Question

How would you balance the following equation: ${I_2} + HN{O_3} \to HI{O_3} + N{O_2} + {H_2}O$?

Answer 1

Hint: The chemical equation in which the number of the atoms in the reactants side is equal to the number of atoms in the products side is known as a balanced chemical equation. It follows the law of conservation of mass.

Complete answer:
In this reaction, iodine molecule $\left( {{I_2}} \right)$ react with nitric acid $\left( {HN{O_3}} \right)$ and produce Iodic acid $\left( {HI{O_3}} \right)$ with nitrogen dioxide $\left( {N{O_2}} \right)$ and water $\left( {{H_2}O} \right)$.
${I_2} + HN{O_3} \to HI{O_3} + N{O_2} + {H_2}O$
In this reaction iodine reacts with nitric acid and produces iodic acid with nitrogen dioxide and water. The steps by which we can balance this chemical reaction are given below:
Step-1: First of all we check the oxidation number of nitrogen atoms. In the reactant side, the oxidation number of nitrogen is $ + 5$, while in the product side the oxidation number of nitrogen is $ + 4$, so the reduction reaction we get is as follows:
${N^{ + 5}} + {e^ - } \to {N^{ + 4}}\;\;\;\;\;\;\;\;\;\;\;\{ equation\;1\} $
Now, the oxidation number of the iodine in reactant side is zero whereas in the product side the oxidation number of iodine is $ + 5$, thus the oxidation reaction we can write as:
${I_2}^0 \to 2{I^{ + 5}} + 10{e^ - }\;\;\;\;\;\;\;\;\;\;\{ equation\;2\} $
Step-2: Now, the following equation we get by the balancing reaction $1$ and $2$:
$10{N^{ + 5}} + {I_2}^0 \to 2{I^{ + 5}} + 10{N^{ + 4}}$
Step-3: After we get the above equation we will similarly multiply $HN{O_3}$ of the given equation with $10$ and $HI{O_3}$ of the equation with $2$ . After this we will balance the hydrogen and oxygen atoms on both sides and at last we will write the final balanced equation.
\[10HN{O_3} + {I_2} \to 2HI{O_3} + 10N{O_2} + {H_2}O\]
Here as we can see the number of hydrogen atoms and oxygen atoms on the reactant side is not equal to the product side. So now we will balance hydrogen atoms and oxygen atoms on both sides. Here in the reactant side there are $10$ hydrogen atoms and $30$ oxygen atoms whereas there are $4$ hydrogen atoms and $27$ oxygen atoms in the product side. Now we will balance them thus we get:
$10HN{O_3} + {I_2} \to 2HI{O_3} + 10N{O_2} + 4{H_2}O$
Thus this is the required balanced equation.

Note:
The basic step to balancing the reaction is to first check the oxidation number of the atoms in the reactant and product side and on the basis of oxidation state, we check whether the reaction is the redox reaction or not.