Question

Which substance/ ion is oxidised and which substance/ion is reduced in the following reaction?
\[3{{N}_{2}}{{H}_{4}}+2Br{{O}_{3}}^{-}\to 3{{N}_{2}}+2B{{r}^{-}}+6{{H}_{2}}O\]

Answer 1

Hint:. Think about what happens during oxidation and reduction. Recall the acronym OIL RIG whose full form is ‘Oxygen is losing, Reduction is gaining’.

Complete step by step answer:
We can write the ionic form of a given reaction by using the ion electron method. We will go through the method required to do this step by step:
Step 1: Write the skeletal equation of the given reaction and also, write the oxidation number of all elements above their respective symbols according to the rules and fixed oxidation numbers. The reaction is as follows:
\[3{{N}^{-2}}_{2}{{H}^{+1}}_{4}+2B{{r}^{+5}}{{O}_{3}}^{-}\to 3{{N}_{2}}^{0}+2B{{r}^{-}}+6{{H}_{2}}O\]

Step 2: Find out the species which have been oxidized and reduced and split the skeletal equation into two half reactions which are as follows:
Oxidation half equation: $2{{N}^{-2}}+4{{e}^{-}}\to \text{ }{{N}_{2}}\text{ (1)}$
Reduction half equation: $B{{r}^{+5}}+6{{e}^{-}}\to \text{ }B{{r}^{-}}\text{ (2)}$
The $Br{{O}_{3}}^{-}$ gets reduced (+5 to -1) to $B{{r}^{-}}$ in the reaction and ${{N}_{2}}{{H}_{4}}$gets oxidized (-2 to 0) to${{N}_{2}}$.

Step 3: Add both equation 1 and 2 by balancing oxidation number by adding electrons and balancing each atom to get the final reaction which is as follows:
\[3{{N}_{2}}{{H}_{4}}+2Br{{O}_{3}}^{-}\to \text{ }3{{N}_{2}}+2B{{r}^{-}}+6{{H}_{2}}O\]

Hence, from above we can conclude that the ${{N}_{2}}{{H}_{4}}$ is the substance which gets oxidized and $Br{{O}_{3}}^{-}$ is the substance which gets reduced in the above given reaction.

Note: In acidic medium, H atoms are balanced by adding hydronium ions to the side deficient in H atoms. Also, you should remember that to verify whether the equation thus obtained is balanced or not, the total charge on either side of the equation must be equal.