Question

Balance the following equation:
$HI + HN{O_3} \to {I_2} + NO + {H_2}O$
A. $6HI + 2HN{O_3} \to 3{I_2} + 2NO + 4{H_2}O$
B. $6HI + 4HN{O_3} \to 3{I_2} + 2NO + {H_2}O$
C. $3HI + 2HN{O_3} \to 2{I_2} + 2NO + 4{H_2}O$
D. None of these

Answer 1

Hint: We can balance this equation using the oxidation number method. First, we write down the oxidation numbers of each element and then balance all elements except hydrogen and oxygen. Then we can equalize the reduction and oxidation numbers. Finally, we can add water molecules to balance the oxygen.

Complete step by step answer:
The first step in balancing redox reactions is to write down the oxidation numbers of each element above the reaction. Hydrogen always has an oxidation number of $ + 1$ and oxygen, $ - 2$. For neutral molecules (with no charge) the sum of oxidation numbers is zero. Since all molecules here are neutral, the sum of oxidation numbers is always zero in this case.
In $HI$, hydrogen has an oxidation number of $ + 1$. Therefore, oxidation number of iodine is obtained by subtracting the oxidation number of hydrogen from zero, that is, $0 - 1 = - 1$
In $HN{O_3}$, the oxidation number of hydrogen is $ + 1$ and that of oxygen is $ - 2$. Since there are three oxygens, their total contribution is $( - 2 \times 3) = - 6$. Hence, oxidation number of nitrogen is obtained by subtracting oxidation numbers of hydrogen and oxygen from zero. Therefore, it is equal to $0 - 1 - ( - 6) = + 5$.
Since ${I_2}$ is an elemental molecule in its normal state, in this molecule the oxidation number of iodine is zero.
In ${H_2}O$, the oxidation number of hydrogen is $ + 1$ and that of oxygen is $ - 2$.
Writing all these down on top of their respective atoms in the equation, we get:
$\mathop H\limits^{ + 1} \mathop I\limits^{ - 1} + \mathop H\limits^{ + 1} \mathop N\limits^{ + 5} \mathop {{O_3}}\limits^{ - 2} \to \mathop {{I_2}}\limits^0 + \mathop N\limits^{ + 2} \mathop O\limits^{ - 2} + \mathop {{H_2}}\limits^{ + 1} \mathop O\limits^{ - 2} $
Our next step is to balance the elements except hydrogen and oxygen. These are iodine and nitrogen. Both sides have one nitrogen each, so we can leave that off. The left side has only one iodine while the right side has two. Therefore, we put a $2$ next to $HI$:
$2HI + HN{O_3} \to {I_2} + NO + {H_2}O$
Coming back to our oxidation numbers, we see that nitrogen’s oxidation number is being reduced from $ + 5$ to $ + 2$. Hence, it is being reduced and each nitrogen atom undergoes a change of $ + 5 - 2 = + 3$.
The iodine is being oxidised (from $ - 1$ to $0$). Thus, the change in oxidation number is $ + 1$ per atom.
We must equalize the changes in oxidation and reduction processes in order to balance the equation. In the reduction process the change is $ + 3$ while in the oxidation process it is $2 \times ( + 1) = + 2$, since two molecules of $HI$ are participating, as obtained from our previous balanced equation. To equalize both processes, we must find the LCM of both the numbers and then multiply each molecule accordingly. The LCM of $3$ and $2$ is $6$. Hence, we multiply the molecules of the oxidation process ($HN{O_3}$ and $NO$) with $2$, since $( + 3) \times 2 = 6$ and multiply the molecules of the oxidation process ($HI$ and ${I_2}$) with $3$, since $( + 2) \times 3 = 6$ . Therefore, we get:
$6HI + 2HN{O_3} \to 3{I_2} + 2NO + {H_2}O$
Notice that the coefficient of $HI$ now becomes $6$, since we already had a $2$ there before.
The last step is to balance the oxygen atoms by adding water molecules wherever necessary. On the left side, we have six oxygens, while on the right side we have three oxygens. Thus, we add three water molecules to the right side. Since there is already a water molecule on the right side, the total number of water molecules on the right side will now be four:
$6HI + 2HN{O_3} \to 3{I_2} + 2NO + 4{H_2}O$
We must now balance the hydrogen atoms. But we see that they are already balanced, since both sides have eight hydrogen atoms each. Since we have balanced all the elements, this is the final balanced equation:
$6HI + 2HN{O_3} \to 3{I_2} + 2NO + 4{H_2}O$

So, the correct answer is Option A .

Note: Besides the oxidation number method, we can also make use of the half-reaction method. In this method, we split the entire reaction into an oxidation half and a reduction half and balance them separately. They are combined in the end to get the final equation.
Note that oxidation is the removal of electrons, while reduction is the gain of electrons.