Write a balanced equation describing the reduction of \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) by \(\mathrm{SO}_{2}\) to produce selenium.

We start by writing the unbalanced redox reaction with the given reactants and products. \[ \mathrm{H}_{2}\mathrm{SeO}_{4} + \mathrm{SO}_{2} \rightarrow \mathrm{Se} + ? \]

First, assign oxidation numbers to each element in the reactants and products: \(H_2SeO_4\): H = +1, Se = +6, and O = -2 \(SO_2\): S = +4 and O = -2 \(Se\): Se = 0 From the oxidation numbers, we can identify the half-reactions: Oxidation half-reaction: \(SO_2\) → S Reduction half-reaction: \(H_2SeO_4\) → Se Now, we balance the atoms in each half-reaction: Oxidation half-reaction: \[ \mathrm{SO}_{2} \rightarrow \mathrm{S} + \mathrm{O}_{2} \] Reduction half-reaction: \[ \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow \mathrm{Se} + 2 \mathrm{H}_{2}\mathrm{O} \]

To balance the charges, we add electrons to the appropriate side of each half-reaction. Oxidation half-reaction: Since S goes from +4 to 0, we need to add 4 electrons to the right side. \[ \mathrm{SO}_{2} \rightarrow \mathrm{S} + \mathrm{O}_{2} + 4 e^{-} \] Reduction half-reaction: Since Se goes from +6 to 0, we need to add 6 electrons to the left side. \[ 6 e^{-} + \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow \mathrm{Se} + 2 \mathrm{H}_{2}\mathrm{O} \]

To combine the half-reactions, we need to multiply each by a factor that makes the number of electrons equal in both half-reactions. Since we have 4 electrons in the oxidation half-reaction and 6 in the reduction half-reaction, we can find the least common multiple, which is 12. Therefore, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. Oxidation half-reaction x 3: \[ 3 \mathrm{SO}_{2} \rightarrow 3\mathrm{S} + 3\mathrm{O}_{2} + 12 e^{-} \] Reduction half-reaction x 2: \[ 12 e^{-} + 2 \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow 2\mathrm{Se} + 4 \mathrm{H}_{2}\mathrm{O} \] Combining the balanced half-reactions: \[ 3 \mathrm{SO}_{2} + 2 \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow 3\mathrm{S} + 3\mathrm{O}_{2} + 2\mathrm{Se} + 4 \mathrm{H}_{2}\mathrm{O} \]

Now, we have the balanced chemical equation for the reduction of H₂SeO₄ by SO₂ to produce selenium: \[ 3 \mathrm{SO}_{2} + 2 \mathrm{H}_{2}\mathrm{SeO}_{4} \rightarrow 3\mathrm{S} + 3\mathrm{O}_{2} + 2\mathrm{Se} + 4 \mathrm{H}_{2}\mathrm{O} \]