Question

When ammonia gas is passed over heated $CuO,$nitrogen gas is evolved from another end. Write a balanced chemical reaction to showcase the above redox change and specify the change of color, if any. Also, specify the oxidizing and reducing agents in the above reaction.

Answer 1

Oxidizing agents: The substances that can easily accept or gain electrons and can oxidize other substances and themselves get reduced.

A few examples of oxidizing agents are Chlorine $\left(Cl\right)$,Fluorine $\left(F\right)$, and Hydrogen peroxide.$\left(H_2{O}_2\right)$.

Reducing agents: The substances that lose or donate electrons and can easily reduce other substances.

A few examples of reducing agents are Lithium $\left(Li\right)$ , Sodium (Na), etc.

When ammonia gas$\left(N{H}_3\right)$ is heated along cupric oxide$\left(CuO\right)$ nitrogen gas is evolved.

The chemical reaction for the above process is as follows:

$\underset{(Copperoxide)}{\underbrace{CuO\left(s\right)}}+\underset{\left(Ammonia\right)}{\underbrace{N{H}_3\left(g\right)}}\to \underset{\left(Copper\right)}{\underbrace{Cu\left(s\right)}}+\underset{\left(Nitrogen\right)}{\underbrace{N_2\left(g\right)}}+\underset{\left(Water\right)}{\underbrace{H_{2}O\left(l\right)}}$

The above reaction is a redox reaction since both oxidation and reduction are taking place.

1. To balance the redox reaction we must divide the entire reaction into oxidation and reduction in half.
2. Then balance atoms except for $H$ and $O$ first.
3. For balancing oxygen atoms at H₂O on the other side and for balancing hydrogen atoms add H+ on the other side.
4. Then balance charges by adding electrons on the side where charges are more.
5. Now at last add both half oxidation and reduction in such a way that all electrons get canceled out.
6. Our balanced redox chemical reaction is obtained.

OXIDATION HALF:

$N{H}_3\left(g\right)\to N_2\left(g\right)$

Multiply $N{H}_3$ by two to balance nitrogen atoms.

$2N{H}_3\left(g\right)\to N_2\left(g\right)+6H^{+}+6e^{-}$

Add $6H^{+}$ on the right side to balance $H$ and six electrons to balance the charge.

REDUCTION HALF:

$\left[CuO\right(s)+2H^{+}+2e^{-}\to Cu(s)+H_{2}O(l)]\times 3$

Add one$H_{2}O$ on the right side to balance oxygen atoms, then ad $2H^{+}$ on the left side to balance $H$ atom and 2 electrons on left side to balance charge.

Now add oxidation and reduction by multiplying reduction half by three to cancel out electrons.

So our balanced redox reaction is as follows;

$\underset{(Copperoxide)}{\underbrace{3CuO\left(s\right)}}+\underset{\left(Ammonia\right)}{\underbrace{2N{H}_3\left(g\right)}}\to \underset{(Coppermetal)}{\underbrace{3Cu\left(s\right)}}+\underset{\left(Water\right)}{\underbrace{3H_{2}O\left(l\right)}}+\underset{(Nitrogengas)}{\underbrace{N_2\left(g\right)\uparrow }}$

Here $CuO$is an oxidizing agent.

And $N{H}_3$ is a reducing agent.

In this reaction, $CuO$ turns from black to reddish-brown as it is reduced to $Cu.$