Question

CrI3=Cr2O7^2-+IO4-Find N- factor​

Answer 1

Answer:

CrI3=Cr2O7^2-+IO4-Find N- factor. 1. See answer. Add answer+5 pts. Log in to add comment. kg1271330

Answer 2

Answer:

The N factor of the reaction $CrI_3 \rightarrow Cr_2 O_7^-^2 +IO^-_4$ is 27.

Explanation:

N factor is the valency factor or conversion factor. For a  reaction, it is considered a change in their oxidation number or a change in their reduction number on both sides of a chemical reaction.

The given reaction in the question is;

$CrI_3 \rightarrow Cr_2 O_7^-^2 +IO^-_4$         (1)

Here in this reaction, only oxidation is taking place. So, it is a double oxidation reaction. We will break the reaction into two parts.

Part A: Oxidation 1

$Cr^+^3\rightarrow Cr_2O_7^-^2$

Here the Cr⁺³ ion is oxidized to Cr⁺⁶. So, the N factor of oxidation reaction will be;

N factor of Cr⁺³$=1|3-6|=3$    (2)

Part B: Oxidation 2

$I_3^-^3 \rightarrow IO_4^-$

Here in this reaction, I⁻ is getting oxidized to I⁺⁷ ion. So, the N factor of this reaction will be;

N factor of I⁻ $=3|-1-7|=24$    (3)

The net N factor of reaction (1) will be the sum of the N factor of reaction (2) and (3) i.e.

The net N factor=3+24=27

Hence, the N factor of the reaction $CrI_3 \rightarrow Cr_2 O_7^-^2 +IO^-_4$ is 27.