Question

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

• A. $Cr^{3 +}$ and $Cr_{2}O_{7}^{2 -}$ are formed
• B. $Cr_{2}O_{7}^{2 -}$ and $H_{2}O$ are formed
• C. $CrO_{4}^{2 -}$ is reduced to +3 state of Cr
• D. $CrO_{4}^{2 -}$ is oxidized to +7 state of Cr.

Answer 1

Answer: (B)

$Cr_{2}O_{7}^{2 -}$ and $H_{2}O$ are formed

Answer 2

Dilute nitric acid converts chromate into dichromate and $H_{2}O$

$2K_{2}CrO_{4} + 2HNO_{3} \rightarrow K_{2}Cr_{2}O_{7} + 2KNO_{3} + H_{2}O$or, $\underset{Yellow}{2CrO_{4}^{2 -}}$ $\mathrm { H } ^ { + }$ $\underset{Orange}{Cr_{2}O_{7}^{2 -}} + H_{2}O$