Selenium is prepared by the reaction of \(\mathrm{H}_{2} \mathrm{SeO}_{3}\) with gascous \(\mathrm{SO}_{2}\) (a) What redox process does the sulfur dioxide undergo? What is the oxidation state of sulfur in the product? (b) Given that the reaction occurs in acidic aqueous solution, what is the formula of the sulfur-containing species? (c) Write the balanced redox equation for the process.

Understanding oxidation states is fundamental in redox reactions. An oxidation state indicates the degree of oxidation (loss of electrons) of an atom in a chemical compound. It helps determine which element is being oxidized and which is being reduced in a reaction.
For example, in \(\text{SO}_2\), the oxidation state of sulfur is +4. This is calculated based on the common oxidation states of oxygen (-2 each) and the overall neutrality of the molecule.
When \(\text{SO}_2\) reacts in an acidic solution, the oxidation state changes as it is converted to \(\text{SO}_4^{2-}\), where sulfur has an oxidation state of +6. This increase in oxidation state indicates that sulfur is losing electrons, which characterizes its role as a reducing agent.

Half-reactions are a way to split a redox reaction into two parts: one for oxidation and one for reduction. This helps in understanding and balancing the reaction.
For instance, to determine the redox changes in the reaction, we write the half-reactions. The reduction half-reaction for \(\text{H}_2\text{SeO}_3 \) is: \[ \text{H}_2\text{SeO}_3 + 4H^+ + 4e^- \rightarrow \text{Se} + 3\text{H}_2\text{O} \]. Here, \( \text{Se} \) gains electrons (reduction).
The oxidation half-reaction for \( \text{SO}_2 \) is: \[ \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow\ \text{SO}_4^{2-} + 4H^+ + 2e^- \]. This shows \( \text{SO}_2 \) losing electrons (oxidation).
Breaking down the steps and writing half-reactions make it easier to see the electron transfer process.

Balancing redox reactions is crucial to ensure the conservation of mass and charge. First, write the individual half-reactions and balance all atoms except hydrogen and oxygen. Next, balance oxygen atoms by adding \( \text{H}_2\text{O} \) molecules and hydrogen atoms by adding \( H^+ \) ions (since the reaction is in acidic solution).
Then, balance the charge by adding electrons ( \( e^- \)). Here, the reduction half-reaction: \[ \text{H}_2\text{SeO}_3 + 4H^+ + 4e^- \rightarrow \text{Se} + 3\text{H}_2\text{O} \] and the oxidation half-reaction: \[ \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4H^+ + 2e^- \].
Combine the half-reactions by ensuring the same number of electrons are transferred. Multiply the reduction half-reaction by 1 and the oxidation half-reaction by 2 to balance the electrons. Combining yields the balanced overall equation: \[ \text{H}_2\text{SeO}_3 + 2\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{Se} + 2\text{SO}_4^{2-} + 4H^+ \]. This shows that both mass and charge are balanced in the final equation.