Question

which component is reduced in the reaction below? 2na + s → na₂s na in na₂s s solid s in na₂s na solid

Explanation:

Step1: Recall reduction definition

Reduction is gain of electrons (or decrease in oxidation state).

Step2: Find oxidation states

• For solid Na: oxidation state is 0. In \( \text{Na}_2\text{S} \), Na has oxidation state \( +1 \) (since S is \( -2 \), and \( 2(+1) + (-2) = 0 \)). So Na is oxidized (loses electrons, oxidation state increases from 0 to +1).
• For solid S: oxidation state is 0. In \( \text{Na}_2\text{S} \), S has oxidation state \( -2 \). So S gains electrons (oxidation state decreases from 0 to -2), meaning S is reduced. The reduced S is now in \( \text{Na}_2\text{S} \), so the component reduced is S in \( \text{Na}_2\text{S} \).

Answer:

S in \( \text{Na}_2\text{S} \) (the option: S in \( \text{Na}_2\text{S} \))