Iodate oxidises chromic hydroxide and gives iodide and chromate in basic medium.

Explanation

a) The ionic skeleton equation is written as
`IO_(3)^(-)+Cr(OH)_(3)overset(OH^(-))rarrI^(-)+CrO_(4)^(2-)`
b) Writing oxidation numbers
`overset(+5)(I)overset(-2)(O_(3)^(-))+overset(+3)(Cr)overset(-2+1)((OH)_(3))rarroverset(-1)(I^(-))+overset(+6)(Cr)overset(-2)(O_(4)^(-2))`
c) Locating atoms undergoing change in oxiation numbers
`overset(+5)(I)O_(3)^(-)+overset(+3)(Cr)(OH)_(3)rarroverset(-1)(I^(-))+overset(+6)(Cr)O_(4)^(-2)`
d) Dividing the reaction into two halve and balancing in acidic medium, separately
Oxidation half - reaction : `Cr(OH)_(3)rarrCrO_(4)^(2-)`
Step 1 : Balance oxygen atoms
`Cr(OH)_(3)+H_(2)OrarrCrO_(4)^(2-)`
Step 2 : Balance hydrogen atoms
`Cr(OH)_(3)+H_(2)O+5OH^(-)rarrCrO_(4)^(2-)+5H_(2)O`
Step 3 : Balance charge
`Cr(OH)_(3)+5OH^(-)rarr3e^(-)+CrO_(4)^(2-)+4H_(2)" ....(a)"`
Reduction half - reaction : `IO_(3)^(-)rarrI^(-)`
Step 1 : Balance oxygen atoms
`IO_(3)^(-)rarrI^(-)+3H_(2)O`
Step 2 : Balance hydrogen atoms
`IO_(3)^(-)+6H_(2)OrarrI^(-)+6OH^(-)+3H_(2)O`
Step 3 : Balance charge
`IO_(3)^()+6H_(2)O+6e^(-)rarrI^(-)+6OH^(-)" ....(b)"`
e) Equalising the elements and adding the two halves
`"eq (a)"xx2+"eq (b)"xx"1, we get"`
`{:(2Cr(OH)_(3)+19OH^(-)rarr6e^(-)+2CrO_(4)^(2-)+8H_(2)O),(IO_(3)^(-0)+6H_(2)O+6e^(-)rarrI^(-)+6OH^(-)),(bar(IO^(-)+2Cr(OH)_(3)+4OH^(-)rarr" ")),(" "I^(-)+2CrO_(4)^(2-)+5H_(2)O):}`
This is the balanced equation.