2$Na_2O_{2(s)} + 2H_2O_{(1)}$→ 4$NaOH_{(s)} +O_{2(g)}$ The substance that is oxidized in the reaction above is

Analyze the question: We need to find which substance loses electrons (its oxidation number increases) in the reaction.

Concept: Oxidation is an increase in oxidation number. Reduction is a decrease in oxidation number.

Step-by-step solution:

The equation as written is not balanced: 2Na₂O₂ + 2H₂O → 4NaOH + O₂ is actually balanced (Na: 4 = 4, O: 4 + 2 = 6 = 4 + 2, H: 4 = 4). Good.

Assign oxidation numbers:

• In Na₂O₂ (a peroxide): Na is +1, so O is -1.
• In H₂O: H is +1, O is -2.
• In NaOH: Na is +1, O is -2, H is +1.
• In O₂: O is 0.

Track the oxygen atoms:

• Oxygen in Na₂O₂ goes from -1 to 0 in O₂. This is an increase, so it is oxidized.
• Oxygen in Na₂O₂ goes from -1 to -2 in NaOH. This is a decrease, so it is reduced.
• Oxygen in H₂O stays at -2 (it ends up in NaOH at -2), so water is neither oxidized nor reduced.

So Na₂O₂ is the only substance that changes in both directions. It undergoes disproportionation, being both oxidized (to O₂) and reduced (to OH⁻). Water does not change its oxidation numbers, so it is not oxidized.

Checking the options:

• A. Na₂O₂: its oxygen is oxidized from -1 to 0. Correct.
• B. NaOH: a product, not a reactant being oxidized.
• C. H₂O: oxidation numbers are unchanged, so it is not oxidized.
• D. O₂: a product.

**Answer: A.

Tip: Peroxides contain oxygen at -1, which sits between 0 and -2. This means peroxides can both gain and lose electrons, so look for disproportionation whenever a peroxide such as H₂O₂ or Na₂O₂ appears in a redox question.