`I_(2)+S_(2)O_(3)^(2-) to I^(-)+S_(4)O_(6)^(2-)`

Explanation

First, separate oxidation and reduction processes
`S_(2) O_(3)^(2-) to S_(4) O_(6)^(2-)` (oxd.)
` I_(2) to I^(-) " "` (red.)
Now , balance the number of each atom like I and S , undergoing reduction and oxidation and then add electrons to the electrically positive side of the equation to balance the charge on both the sides of the equation .
`2 S_(2) O_(3)^(2-) to S_(4) O_(6)^(2-) + 2e`
`I_(2) + 2e to 2 I^(-)`
Each balanced half-reaction involves two electrons . Add them . Electrons get cancelled .
`I_(2) + 2 S_(2) O_(3)^(2-) = S_(4) O_(6)^(2-) + 2I^(-)`
The formula-unit equation may now be written as
`I_(2) + 2 Na_(2)S_(2)O_(3) = Na_(2) S_(4) O_(6) + 2 NaI`
Here , `Na^(+)` ions are .spectator . ions.