What would happen when as solution of potassium 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 -}

ang

H_{2} O

are formed

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C r O_{4}^{2 -}

is reduced to

C r^{3 +}

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None of the above

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Solution

The correct option is B $C r_{2} O_{7}^{2 -}$ ang $H_{2} O$ are formed
Reaction:
$C r O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O$

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Similar questions

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

H_{2} S O_{3} + C r_{2} O_{7}^{2 -} \to H_{2} S O_{4} + C r^{3 +} + H_{2} O

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

C r^{3 +}

K_{2} C r_{2} O_{7}

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e \to 2 C r^{3 +} + 7 H_{2} O

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

0.136

N_{2} H_{5}^{+}

N_{2} H_{5}^{+} + C r_{2} O_{7}^{2 -} \to N_{2} + C r^{3 +} + H_{2} O

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