Question

`2"AgNO"_(3"(aq)") + ("NH"_4)_2"Cr"_2"O"_(7"(aq)")-> 2"NH"_4"NO"_(3"(aq)") + "Ag"_2"Cr"_2"O"_(7"(s)")` Find the ionic equation of formation of silver dichromate ?

Answer 1

The net ionic equation is `"2Ag"^"+""(aq)" + "Cr"_2"O"_7^"2-""(aq)" → "Ag"_2"Cr"_2"O"_7"(s)"`

Explanation:

The "molecular equation" is

`"2AgNO"_3"(aq)" + ("NH"_4)_2"Cr"_2"O"_7"(aq)" → "2NH"_4"NO"_3"(aq)" + "Ag"_2"Cr"_2"O"_7"(s)"`

To get the ionic equation, we write the soluble ionic compounds as ions.

`"2Ag"^"+""(aq)" + "2NO"_3^"-""(aq)" + "2NH"_4^"+""(aq)" + "Cr"_2"O"_7^"2-""(aq)" → "2NH"_4^"+""(aq)" + "2NO"_3^"-""(aq)" + "Ag"_2"Cr"_2"O"_7"(s)"`

To get the net ionic equation, we cancel the spectator ions: the ions that appear on both sides of the reaction arrow.

`"2Ag"^"+""(aq)" + color(red)(cancel(color(black)("2NO"_3^"-""(aq)"))) + color(red)(cancel(color(black)("2NH"_4^"+""(aq)"))) + "Cr"_2"O"_7^"2-""(aq)" → color(red)(cancel(color(black)("2NH"_4^"+""(aq)"))) + color(red)(cancel(color(black)("2NO"_3^"-""(aq)"))) + "Ag"_2"Cr"_2"O"_7"(s)"`

The net ionic equation is

`"2Ag"^"+""(aq)" + "Cr"_2"O"_7^"2-""(aq)" → "Ag"_2"Cr"_2"O"_7"(s)"`