91Ó°ÊÓ

The reactants are given as \(\mathrm{H}_{2}\mathrm{SeO}_{4}\) (selenic acid) and \(\mathrm{SO}_{2}\) (sulfur dioxide). The product is selenium. Our initial unbalanced equation looks like this: \[\mathrm{H}_{2}\mathrm{SeO}_{4} + \mathrm{SO}_{2} \rightarrow \mathrm{Se}\]

Let's determine the oxidation states of the elements in the reactants and products: 1. In \(\mathrm{H}_{2}\mathrm{SeO}_{4}\): - Hydrogen: +1 (Since it's in a compound with nonmetals) - Selenium: Let's call it x - Oxygen: -2 (As it's the most electronegative element) Using the rule that the sum of oxidation states in a compound is zero, we have: \[2(+1) + x + 4(-2) = 0\] Solving for x gives the oxidation state of Selenium: \[x = +6\] 2. In \(\mathrm{SO}_{2}\): - Sulfur: Let's call it y - Oxygen: -2 (still the most electronegative element) Using the same rule for the sum of oxidation states in a compound: \[y + 2(-2) = 0\] Solving for y gives the oxidation state of Sulfur: \[y = +4\] 3. In the Selenium product, the oxidation state is 0, as it's in its elemental form.

Now that we know the oxidation states of the elements involved, let's determine their change in oxidation states during the reaction: 1. Selenium: From +6 in \(\mathrm{H}_{2}\mathrm{SeO}_{4}\) to 0 in the elemental form. - Change in oxidation state: -6 2. Sulfur: From +4 in \(\mathrm{SO}_{2}\) to an unknown oxidation state as there might be other sulfur containing products. Since selenium is reduced by 6 units, we should look for a product where sulfur is oxidized by 6 units to balance the oxidation states. In this case, \(\mathrm{SO}_{4}^{2-}\) (sulfate ion) would be that product, as sulfur has an oxidation state of +6 in it. Our unbalanced equation now looks like this: \[\mathrm{H}_{2}\mathrm{SeO}_{4} + \mathrm{SO}_{2} \rightarrow \mathrm{Se} + \mathrm{SO}_{4}^{2-}\]

To balance the chemical equation, we need to ensure the same number of atoms of each element on both sides of the equation. Looking at the unbalanced equation, we notice that we need to add \(\mathrm{H}^{+}\) and \(\mathrm{H}_{2}\mathrm{O}\) to balance hydrogen and oxygen. Balancing the hydrogen and oxygen atoms: \[\mathrm{H}_{2}\mathrm{SeO}_{4} + \mathrm{SO}_{2} + 2\,\mathrm{H}^+ \rightarrow \mathrm{Se} + \mathrm{SO}_{4}^{2-} + 2\,\mathrm{H}_{2}\mathrm{O} \] Now, our balanced chemical equation for the reduction of \(\mathrm{H}_{2}\mathrm{SeO}_{4}\) by \(\mathrm{SO}_{2}\) to produce selenium is: \[\mathrm{H}_{2}\mathrm{SeO}_{4} + \mathrm{SO}_{2} + 2\,\mathrm{H}^+ \rightarrow \mathrm{Se} + \mathrm{SO}_{4}^{2-} + 2\,\mathrm{H}_{2}\mathrm{O} \]