Question

Identify the oxidant and the reductant in the following reaction respectively:
$I_2+N{a}_2{S}_2{O}_3\to N{a}_2{S}_4{O}_6+NaI$

• A. $I_2$ $\&$ $N{a}_2{S}_2{O}_3$
• B. $N{a}_2{S}_2{O}_3$ $\&$ $I_2$
• C. $I_2$ $\&$ $NaI$
• D. $NaI$ $\&$ $I_2$

Answer 1

The correct option is A $I_2$ $\&$ $N{a}_2{S}_2{O}_3$

$\stackrel{\left(0\right)}{I_2}+\stackrel{(+2)}{N{a}_2{S}_2{O}_3}\to \stackrel{(+2.5)}{N{a}_2{S}_4{O}_6}+\stackrel{(-1)}{NaI}.$
$\stackrel{\left(0\right)}{I_2}\left(\text{oxidant}\right)\to \stackrel{(-1)}{NaI}\left(\text{reduction half}\right).$
$\stackrel{(+2)}{N{a}_2{S}_2{O}_3}\text{(reductant)}\to \stackrel{\left(2.5\right)}{N{a}_2{S}_4{O}_6}\left(\text{oxidation half}\right).$

A species which gain electron in a redox reaction is called as oxidising agent.
$\left(I_2\right)$ gets converted into nitric oxide $\left(I^{-}\right)$. Iodine is getting reduced because it is gaining electrons, so $I_2$ is the oxidizing agent.

A species which looses electron in a redox reaction is called as reducing agent.
$N{a}_2{S}_2{O}_3$ gets converted to $N{a}_2{S}_4{O}_6$ $N{a}_2{S}_2{O}_3$ is getting oxidized since there is loss of electrons , so it is the reducing agent.

$I_2$ and $N{a}_2{S}_2{O}_3$ in this order is the correct answer.