Question

After balancing the following reaction, what is the coefficient of $ Br{O_3}^{ - 1} $ to $ B{r^{ - 1}} $ .
 $ Br{O_3}^{ - 1}(aq.){\text{ + B}}{{\text{r}}^{ - 1}}(aq.){\text{ + }}{{\text{H}}^ + }(aq.){\text{ }} \to {\text{ B}}{{\text{r}}_2}(l){\text{ + }}{{\text{H}}_2}{\text{O}} $
 $ (i){\text{ 1:5}} $
 $ (ii){\text{ 1:3}} $
 $ (iii){\text{ 1:2}} $
 $ (iv){\text{ 1:1}} $
 $ (v){\text{ 2:3}} $

Answer 1

Hint: We are given with the redox reaction in which there is reduction and oxidation bromine takes place. The reaction takes place in an acidic medium as we can see there is involvement of $ {H^ + } $ . We will first balance the reaction and then find the ratio of the coefficient of $ Br{O_3}^{ - 1} $ to $ B{r^{ - 1}} $ .

Complete answer:
A redox reaction is said to be balanced when the total net charge of reactants is equal to total net charge of products. We will balance the given reaction by separating oxidation and reduction reactions separately. Then we will add both the reactions and thus we get the complete balanced reaction. Therefore we will balance them separately as:
For oxidation reaction:
 $ B{r^{ - 1}}{\text{ }} \to {\text{ B}}{{\text{r}}_2} $
This is an oxidation reaction since we can see that the oxidation number of bromine increases from $ - 1 $ to $ 0 $ . Firstly we will balance the atomicity,
 $ 2B{r^{ - 1}}{\text{ }} \to {\text{ B}}{{\text{r}}_2} $
Now balance the charge on both sides. There is a net charge of $ - 2 $ at the reactant side but no charge at the product side. So, add $ 2{e^{ - 1}} $ at the product side.
 $ 2B{r^{ - 1}}{\text{ }} \to {\text{ B}}{{\text{r}}_2}{\text{ + 2}}{{\text{e}}^{ - 1}} $ _____________ $ (1) $
Hence it is a balanced oxidation reaction.
For reduction reaction:
 $ Br{O_3}^{ - 1}{\text{ }} \to {\text{ B}}{{\text{r}}_2} $
This is a reduction reaction as the oxidation state of bromine decreases from $ + 5 $ to $ 0 $ . Now firstly balance the atomicity of bromine atoms.
 $ 2Br{O_3}^{ - 1}{\text{ }} \to {\text{ B}}{{\text{r}}_2} $
Now the atomicity of oxygen is six at product side but there is no oxygen at product side. We have to add six moles of water to the products.
 $ 2Br{O_3}^{ - 1}{\text{ }} \to {\text{ B}}{{\text{r}}_2}{\text{ + 6}}{{\text{H}}_2}{\text{O}} $
But now there is no hydrogen present at reactant, but we have twelve hydrogen atoms in products. So we have to add $ 12{H^ + } $ at the reactant side. Also for balancing the charge then we have to add ten electrons at the reactant side.
 $ {\text{ 10}}{{\text{e}}^{ - 1}}{\text{ + 12}}{{\text{H}}^ + }{\text{ + }}2Br{O_3}^{ - 1}{\text{ }} \to {\text{ B}}{{\text{r}}_2}{\text{ + 6}}{{\text{H}}_2}{\text{O}} $ _________ $ (2) $
Thus we balanced both the reactions. Now before adding them we have to remove the electrons. Therefore multiply $ (1) $ by $ 5 $ and multiply $ (2) $ by $ 1 $ . Now adding both of them we get,
 $ 2Br{O_3}^{ - 1}(aq.){\text{ + 10B}}{{\text{r}}^{ - 1}}(aq.){\text{ + 12}}{{\text{H}}^ + }(aq.){\text{ }} \to {\text{ 6B}}{{\text{r}}_2}(l){\text{ + 6}}{{\text{H}}_2}{\text{O}} $
Therefore it is a completely balanced redox reaction.
Now comparing coefficient of $ Br{O_3}^{ - 1} $ to $ B{r^{ - 1}} $ we get,
 $ \dfrac{2}{{10}}{\text{ = 1:5}} $
Hence the correct option is $ (i){\text{ 1:5}} $ .

Note:
A reaction is said to be complete when the atomicity of atom and total net charge is equal at both reactant side and product side. It is mandatory that while adding the reactions the electrons which we have added must be cancelled out. There should not be any transition of electrons in a balanced chemical equation.