2Na₂O_2(s) + 2H₂O_2(I) → 4NaOH_(s) + O_2(s). Which substance is oxidized?

Correct answer: H₂O₂ (Option C as intended).

Why it’s not A: The equation is 2Na₂O₂ + 2H₂O₂ → 4NaOH + O₂. Option A lists 2NaO₂ (sodium superoxide), which does not appear in the reaction. The peroxide present is Na₂O₂, and in the process its oxygen goes from −1 (in peroxide) to −2 (in NaOH), so that species is actually being reduced, not oxidized.

Oxidation-number check:

• H₂O₂ (hydrogen peroxide): O = −1 → O = 0 in O₂ (increase) ⇒ oxidation.
• Na₂O₂ (sodium peroxide): O = −1 → O = −2 in NaOH (decrease) ⇒ reduction.

Note: If your option C is printed as “H₂O(l)”, that’s almost certainly a typo for H₂O₂(l). Choose that option as the oxidized substance.