Question

4. For the reaction shown in equation 2, identify the atom oxidized, the atom reduced, the oxidizing agent and the reducing agent. 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g) (2)

Answer 1

In the reaction 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g), sodium is oxidized, hydrogen is reduced, water (H2O) is the oxidizing agent, and sodium (Na) is the reducing agent.

Step-by-step explanation:

In the reaction 2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g), to determine which atom is oxidized and which is reduced, we examine the changes in oxidation states. Sodium (Na) starts as a neutral atom and becomes part of the compound NaOH, where it has an oxidation state of +1.

Hence, sodium is oxidized because it loses electrons. On the other side, oxygen (O) in water starts with an oxidation state of -2 and ends with the same state in NaOH. However, hydrogen (H2O) changes from +1 in water to 0 in molecular hydrogen gas (H2), so hydrogen is reduced as it gains electrons.

The oxidizing agent is the species that gets reduced, which in this case is water (H2O), while the reducing agent is the species that gets oxidized, which is sodium (Na).