JEE MAIN - Chemistry (2003 - No. 45)
What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?
$$Cr_2O^{2-}_7$$ and H2O are formed
$$CrO^{2-}_4$$ is reduced to +3 state of Cr
$$CrO^{2-}_4$$ is oxidized to +7 state of Cr
Cr3+ and $$Cr_2O^{2-}_7$$ are formed
Explanation
When a solution of potassium chromate is treated with an excess of dilute nitric acid. Potassium dichromate and $${H_2}O$$ are formed.
$$2{K_2}Cr{O_4} + 2HN{O_3} \to {K_2}C{r_2}{O_7} + 2KN{O_3} + {H_2}O$$
Hence $$C{r_2}{O_7}^ - $$ and $${H_2}O$$ are formed.
$$2{K_2}Cr{O_4} + 2HN{O_3} \to {K_2}C{r_2}{O_7} + 2KN{O_3} + {H_2}O$$
Hence $$C{r_2}{O_7}^ - $$ and $${H_2}O$$ are formed.
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