<img src="https://tex.z-dn.net/?f=I_%7B2%7D" id="TexFormula1" title="I_{2}" alt="I_{2}" align="absmiddle" class="latex-formula">

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Leni [432]

3 years ago

+ $2 S_{2} O_{3} ^{-2}$ ----> $2 I^{-}$ + $S_{4} O_{6}^{-2}$
(Find the Oxidant and reductant)

Chemistry

1 answer:

gogolik [260]3 years ago

7 0

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

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.

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$\boxed{\text{exhaust stroke}}$

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