Final answer:
In the reaction SiO₂ + 2H₂O → HSiO₄, oxidation numbers indicate that silicon (Si) and hydrogen (H) maintain their oxidation states, and thus neither are oxidized nor reduced.
Step-by-step explanation:
To determine the oxidation numbers and identify the elements that are oxidized or reduced in the reaction SiO₂ (s) + 2H₂O(l) → HSiO₄(aq), we first assign oxidation numbers to each element. In SiO₂, silicon (Si) has an oxidation number of +4 (since oxygen (O) is usually -2 and there are two oxygens, giving -4 total, which must balance with Si for a neutral molecule). In H₂O, hydrogen (H) is +1 (since there are two H for a total of +2, balancing with one -2 O). In HSiO₄, silicon remains +4, hydrogen is +1, and oxygen is -2. Since the oxidation number of silicon and hydrogen remain unchanged, they are neither oxidized nor reduced. Oxygen also retains the same oxidation number in both reactants and products, and therefore is neither oxidized nor reduced. Therefore, the answer to the student's question is that none of the elements are oxidized or reduced in the given reaction, making all answer choices A, B, C, and D incorrect.