Question

Sulfur undergoes oxidation to form sulfur trioxide (SO$_3$) and sulfur dioxide (SO$_2$) in the following reactions:
1. \( \text{S} + \text{O}_2 \to \text{SO}_2 \)
2. \( 2\text{SO}_2 + \text{O}_2 \to 2\text{SO}_3 \)
Given that sulfur (S) is oxidized to SO$_2$, and SO$_2$ is further oxidized to SO$_3$, calculate the change in oxidation states of sulfur in both reactions.

Hint:

When calculating changes in oxidation states, remember that oxidation involves an increase in the oxidation state, while reduction involves a decrease in the oxidation state.

Answer 1

- In the first reaction, sulfur (S) combines with oxygen (O) to form sulfur dioxide (SO$_2$). In SO$_2$, sulfur has an oxidation state of +2. Since sulfur in its elemental form (S) has an oxidation state of 0, sulfur is oxidized from 0 to +2. - In the second reaction, sulfur dioxide (SO$_2$) reacts with oxygen to form sulfur trioxide (SO$_3$). In SO$_3$, sulfur has an oxidation state of +6. Therefore, sulfur is further oxidized from +2 to +6. Thus, sulfur is oxidized from 0 to +2 in the first reaction, and from +2 to +6 in the second reaction.