Question

For the reaction, $A{s_2}{S_3} \to A{s^{ + 5}} + S{O_4}^{ - 2}$, the n-factor is:
(A) 11
(B) 28
(C) 61
(D) 5/3

Answer 1

Hint: First break $A{s_2}{S_3}$ into its constituent atoms and see their individual oxidation states. Then see the oxidation states of the products obtained and count the total number of electrons lost or gained. These lost or gained electrons form the n-factor of the reaction.

Complete step by step answer:
-The given reaction is the oxidation reaction of $A{s_2}{S_3}$. First let us see the complete reaction:
$A{s_2}{S_3} \to 2As{O_4}^{ - 3} + 3S{O_4}^{ - 2}$
Here in $As{O_4}^{ - 3}$ the oxidation state of As is (+5). So, in short this reaction can also be written as:
$A{s_2}{S_3} \to A{s^{ + 5}} + S{O_4}^{ - 2}$
Balanced reaction is: $A{s_2}{S_3} \to 2A{s^{ + 5}} + 3S{O_4}^{ - 2}$
-Now let us break the compound $A{s_2}{S_3}$ into 2 parts and see the oxidation states of As and S individually.
So, $A{s_2}{S_3}$ can also be written as a combination of $A{s_2}^{ + 6}$ and ${S_3}^{ - 6}$.
-Talking about $A{s_2}^{ + 6}$ we know that for 1 As atom the oxidation state was (+3) and so for 2 As atoms the oxidation state is (+6).
In $2A{s^{ + 5}}$ also the oxidation state of 1 As atom was (+5) but for 2 As atoms it is (+10).
After oxidation $A{s_2}^{ + 6}$ was converted into $2A{s^{ + 5}}$. We can also say that the oxidation state changed from (+6) to (+10). This means that there was a loss of 4 electrons.
This can be written in the form of reaction also:
$A{s_2}^{ + 6} \to 2A{s^{ + 5}} + 4{e^ - }$
-Now let’s talk about${S_3}^{ - 6}$. Here we know that the oxidation state of 1 S atom is (-2) and that for 3 S atoms is (-6).
In $3S{O_4}^{ - 2}$ the oxidation state of 1 S atom is (+6) and that of 3 S atoms is (+18).
After oxidation ${S_3}^{ - 6}$ was converted to $3S{O_4}^{ - 2}$. We can also say that its oxidation state changed from (-6) to (+18). This means that there was a loss of 24 electrons.
This can be written in the form of reaction also:
 ${S_3}^{ - 6} \to 3S{O_4}^{ - 2} + 24{e^ - }$
-The n-factor of the reaction is calculated by adding the total no of electrons lost which is:
= 4 + 24
= 28 electrons
So, the n-factor of the reaction is 28.

The correct option is: (B) 28.

Note: While calculating the oxidation states always check whether that state is for 1 atom or more atoms. Also check whether in that reaction those electrons are being lost or gained, that is that element is being reduced or oxidised.