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Q.

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+3H2O

KIO3 (Oxidizing agent) reduces to I2
KI  (reducing agent)  oxidizes to I2 

This is a comproportion reaction Oxidation number of I change from +5 to zero and -1 to zero

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