Question

Metal $\mathrm{A}$ is found in the earth's crust and on exposure to moist air, it forms a reddish-brown flaky substance. When a container made up of metal $\mathrm{A}$ is used to store a blue coloured solution of $\mathrm{B}$, the blue colour changes to pale green and reddish-brown metal $\mathrm{C}$ is formed. Identify $\mathrm{A},\mathrm{B},\mathrm{C}$ and write a balanced equation for the reaction. Also name the type of reaction.

Answer 1

Given metal $\mathrm{A}$ is found in the earth's crust and on exposure to moist air, it forms a reddish-brown flaky substance.

It should be iron because it reacts with air and moisture and forms rust which is reddish-brown in colour.

Corrosion- Definition

It is a phenomenon in which oxidation of a substance takes place in presence of air and moisture and the substance turns into an undesirable product.

It is generally a redox reaction in which oxidation(loss of electrons) and reduction(gain of electrons) take place simultaneously.

Rusting of iron
Iron gets rusted in presence of air and moisture.
$4\mathrm{Fe}\left(\mathrm{s}\right)+3\mathrm{O}{}_2(\mathrm{g})+\mathrm{xH}{}_2\mathrm{O}(\mathrm{l})\to 2\mathrm{Fe}{}_2\mathrm{O}{}_3.\mathrm{xH}{}_2\mathrm{O}(\mathrm{s})$
Iron oxygen water Rust

Here oxidation state of $\mathrm{Fe}$ in $\mathrm{Fe}$ is $0$ and that of $\mathrm{O}$ in $\mathrm{O}{}_2$ is $0$ at the reactant side but the oxidation state of $\mathrm{Fe}$ in $\mathrm{Fe}{}_2\mathrm{O}{}_3$ is $+3$ and that of $\mathrm{O}$ in $\mathrm{Fe}{}_2\mathrm{O}{}_3$ is $-2$ at the product side.
⦁ So, we can say that $\mathrm{Fe}$ is getting oxidised since its oxidation number is increasing from $0$ to $+3$ and $\mathrm{O}{}_2$ is getting reduced since its oxidation number is decreasing from $0$ to $-2$ .
So, iron gets oxidised and oxygen gets reduced.

Here rust formed is an undesirable product.

Also given when a container made up of metal $\mathrm{A}$ is used to store a blue coloured solution of $\mathrm{B}$, the blue colour changes to pale green and reddish-brown metal $\mathrm{C}$ is formed.

When blue coloured solution which is ${\mathrm{CuSO}}_4$ solution is placed in $\mathrm{Fe}$ container displacement reaction takes place and the formation of ferrous sulphate takes place which is a green colour solution and on the iron deposition of copper takes place which turns grey iron to reddish-brown in colour.

Reaction

$\mathrm{CuSO}{}_4(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})\to \mathrm{FeSO}{}_4(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{s}\right)$

Copper(II)sulphate Iron Iron sulphate Copper

(Blue ) (Grey) (Green) (Brown)

Since $\mathrm{Fe}$ is more reactive than $\mathrm{Cu}$ displacement takes place.

Reactivity series

$\mathrm{K}>\mathrm{Na}>\mathrm{Ca}>\mathrm{Mg}>\mathrm{Al}>\mathrm{Zn}>\mathrm{Fe}>\mathrm{Ni}>\mathrm{Sn}>\mathrm{Pb}>\mathrm{H}>\mathrm{Cu}>\mathrm{Hg}>\mathrm{Ag}>\mathrm{Au}>\mathrm{Pt}$

Here potassium is the most reactive metal and platinum is the least reactive metal in this series.

So, from the reactivity series, we can see $\mathrm{Fe}$ is more reactive than $\mathrm{Cu}$ so it can displace copper from its solution and therefore the blue colour is destroyed.
This is a displacement reaction.

So $\mathrm{A}$ is $\mathrm{Fe}$ metal , $\mathrm{B}$ is ${\mathrm{CuSO}}_4\left(\mathrm{aq}\right)$ solution and $\mathrm{C}$ is $\mathrm{Cu}$ metal.