Question

In the ring test of $N{O_3}^ - $ ion, $F{e^{2 + }}$ ions reduces nitrate ion to nitric oxide, which combines with $F{e^{2 + }}(aq)$ ion to form a brown complex. Write the reaction involved in the formation of the brown ring.

Answer 1

Hint: A nitrate test is used to determine the presence of nitrate ion in solution. Testing for the presence of nitrate is generally difficult compared with testing for other anions, as almost all nitrates are soluble in water.

Complete answer:
Brown ring test is a qualitative test and can be performed by adding iron sulfate to a solution of a nitrate, then slowly adding concentrated sulfuric acid such that the acid forms a layer below the aqueous solution. A brown ring will form at the junction of the two layers, indicating the presence of the nitrate ion.
$
  N{O_3} + 3F{e^{2 + }} + 4{H^ + } \to NO + 3F{e^{3 + }} + 2{H_2}O \\
  {[Fe{({H_2}O)_6}]^{2 + }} + NO \to {[Fe{({H_2}O)_5}(NO)]^{2 + }} + {H_2}O \\
$
In the first reaction, ferrous ions react with nitrate ions in the presence of protons to form ferric ions and nitric oxide. In the first reaction, the oxidation state of nitrogen decreases by $3$.
In the second reaction nitric oxide reacts with hydrated ferrous ions to form a brown ring complex. Also there is no change in oxidation of ferrous ions.

Note:
Note that the presence of nitrite ions will interfere with this test. The nitrate anion is an oxidizer, and many tests for the nitrate anion are based on this property. Unfortunately, other oxidants present in the analyte may interfere and give erroneous results. Nitrate can also be detected by first reducing it to the more reactive nitrite ion and using one of many nitrite tests.