Balance the following reaction by oxidation number method:
`CrO_(4)^(2-)+I^(ө)rarrCr^(3+)+IO_(3)^(ө)` ( in alkaline or basic medium)

Explanation

Proceeding on the pattern of last illustration
a. `overset(+6)CrO_(4)^(2-)+overset(-1)(I^(ө))rarroverset(+3)Cr^(3+)+overset(+5)(IO_(3^(ө)))`
b. i. `Cr^(6+)rarrCr^(3+)`(reduction: decrease in oxidation state)
ii. `I^(ө)rarrI^(5+)` (oxidation increase in oxidation state)
c. i. `Cr^(6+)+3e^(-)rarrCr^(3+)`
ii. `I^(ө)rarrI^(5+)6e^(-)`
d. i. `[Cr^(6+)+3e^(-)rarrCr^(3+)]xx2`
ii. `[I^(ө)rarrI^(5+)+6e^(-)]xx1`
On adding (i) and (ii), we get
`2Cr^(6+)+I^(ө)rarr2Cr^(3+)+I^(5+)`
On comparing with given reaction
`2CrO_(4)^(2-)+I^(ө)rarr2Cr^(3+)+IO_(3)^(ө)`
Note that give reaction is ionic, so first of all, the charge needs to be balanced. Balanced charge by adding `oveset(ө)OH` ions ( in basic medium). A charge of `-5` on the left and `+5` on the right side means add `10overset(ө)OH` ions to the right side of the equation.
`2CrO_(4)^(2-)+I^(ө)rarr2Cr^(3+)+IO_(3)^(ө)+10overset(ө)OH`
finally add `5H_(2)O` molecules on the left to balance `H` and `O` and get the balanced equation.
`2CrO_(4)^(2-)+I^(ө)+5H_(2)Orarr2Cr^(3+)+IO_(3)^(ө)+10overset(ө)OH`