Question

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

Short answer

The balanced equation for the reduction of \(H_{2}SeO_{4}\) by \(SO_{2}\) to produce selenium is: \(H_{2}SeO_{4} + 3SO_{2} + 3H_{2}O → Se + 6H^{+} + 1.5SO_{4}^{2-}\)

Step-by-step solution

1. Determine the initial oxidation states of the elements in the compounds

First, we must determine the oxidation states of the elements in both compounds. For H2SeO4: - The oxidation state of hydrogen is +1. - The oxidation state of oxygen is -2. - Therefore, the oxidation state of selenium should be +6 (as 2(+1) + x + 4(-2) = 0). For SO2: - The oxidation state of oxygen is -2. - Hence, the oxidation state of sulfur should be +4 (as x + 2(-2) = 0).

2. Determine the change in oxidation states during the reaction

Next, we need to determine the change in oxidation states of the two reactants during the reaction to produce selenium. The reduction of H2SeO4 means that selenium will lose electrons and change its oxidation state from +6 to 0 (as selenium is in its elemental form). The SO2 will facilitate this reaction by providing the electrons needed for the reduction to occur. Thus, the oxidation state of sulfur will change from +4 to +6.

3. Balance the equation

Now we'll balance the equation by ensuring that the total number of electrons gained equals the total number of electrons lost. Reduction half-reaction: - H2SeO4 → Se + H2O (unbalanced) - To balance Se: H2SeO4 → Se + 2H2O - To balance O: H2SeO4 → Se + 2H2O + 6e⁻ (Se loses 6 electrons in this reduction half-reaction) Oxidation half-reaction: - SO2 → SO4⁻² (unbalanced) - To balance S: 2SO2 → SO4⁻² - To balance O: 2SO2 + 2H2O → SO4⁻² - To balance H: 2SO2 + 2H2O → SO4⁻² + 4H⁺ - 2SO2 + 2H2O → SO4⁻² + 4H⁺ + 4e⁻ (S gains 4 electrons in this oxidation half-reaction) To make sure the total number of electrons lost equals the total number of electrons gained, we'll multiply the oxidation half-reaction by 1.5 so that both half-reactions have the same number of electrons exchanged: - 3SO2 + 3H2O → 1.5SO4⁻² + 6H⁺ + 6e⁻ Finally, add the two balanced half-reactions: H2SeO4 + 6e⁻ + 3SO2 + 3H2O → Se + 6H⁺ + 1.5SO4⁻² + 6e⁻ The 6e⁻ on both sides of the equation cancel out, leaving:

4. Final balanced equation

H2SeO4 + 3SO2 + 3H2O → Se + 6H⁺ + 1.5SO4⁻² This is the final balanced equation for the reduction of H2SeO4 by SO2 to produce selenium.