Question

The equivalent weight of $N{a}_2{S}_2{O}_3$ as reductant in the reaction, $N{a}_2{S}_2{O}_3+H_{2}O+C{l}_2\to N{a}_{2}S{O}_4+2HCl+S$ is:
[Given: Molecular weight of $N{a}_2{S}_2{O}_3$=M]
A. M/1
B. M/2
C. M/6
D. M/8

Answer 1

Hint- We will let the molecular weight be M as it is given in the question. Wherever the value of molecular mass will be needed, we will just write M and we will get our answer.

Complete answer:
The equation given to us by the question is-
$\Rightarrow N{a}_2{S}_2{O}_3+H_{2}O+C{l}_2\to N{a}_{2}S{O}_4+2HCl+S$
​$N{a}_2{S}_2{O}_3$ acts as a reductant to form$N{a}_{2}S{O}_4$ .
We will first find out the oxidation state of S in $N{a}_2{S}_2{O}_3$, we get-
$$\begin{gathered} \Rightarrow 2\left(+1\right)+2\left(S\right)+3\left(-2\right)=0 \\ \Rightarrow 2+2S-6=0 \\ \Rightarrow 2+2S=6 \\ \Rightarrow 2S=4 \\ \Rightarrow S=+2 \end{gathered}$$
Oxidation state of $S$ in $N{a}_2{S}_2{O}_3$​ is +2
Now we will find out the oxidation state of S in $N{a}_{2}S{O}_4$, we get-
$$\begin{gathered} \Rightarrow -6+S=0 \\ \Rightarrow S=+6 \end{gathered}$$
Oxidation state of $S$ in $N{a}_{2}S{O}_4$ ​ is +6
As we can see from the above values, the change in oxidation state of S in both the compounds is-
∴ Change in oxidation state of S=+4
So, change in oxidation state of $N{a}_2{S}_2{O}_3$ will be-
∴ Change in oxidation state of $N{a}_2{S}_2{O}_3$​=8
So, Equivalent weight-
$$\begin{gathered} \Rightarrow {\displaystyle \frac{8}{mol.wt.}} \\ \Rightarrow {\displaystyle \frac{8}{M}} \end{gathered}$$
Thus, option D is the correct option.

Note: Oxidation is defined as the gaining of oxygen. The chemical substance changes due to the addition of oxygen into it. Oxidation occurs when an atom, molecule, or ion loses one or more electrons in a chemical reaction.