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Chapter 20: Problem 61

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

Short Answer

Expert verified
The balanced equation for the reduction of \(\mathrm{H}_{2}\mathrm{SeO}_{4}\) by \(\mathrm{SO}_{2}\) to produce selenium is: \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 2\mathrm{H}_{2}\mathrm{O} + 3\mathrm{SO}_{2} \rightarrow Se + 5\mathrm{H}_{2}\mathrm{O} + 6e^{-}\).

Step by step solution

01

Write the unbalanced chemical equation.

The equation which includes the reactants and the product is given as \(\mathrm{H}_{2}\mathrm{SeO}_{4} + \mathrm{SO}_{2} \rightarrow Se\) . Note that at this point, the equation is not yet balanced.
02

Balance atoms other than H and O in the equation.

Selenium atoms are already balanced. The sulfur atoms are not balanced. Ideally, we would prefer to have the sulfur atom of \(\mathrm{SO}_{2}\) (which is undergoing oxidation) furnishing the electrons needed for the reduction. However, the oxidation state change of selenium is by 6 and that of sulfur is by 2. To match both, there must be three \(\mathrm{SO}_{2}\) for each \(\mathrm{H}_{2}\mathrm{SeO}_{4}\) . The equation now becomes \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 3\mathrm{SO}_{2} \rightarrow Se\) .
03

Balance the oxygen atoms.

The right side of the equation is missing six oxygen atoms. To balance this, add three water molecules (\(\mathrm{H}_{2}\mathrm{O}\)) on the right side to deliver these oxygen atoms. The equation becomes \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 3\mathrm{SO}_{2} \rightarrow Se + 3\mathrm{H}_{2}\mathrm{O}\) .
04

Balance the hydrogen atoms.

There are four hydrogen atoms on the left side and six hydrogen atoms on the right side of the equation, so two hydrogen atoms need to be added on the left. This is achieved by adding two molecules of H\(_2\)O on the left side of the equation. The equation now becomes \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 2\mathrm{H}_{2}\mathrm{O} + 3\mathrm{SO}_{2} \rightarrow Se + 5\mathrm{H}_{2}\mathrm{O}\) .
05

Balance charge in the atoms.

The oxidation state change of sulfur is from +4 to +6, so it loses two electrons. Therefore, six electrons (for three \(\mathrm{SO}_{2}\) molecules) must be added to the right side to balance this. The equation now becomes \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 2\mathrm{H}_{2}\mathrm{O} + 3\mathrm{SO}_{2} \rightarrow Se + 5\mathrm{H}_{2}\mathrm{O} + 6e^{-}\) .
06

Final check for complete balancing.

Check if all types of atoms are balanced and also the charges. When this is confirmed, the resulting balanced equation is \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 2\mathrm{H}_{2}\mathrm{O} + 3\mathrm{SO}_{2} \rightarrow Se + 5\mathrm{H}_{2}\mathrm{O} + 6e^{-}\) .

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