What happens when a solution of pottasium chromate is treated with an excess of dilute nitric acid?

C r^{3 +}

and

C r_{2} O_{7}^{2 -}

are formed

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C r_{2} O_{7}^{2 -}

and

H_{2} O

are formed

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C r O_{4}^{2 -}

is reduced to

+ 3

state of

C r

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C r O_{4}^{2 -}

is reduced to 0 state of

C r

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Solution

The correct option is B $C r_{2} O_{7}^{2 -}$ and $H_{2} O$ are formed
When a solution of potassium chromate is treated with an excess of dilute nitric acid, dichromate and water are formed.
$\begin{matrix} 2 C r O_{4}^{2 -} & + & 2 H^{+} & ⟶ & C r_{2} O_{7}^{2 -} & + & H_{2} O Y e l l o w & O r a n g e \end{matrix}$

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Similar questions

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

{C r_{2} O_{7}}^{2 -}

100 m L

0.01 M {C r_{2} O_{7}}^{2 -}

C r^{3 +}

F e^{2 +}

C O_{2}

C r O_{4}^{2 -}

C r_{2} O_{7}^{2 -}

C r O_{4}^{2 -} (a q) C O_{2} - - \to C r_{2} O_{7}^{2 -}

What happens when K₂Cr₂O₇ is treated with KI?

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