Question

Balance the following redox reactions in acidic solutions: $BrO_{3}^{-}+{{N}_{2}}{{H}_{4}}\to B{{r}^{-}}+{{N}_{2}}.$
A.$2BrO_{3}^{-}+3{{N}_{2}}{{H}_{4}}\to 2B{{r}^{-}}+3{{N}_{2}}\text{ +}6{{H}_{2}}O$
B.$3BrO_{3}^{-}+3{{N}_{2}}{{H}_{4}}\to 2B{{r}^{-}}+3{{N}_{2}}+4{{H}_{2}}O$
C.$3BrO_{3}^{-}+3{{N}_{2}}{{H}_{4}}\to 2B{{r}^{-}}+3{{N}_{2\text{ }}}+6{{H}_{2}}O$
D.None of these

Answer 1

We know that from the reaction, we can identify that Lead oxide is reduced to lead and ammonia is being oxidized to nitrogen. Oxidation and reduction reaction occurs Identification of loss or gain of electrons helps to detect which element is being oxidized or reduced.

Complete answer:
The first step in balancing any redox reaction is deciding if it is even an oxidation-decrease reaction, which requires that species display changing oxidation states during the reaction. To keep up charge neutrality in the example, the redox reaction will involve both a reduction segment and oxidation segments and is regularly isolated into two hypothetical half-reactions to help in understanding the reaction. This requires recognizing which component is oxidized and which element is reduced. Let us look onto the steps to be followed to balance the above unbalanced equation.
1.Unbalanced chemical equation
$BrO_{3}^{-}+{{N}_{2}}{{H}_{4}}\to B{{r}_{2}}+{{N}_{2}}$
2.Making a separate list of both reactant and product sides.
$BrO_{3}^{-}+{{N}_{2}}{{H}_{4}}\to B{{r}_{2}}+{{N}_{2}}$
3.Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.
$5{{N}_{2}}{{H}_{4}}+4BrO_{3}^{-}+20{{e}^{-}}+24{{H}^{+}}\to 5{{N}_{2}}+2B{{r}_{2}}+20{{e}^{-}}+12{{H}_{2}}O+20{{H}^{+}}$
4.Finally, always check to see that the equation is balanced. First, verify that the equation contains the same type and number of atoms on both sides of the equation.

Element	Reactant(Left)	Product(Right)	Difference
$N$	$5\times 2$	$5\times 2$	0
$H$	$\left( 5\times 4 \right)+\left( 4\times 1 \right)$	$12\times 2$	0
$Br$	$4\times 1$	$2\times 2$	0
$O$	$4\times 3$	$12\times 1$	0

Since the sum of individual atoms on the left side of the equation matches the sum of the same atoms on the right side, and since the charges on both sides are equal we can write a balanced equation.
$4BrO_{3}^{-}+5{{N}_{2}}{{H}_{4}}+4{{H}^{+}}\to 2B{{r}_{2}}+5{{N}_{2}}+12{{H}_{2}}O$
Therefore the correct answer is option A.

Note:
Remember that adjusting redox reactions initially requires splitting the condition into the two half-reactions of reduction and oxidation. All atoms aside from oxygen and hydrogen ought to be balanced first in acidic conditions, the oxygen atoms ought to be balanced with water, while hydrogen atoms ought to be balanced.