KIO3 + 5KI + 6HCl → 3I2 + 6KCl + 3H2O Correct statement of this reaction is
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a
I- is reduced to I2
b
IO-3 is oxidized to I2
c
IO-3 is reduced to I2
d
Oxidation number of I decreases from -1 (in KI) to zero (in I2)
answer is C.
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Detailed Solution
KIO3+5 KI+6HCl→3I20+6KCl+3H2OKIO3 (Oxidizing agent) reduces to I2KI (reducing agent) oxidizes to I2 This is a comproportion reaction Oxidation number of I change from +5 to zero and -1 to zero