I2+10HNO3→2HIO3+10NO2+4H2O
Iodine gets oxidized to IO3− when it reacts with an oxidizing agent (HNO3). The oxidation number of I in product will be
Oxidation: I2−IO3−
−1=x+3×(−2)
−1=x−6
x=−1+6=+5
Reduction: concentrated HNO3 gives NO2.
Iodine reacts with concentrated HNO3 to yield Y along with other products. The oxidation state of iodine in Y , is:
Held on 12 Jan 2019 · Verified 6 Jul 2026.
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