Question

Identify the oxidant and the reductant in the following reaction respectively:
$3N{a}_{2}HAs{O}_3+KBr{O}_3+6HCl\to 6NaCl+KBr+3H_{3}As{O}_4$

• A. $N{a}_{2}HAs{O}_3$ $\&$ $KBr{O}_3$
• B. $KBr{O}_3$ $\&$ $N{a}_{2}HAs{O}_3$
• C. $HCl$ $\&$ $KBr$
• D. $KBr$ $\&$ $HCl$

Answer 1

The correct option is B $KBr{O}_3$ $\&$ $N{a}_{2}HAs{O}_3$

$\stackrel{(+3)}{3N{a}_{2}HAs{O}_3}+\stackrel{(+5)}{KBr{O}_3}+6HCl\to 6NaCl+\stackrel{(-1)}{KBr}+3\stackrel{(+5)}{H_{3}As{O}_4}.$
It is a redox reaction.
Oxidant are compound which get reduced and oxide the other compound.
Reductant are compound which get oxidised and reduce the others.
Here $A{s}^{+3}$ lose 2 electron and get oxidised to +5 oxidation state. Hence $N{a}_{2}HAs{O}_3$ is an reducing agent or reductant.
$\stackrel{(+3)}{N{a}_{2}HAs{O}_3}\left(\text{reductant}\right)\to \stackrel{(+5)}{H_{3}As{O}_4}\left(\text{oxidation half}\right).$

Also $B{r}^{+}5$ gains 6 electron and get reduced to -1 oxidation state. Hence $KBr{O}_3$ is the oxidising agent or oxidant.
$\stackrel{(+5)}{KBr{O}_3}\text{(oxidant)}\to \stackrel{(-1)}{KBr}\left(\text{reduction half half}\right).$