Question

Name the process of conversion \[{K_2}C{r_2}{O_7}\] to \[C{r_2}{(S{O_4})_3}\] ?

Answer 1

Hint: To find the process of conversion of any equation, we first need to know about the balanced equation of the conversion. As this will help us to know about the proper charges on each molecule in the equation.

Complete answer:
To proceed with this question, let’s recall briefly about the type of reactions.
We know there are mint types of reactions happening around us and in laboratories. Some of them are oxidation reactions, some are reduction reactions, disproportionation (redox), displacement reactions and many more.
To find out what kind of reaction is happening, we have to focus on their oxidation states.
So, oxidation state is the total number of electrons that a particular molecule or atom can lose or gain in order to make a bond with another molecule.
So, let’s begin with our equation
Conversion of \[{K_2}C{r_2}{O_7}\] to \[C{r_2}{(S{O_4})_3}\]
First write the balanced chemical equation
\[{K_2}C{r_2}{O_7} + {H_2}S{O_4} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + 3\left[ O \right]\]
Now finding the oxidation state of chromium in \[{K_2}C{r_2}{O_7}\]
Let’s begin, it will be –
$ 2 + 2x - 14 = 0 $
$ 2x - 12 = 0 $
$ x = + 6 $
Hence the oxidation state of chromium in \[{K_2}C{r_2}{O_7}\] is \[ + 6......(i)\]
Now, we will find the oxidation state of chromium in \[C{r_2}{(S{O_4})_3}\]
Also we know that \[{(S{O_4})_3} = - 2\]
So,
$ 2x + 3( - 2) = 0 $
$ x = + 3 $
Hence the oxidation state of chromium in \[C{r_2}{(S{O_4})_3}\] is \[ + 3......(ii)\]
Now compare both the equation \[(i)and(ii)\],
The oxidation state of chromium is changing from \[ + 6{\text{ to + 3}}\] , hence, this is clearly a reduction reaction.
So, the answer to our question that the process of conversion \[{K_2}C{r_2}{O_7}\] to \[C{r_2}{(S{O_4})_3}\] is called reduction reaction.

Note:
There are also reactions that undergo oxidation and reduction at the same time like the cannizzaro reaction where two molecules of aldehyde react with alkali and one molecule reduces to alcohol and the other gets oxidized to carboxylic acid. Such reactions are known as disproportionation reactions or redox reactions.