Animated Solution for Chemistry - d and f-Block Elements: What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

A Tale of pH and Colors

Imagine a beaker filled with a bright, vibrant yellow solution. This is our starting point: a solution of potassium chromate ($K_{2}Cr{O}_4$). When dissolved in water, it completely dissociates to give potassium ions and the star of our show, the chromate ions ($Cr{O}_4^{2-}$). It is these chromate ions that are responsible for the striking yellow color of the solution.

Now, the problem asks us to treat this solution with an excess of dilute nitric acid ($HN{O}_3$). Nitric acid is a strong acid, which means it completely ionizes in water to release a flood of $H^{+}$ ions. This drastically lowers the pH, making the medium highly acidic.

Here is where the magic of transition metal chemistry happens. Chromate ions are highly sensitive to changes in pH. In an acidic medium, they undergo a condensation reaction. Two chromate ions combine, reacting with the $H^{+}$ ions, to form a new, larger species: the dichromate ion ($C{r}_2{O}_7^{2-}$), along with a molecule of water ($H_{2}O$).

Visually, this chemical transformation is stunning. As the dichromate ions form, the solution shifts from its original bright yellow to a deep, striking orange!

Notice the equilibrium arrows in our balanced equation. This reaction is completely reversible! If we were to add a base instead, providing $O{H}^{-}$ ions, they would neutralize the $H^{+}$ ions to form water. According to Le Chatelier's principle, the system would try to counteract this change by shifting the equilibrium back to the left. The orange dichromate would revert right back to the yellow chromate.

A very common trap in this topic is assuming that because the color changed so drastically, a redox reaction must have occurred. Let's verify this by checking the oxidation states.

In the chromate ion ($Cr{O}_4^{2-}$), oxygen is $-2$. So, $x+4(-2)=-2$, which gives $x=+6$. The oxidation state of chromium is $+6$.

In the dichromate ion ($C{r}_2{O}_7^{2-}$), oxygen is again $-2$. So, $2x+7(-2)=-2$, which gives $2x=+12$, and $x=+6$. The oxidation state of chromium is still $+6$!

No electrons are exchanged. This is purely an acid-base driven structural change, not a redox reaction. Therefore, the chromate ion is not reduced to $C{r}^{3+}$.

Looking at our balanced equation, the products formed in the acidic medium are the dichromate ion ($C{r}_2{O}_7^{2-}$) and water ($H_{2}O$). Comparing this with our given options, we can confidently conclude that option (b) is the correct answer.