Question

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

(Find the Oxidant and reductant)

Answer 1

Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

Step-by-step explanation:

An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.

A Reductant thus exactly the opposite.

Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:

I2 --> 2I-.

The oxidation number reduced from 0 to -1.

In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.

Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.