usual calculation, the oxidation state of iodine on reactant side is +5, because K has oxidation of +1 and O3 together has -6.
On the product side, in unreacted state the oxidation state of 0.
So the change is +5--->0 which is reduction.
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