`I^(-)` reduces `IO_(3)^(-)` and `I_(2)` and itself oxidised to `I_(2)` in acidic medium. Thus, final reaction is

Options

A. `I^(-)+IO_(3)^(-)+6H^(+) rarr I_(2)+3H_(2)O`
B. `5I^(-)+IO_(3)^(-)+6H^(+) rarr 3I_(2)+3H_(2)O`
C. `I^(-)+IO_(3)^(-) rarr I_(2)+O_(3)`
D. None of them

Answer

B

Explanation

To balance oxidation number, cross-multiply by change in oxidation number
`5I^(-)+IO_(3)^(-) rarr 3I_(2) rarr 3I_(2)+3H_(2)O` [balance (`O`) by `H_(2)O+6H^(+)`]
balance (`H`) by `H^(+)`
Thus, `5I^(-)+IO_(3)^(-)+6H^(+) rarr 3I_(2)+3H_(2)O`