# Oxidizing agent, reducing agent, acid or base

In concrete reactions, how can you define whether Ammonia reacts as an oxidizing agent, reducing agent, an acid or a base? I'm confused because in the examples given it's not an acid/ base when an $\ce {H+}$ is donated/ accepted.

$$\ce {2NH3 + 3Cl2 -> N2 + 6HCl}$$ (oxidizing agent)

$$\ce {Na + NH3 -> NaNH2 + \frac{1}{2}H2}$$ (reducing agent)

$$\ce{NH3 + H2O + CO2 -> NH4HCO3}$$ (base)

$$\ce{NH3 + LiCH3 -> LiNH2 + CH4}$$ (acid)

• In the example reactions, ammonia is incorrectly stated to be an oxidizing agent in the reaction with chlorine (here it acts as a reducing agent); when reacting with sodium metal, ammonia is an oxidizing agent and not a reducing agent as shown. Not sure that helps address your question. – iad22agp Jun 17 '17 at 12:11

The use of the terms "oxidizing agent" and "reducing agent" in this example apparently refer to the overall reaction. Ammonia certainly can act as an acid in the process of reacting with sodium metal, and as a (Lewis) base during the reaction with chlorine. However, the net effect of the reaction is a reduction or oxidation, rather than a simple proton transfer.

I'm basing this answer from translation of the original question (in German) which I believe is supposed to read:

Ammonia can exist in oxidised form, reduced form, acid or base. Name the form in which ammonia exist in the following: (correct me if I am wrong)

Focusing on the acid/base forms (3) and (4):

• ammonia acts as an base by accepting a proton $\ce{H+}$ from $\ce{H2O}$.

Recall the Brønsted-Lowry theory of acids and bases which states: An acid is a proton (hydrogen ion) donor, and a base is a proton (hydrogen ion) acceptor.

• similarly on the last equation $\ce{NH3}$ donates a proton to $\ce{LiCH3}$ to form lithium amide $\ce{Li^+ NH2^-}$ ($\ce{LiNH2}$) and $\ce{CH4}$