Question

What is the change in oxidation state of chromium when potassium dichromate is treated with dilute sulphuric acid?

Answer 1

Hint: Potassium dichromate is a potassium salt of dichromic acid. It is an inorganic chemical compound with chemical formula \[{K_2}C{r_2}{O_7}\] . It has a red-orange crystal appearance. Potassium dichromate is most commonly used as an oxidizing agent in various chemical reactions.

Complete answer:
The oxidizing agent potassium dichromate is acidified with dilute sulphuric acid.
\[{K_2}C{r_2}{O_7} + 4{H_2}S{O_4} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + 4{H_2}O + 3(O)\]
From the above equation, we can see potassium dichromate reacts with four moles of dilute sulphuric acid to give potassium sulphate, chromium sulphate, four moles of water and three atoms of oxygen. These three atoms of oxygen are the reason that potassium dichromate is considered a strong oxidising agent.
Let us see the electron half reaction
\[C{r_2}{O_7}^{2 - } + 14{H^ + } + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O\]
Dichromate ion reacts with the Hydrogen ion given by sulphuric acid to form Chromium (III) ion and water. Electrons are added to the left side to balance the charge in the reaction.
Calculating the oxidation state of chromium in dichromate ion, for now we assume it to be x.
\[2x + 7( - 2) = - 2\]
\[2x = 12\]
on solving above equation, we get the value of x
\[x = + 6\]
oxidation state of the chromium (III) ion is \[ + 3\]
Change in oxidation state is given by subtracting the initial oxidation state from the final oxidation state of the element. Now calculating the change in the oxidation state of the Chromium
Change in oxidation state: \[3 - 6 = - 3\]

Note:
Chromium undergoes reduction as its oxidation state is decreased. The orange colour of the solution containing chromium (VI) ions changes to the green colour containing chromium (III) ions. Oxygen liberated in the reaction is known as Nascent oxygen, it is a monatomic element and it is very reactive.