What would happen when solution of potassium chromate is treated with the excess of nitric acid?

$C r_{2} O_{7}^{- 2} a n d H_{2} O$ are formed

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$C r_{2} O_{4}^{2 -}$ is reduced to $C r^{+ 3}$

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$C r^{+ 3}$ is oxidized to $C r^{+ 6}$

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$C r^{+ 3} a n d C r_{2} O_{7}^{- 2}$ is formed

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Solution

The correct option is A
$C r_{2} O_{7}^{- 2} a n d H_{2} O$ are formed

$2 C r O_{4}^{- 2} + 2 H^{+} \to C r_{2} O_{7}^{- 2} + H_{2} O$
Hence $C r_{2} O_{7}^{- 2}$ and $H_{2} O$ are formed.

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Similar questions

C r_{2} O_{7}^{- 2} + I^{-} + H^{+} \to C r^{+ 3} + I_{2} + H_{2} O

C r_{2} O_{7}^{- 2} + F e^{+ 2} \to C r^{+ 3} + H_{2} O

What happens when K₂Cr₂O₇ is treated with KI?

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