# Identify the substance oxidized, the substance reduced, the oxidizing agent, and the reducing agent? 2HNO3 + 3H3AsO3 -----> 2NO + 3H3AsO4 + H2O

Mar 27, 2015

the substance oxidized is H3AsO3, since the oxidation number of As goes from +3 to +5, the substance reduced is HNO3, with the oxidation number of N going from +5 to +2, the oxidizing agent is the substance reduced ( HNO3) and the reducing agent is the substance oxidized (H3AsO3)

Mar 27, 2015

Nitric acid, $H N {O}_{3}$, is the substance reduced to NO, while arsenous acid, ${H}_{3} A s {O}_{3}$, is oxidized to arsenic acid, ${H}_{3} A s {O}_{4}$.

There are three ways to determine this.

a) The easiest one is to observe the oxygen/non-metal ratio: 3:1 in nitric acid and 1:1 in NO.

So nitrogen undergoes an oxygen loss, that is, the same as a reduction, while arsenic gains an extra oxygen, so it is oxidized.

b) The second way is to balance half-reactions to deduce which is the substance that "absorbs" electrons (the oxidant that is getting reduced) and which one donates electrons (while getting oxidized).

The nitric acid half-reaction is:

$H N {O}_{3} + 3 {H}^{+} + 3 {e}^{-}$$N O + 2 {H}_{2} O$

The three electrons were needed to balance electrical charge on the left side, so this is a reduction half-reaction.

The arsenous acid half-reaction is:

${H}_{3} A s {O}_{3} + {H}_{2} O \to {H}_{3} A s {O}_{4} + 2 {H}^{+} + 2 {e}^{-}$

The electrons were donated, so the half-reaction is an oxidation.

c) The third way consists in "monitoring" the oxidation number (o. n.) of all the elements before and after the reaction.

Once again we have an increase of oxidation number of arsenic, the oxidized element, and a decrease in the o. n. of nitrogen, the reduced element.