Question

What are the coefficients in front of $N{O_3}(aq)$ and $Cu(s)$ when the following redox equation is balanced in an acidic solution?
\\_\\_NO_3^ - (aq) + \\_\\_Cu(s) \to \\_\\_NO(g) + \\_\\_C{u^{2 + }}(aq)
(A) $2,6$
(B) $3,6$
(C) $3,4$
(D) $2,3$

Answer 1

Hint : A redox reaction is the reaction in which both oxidation and reduction takes place simultaneously. To solve this question, we have to balance the given reaction. Before balancing, check the oxidation state of each element involved in the reaction to find out which element is going into oxidation and which element is going into reduction. First try to balance the half reaction of oxidation and reduction and then to balance the complete reaction. Use water molecules and ${H^ + }$ atoms to balance oxygen.

Complete Step By Step Answer:
Here, an aqueous solution of nitric acid reacts with copper metal in an acidic medium to produce nitric oxide and copper ions.
First, we will write down the current oxidation number of each element involved in the reaction.
In $NO_3^ -$ , the oxidation number of nitrogen is $+ 5$ and the oxidation number of oxygen is $- 2$ .
The oxidation number of copper metal $(Cu)$ is $0$ as it is in its natural state.
In $NO$ , the oxidation state of nitrogen is $+ 2$ while the oxidation state of oxygen is $- 2$ .
For easy observation, let us write the oxidation number of each element just above them:
$\mathop N\limits^{ + 5} \mathop {O_3^ - }\limits^{ - 2} (aq) + \mathop {Cu}\limits^0 (s) \to \mathop N\limits^{ + 2} \mathop O\limits^{ - 2} (g) + C{u^{2 + }}(aq)$
From the above equation, we can observe that the oxidation number of nitrogen is reduced from $+ 5$ to $+ 2$ while copper is oxidized from an oxidation state of $0$ to $+ 2$ .
Hence, we can say that nitrogen undergoes reduction while copper undergoes oxidation.
$\mathop N\limits^{ + 5} + 3{e^ - } \to \mathop N\limits^{ + 2} \to$ reduction
$\mathop {Cu}\limits^0 \to \mathop {Cu}\limits^{ + 2} + 2{e^ - } \to$ oxidation
Before going any further we must balance the above half reactions to get an equal number of electron transfers in each case. To balance them, simply multiply the reduction equation by $3$ and oxidation equation by $2$ to make the total number of electrons involved equal to $6$ in both the equations.
After multiplication, we get,
$\mathop {2N}\limits^{ + 5} + 6{e^ - } \to 2\mathop N\limits^{ + 2}$
$\mathop {3Cu}\limits^0 \to \mathop {3Cu}\limits^{ + 2} + 6{e^ - }$
Let's write the changes of half-reaction in the main reaction.
$2NO_3^ - (aq) + 3Cu(s) \to 2NO(g) + 3C{u^{2 + }}(aq)$
Now, let balance the oxygen in the chemical equation. Here we see that there are $6$ oxygen atoms on the reactants’ side and $2$ oxygen atoms on the products’ side. Hence to balance the oxygen atoms, we will simply add $4$ ${H_2}O$ molecules on the products’ side.
Hence our chemical equation becomes:
$2NO_3^ - (aq) + 3Cu(s) \to 2NO(g) + 3C{u^{2 + }}(aq) + 4{H_2}O(l)$
At last, just balance the hydrogen atoms by adding $8{H^ + }$ atoms on the reactants’ side.
$8{H^ + } + 2NO_3^ - (aq) + 3Cu(s) \to 2NO(g) + 3C{u^{2 + }}(aq) + 4{H_2}O(l)$
Hence, the coefficients in front of $N{O_3}(aq)$ and $Cu(s)$ is $2$ and $3$ . Thus, option D is correct.

Note :
The term oxidation state and oxidation number are both the same and are used interchangeably. To find the oxidation state of nitrogen, just consider its oxidation state to be an arbitrary value $x$ , add $x$ to the oxidation state of oxygen $( - 2)$ and put it equal to the total charge of the molecule. If more than one oxygen atom is present, then multiply the oxidation state of oxygen by the number of times to atoms of oxygen present in the molecule.