During the disproportionation of `I_(2)` to iodide and iodate ions, the ratio of iodate and iodide ions formed in alkaline medium is

Options

A. `1 : 5`
B. `5 : 1`
C. `3 : 1`
D. `1 : 3`

Answer

A

Explanation

Reduction half-cell
(`i`) `I_(2) rarr 2I^(-)+IO^(3-)`
(`ii`) `I_(2)+2e^(-) rarr 2I^(-)` ….(`1`)
Oxidation half-cell
(`i`) `I_(2) rarr 2I^(-)+IO^(3-)`
(`ii`) `I_(2) rarr 2IO^(3-)+10e^(-)`
(`iii`) `I_(2)+12^(-)OH rarr 2IO^(3-)+10e^(-)`
(`iv`) ` I_(2)+12^(-)OH rarr 2IO^(3-)+10e^(-)+6H_(2)O` .....(`2`)
Multiply eqn. `(1)xx5+eqn.(2)`
`6I_(2)+12^(-)OH rarr 10I^(-)+2IO_(3)^(-)+6H_(2)O`
Ratio of `(IO_(3)^(-))/(I^(-))=(2)/(10)=1 : 5`