What would happen when a 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
$C r O_{4}^{2 -}$ is reduced to +3 state of Cr
$C r_{2} O_{7}^{2 -}$ and $H_{2} O$ are formed
$C r O_{4}^{2 -}$ is oxidized to +7 state of Cr

Question

What would happen when a 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
$C r O_{4}^{2 -}$ is reduced to +3 state of Cr
$C r_{2} O_{7}^{2 -}$ and $H_{2} O$ are formed
$C r O_{4}^{2 -}$ is oxidized to +7 state of Cr

Toppr Admin · Accepted Answer

A solution of potassium-xA0-chromate- when-xA0-treated with an excess of dilute nitric acid-xA0- Cr2O2-x2212-7 and H2O are formed-2K2CrO4-2HNO3-x2192-K2Cr2O7-2KNO3-H2OHence-xA0-Cr2O2-x2212-7 and H2O are formed-

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?Cr3+ and Cr2O2−7 are formedCrO2−4 is reduced to +3 state of CrCr2O2−7 and H2O are formedCrO2−4 is oxidized to +7 state of Cr

A solution of potassium chromate, when treated with an excess of dilute nitric acid, Cr2O2−7 and H2O are formed.2K2CrO4+2HNO3→K2Cr2O7+2KNO3+H2OHence, Cr2O2−7 and H2O are formed.

What would happen when a 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
$C r O_{4}^{2 -}$ is reduced to +3 state of Cr
$C r_{2} O_{7}^{2 -}$ and $H_{2} O$ are formed
$C r O_{4}^{2 -}$ is oxidized to +7 state of Cr

A solution of potassium chromate, when treated with an excess of dilute nitric acid, $C r_{2} O_{7}^{2 -}$ and $H_{2} O$ are formed.

$2 K_{2} C r O_{4} + 2 H N O_{3} \to K_{2} C r_{2} O_{7} + 2 K N O_{3} + H_{2} O$

Hence, $C r_{2} O_{7}^{2 -}$ and $H_{2} O$ are formed.