HOI + NO = I2 + NO3-1 balance the following reaction in acid
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HOI + NO = I2 + NO3-1 balance the following reaction in acid
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120 5 HOI NO I2 NO31Here the oxidation number of I decreases from 1 to 0 theoxidation number of N increases from 2 to 5So the reduction half reaction is HOI I2 the oxidation
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