Final answer:
Sulfur in the reaction H2S + 2 NaOH → Na2S + 2 H2O is neither oxidized nor reduced because its oxidation state remains at -2 throughout the reaction.
Step-by-step explanation:
In the chemical reaction H2S + 2 NaOH → Na2S + 2 H2O, sulfur (S) is neither being reduced nor oxidized. During this reaction, sodium hydroxide (NaOH) reacts with hydrogen sulfide (H2S) to produce sodium sulfide (Na2S) and water (H2O). In this reaction, the oxidation state of sulfur in hydrogen sulfide is -2, which remains the same as in sodium sulfide. This tells us that sulfur has not undergone a change in oxidation state, and therefore it has not been oxidized or reduced.
Oxidation and reduction refer to the gain or loss of electrons in a substance, respectively. Oxidizing agents donate oxygen or remove electrons, while reducing agents remove oxygen or add electrons. Since the sulfur in H2S is already in a low oxidation state and doesn't change in this reaction, it is acting neither as an oxidant nor as a reductant.