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Leni [432]
3 years ago
13

+ 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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If we will have <u>interactions</u> we will need more <u>energy</u> to break them in order to go from liquid to gas. If we need more <u>energy</u>, therefore, the <u>temperature will be higher</u>.

In this case, we can discard the <u>propanone</u> because this molecule don't have the ability to form <u>hydrogen bonds</u>. (Let's remember that to have hydrogen bonds we need to have a hydrogen bond to a <u>heteroatom</u>, O, N, P or S).

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