Question 21.3: Nitrate ion can be reduced to ammonia by enzymes called nitr...

Collect, Organize, and Analyze We need to assign oxidation numbers according to the guidelines in Section 4.9. We know that the oxidation numbers of nitrogen and oxygen in each polyatomic ion must add up to the charge on the ion. Oxygen in compounds usually has an oxidation number of –2.

Solve The oxidation number of nitrogen is unknown, so we call it x. The oxidation number of nitrogen in NO_{3}^{-} is

x + 3(-2) = -1
x = +5

The oxidation number of nitrogen in NO_{2}^{-} is

x + 2(-2)= -1
x =+3

The half-reaction is

NO_{3}^{-}(aq) → NO_{2}^{-}(aq)

The nitrogen is balanced; we balance oxygen by adding water:

NO_{3}^{-}(aq) → NO_{2}^{-}(aq)+H_{2}O(\ell)

Then we balance hydrogen by adding hydrogen ions:

2 H^{+}(aq) +NO_{3}^{-}(aq) → NO_{2}^{-}(aq)+H_{2}O(\ell)

We balance charge by adding electrons:

2 e^{-} + 2 H^{+}(aq) +NO_{3}^{-}(aq) → NO_{2}^{-}(aq)+H_{2}O(\ell)

We switch to a basic solution by adding the same number of OH^{-} ions to both sides of the equation:

2 OH^{-}(aq) + 2 e^{-} + 2 H^{+}(aq) +NO_{3}^{-}(aq) → NO_{2}^{-}(aq)+H_{2}O(\ell)+2 OH^{-}(aq)

The hydrogen ions combine with the hydroxide ions to form water, and we cancel the species that are the same on both sides:

2 e^{-}+\sout{2} H_{2}O(\ell) + NO_{3}^{-}(aq) → NO_{2}^{-}(aq)+ \sout{H_{2}O(\ell)}+2 OH^{-}(aq)

This gives us a final equation for the reduction half-reaction:

2 e^{-} + H_{2}O(\ell)+NO_{3}^{-}(aq) →NO_{2}^{-}(aq)+2 OH^{-}(aq)

Think About It Assigning oxidation numbers is a convenient way of identifying which element is reduced or oxidized in a half-reaction and of determining how many electrons are gained or lost. The balanced half-reaction confirms that electrons are added to nitrate to reduce it to nitrite ion.