Question

The species acting as oxidant and reductant respectively, are:
(A) $BrO_3^ - ,{N_2}{H_4}$
(B) ${N_2}{H_4},BrO_3^ -$
(C) ${N_2},BrO_3^ -$
(D) $B{r^ - },{N_2}{H_4}$

Answer 1

First write the balanced redox reaction of ${N_2}{H_4}$ reacting with $BrO_3^ -$. Then assign oxidation state to each of the species involved in the reaction. Identify the reducing and oxidising agents in the reaction. Reducing agents are also called as reductants and oxidising agents are also called as oxidants.

Complete step by step solution:
From the options, it is clear that we need to write the reaction of ${N_2}{H_4}$ with $BrO_3^ -$ to find the correct answer.
The required balanced redox reaction is as follows:
$3{N_2}{H_4} + 2BrO_3^ - \to 3{N_2} + 2B{r^ - } + 6{H_2}O$
Oxidant: The reagent which undergoes reduction or increases the oxidation state of an element in the given substance. These reagents are also called oxidising agents.
Reductant: The reagent undergoes oxidation or decreases the oxidation state of an element in the given substance. These reagents are also called as reducing agents.
Now, calculating the oxidation state of elements in the redox reaction:
Oxidation state of N in ${N_2}{H_4}$ = -2
Oxidation state of Br in $BrO_3^ -$ = +5
Oxidation state of N in ${N_2}$ = 0
Oxidation state of Br in $B{r^ - }$= -1
The species whose oxidation state increases during the reaction, is getting oxidised or undergoing oxidation and the species whose oxidation state decreases, is getting reduced or undergoing reduction. Since, the oxidation state of N is changing from -2 to 0 in the reaction, therefore, N is getting oxidised and hence ${N_2}{H_4}$ acting as reductant or reducing agent. Oxidation state of Br is changing from +5 to 0 in the reaction, hence $BrO_3^ -$ is acting as an oxidant or oxidising agent.

Hence, option A is the correct answer.

Note: The chemical reactions in which both oxidation and reduction occur simultaneously are termed as redox reactions. Here, in the redox reaction of ${N_2}{H_4}$ and $BrO_3^ -$, oxidation of ${N_2}{H_4}$ while reduction of $BrO_3^ -$ is taking place.