Use the chemical equation to answer the question.WO3(s) + 3H2(g) → W(s) + 3H2O(g)Which statement correctly describes the reaction?(1 point)ResponsesTungsten (W) changes oxidation numbers from +6 to zero, so it undergoes reduction.Tungsten (W) changes oxidation numbers from +6 to zero, so it undergoes reduction.Oxygen (O) changes oxidation numbers from –2 to zero, so it undergoes oxidation.Oxygen (O) changes oxidation numbers from –2 to zero, so it undergoes oxidation.Oxygen (O) changes oxidation numbers from –2 to zero, so it undergoes reduction.Oxygen (O) changes oxidation numbers from –2 to zero, so it undergoes reduction.Tungsten (W) changes oxidation numbers from +6 to zero, so it undergoes oxidation.

Answer 1

To analyze the chemical reaction \( \text{WO}_3(s) + 3\text{H}_2(g) \rightarrow \text{W}(s) + 3\text{H}_2\text{O}(g) \), we need to determine the changes in oxidation states.

1. Tungsten (W) in \( \text{WO}_3 \) has an oxidation state of +6. In the products, tungsten (W) exists in its elemental form, which has an oxidation state of 0. Therefore, tungsten is reduced, as its oxidation state decreases from +6 to 0.

2. Oxygen (O) in \( \text{H}_2\text{O} \) has an oxidation state of -2. In the reactants, oxygen in \( \text{WO}_3 \) has an oxidation state of -2 as well, but in its elemental form (if we were to consider any \( O_2 \) or separate elemental oxygen, which is not in this direct reaction), it would have an oxidation state of 0. No change in oxidation state is needed for oxygen in \( \text{H}_2\text{O} \) since it doesn't appear in elemental form here like tungsten does.

Based on these points, the correct statement is:

Tungsten (W) changes oxidation numbers from +6 to zero, so it undergoes reduction.