Question

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

• A. $C{r}_2{O}_7^{2-}$ and $H_{2}O$ are formed.
• B. $Cr{O}_7^{2-}$ is reduced to $+3$ state of $Cr$.
• C. $Cr{O}_7^{2-}$ is oxidized to $+7$ state of $Cr$.
• D. $C{r}^{3+}$ and $C{r}_2{O}_7^{2-}$ are formed.

Answer 1

Answer: (A), (B)

The correct options are
A $C{r}_2{O}_7^{2-}$ and $H_{2}O$ are formed.
B $Cr{O}_7^{2-}$ is reduced to $+3$ state of $Cr$.

When potassium chromate is treated with an excess of dilute nitric acid, the following reaction takes place:
$2K_{2}Cr{O}_4+2HN{O}_3\to K_{2}C{r}_2{O}_7+2KN{O}_3+H_{2}O$
Oxidation state of $Cr$ in $K_{2}Cr{O}_4$ and $K_{2}C{r}_2{O}_7$ is +6.
$N$ in $HN{O}_3$ can not be oxidised as it is already present in its highest oxidation state.