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 \(\mathrm{H}_{2}\mathrm{SeO}_{4}\) 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

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.

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 \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 3\mathrm{SO}_{2} \rightarrow Se\) .

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 \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 3\mathrm{SO}_{2} \rightarrow Se + 3\mathrm{H}_{2}\mathrm{O}\) .

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 \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 2\mathrm{H}_{2}\mathrm{O} + 3\mathrm{SO}_{2} \rightarrow Se + 5\mathrm{H}_{2}\mathrm{O}\) .

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 \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 2\mathrm{H}_{2}\mathrm{O} + 3\mathrm{SO}_{2} \rightarrow Se + 5\mathrm{H}_{2}\mathrm{O} + 6e^{-}\) .

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 \(\mathrm{H}_{2}\mathrm{SeO}_{4} + 2\mathrm{H}_{2}\mathrm{O} + 3\mathrm{SO}_{2} \rightarrow Se + 5\mathrm{H}_{2}\mathrm{O} + 6e^{-}\) .

Key concepts

Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants, the substances that start a reaction, and the products, the substances formed by the reaction.
In the given exercise, we start with \(\mathrm{H}_2 \mathrm{SeO}_4\) and \(\mathrm{SO}_2\) as our reactants and end with selenium and water as our products. The arrow \((\rightarrow)\) separates reactants from products, indicating the direction of the reaction.
Writing a chemical equation is essential as it helps visualize what is happening at the molecular level, providing a clear picture of the transformation involved in a reaction.

Balancing Equations

Balancing equations ensures that the number of atoms for each element is the same on both sides of the equation, reflecting the law of conservation of mass.
In balancing equations, we often need to adjust the coefficients (numbers in front of molecules) rather than changing the subscripts (numbers as part of chemical formulas).

• First, balance elements other than hydrogen and oxygen.
• Next, balance oxygen molecules, typically by adding water.
• Finally, adjust the hydrogen atoms.

The given equation becomes balanced by making sure the selenium, oxygen, and hydrogen atoms, along with any charges, are equal on both sides. This results in a correct and balanced chemical equation for the redox reaction.

Oxidation States

Oxidation states, also known as oxidation numbers, are used to track the transfer of electrons in chemical reactions.
They can be thought of as a "bookkeeping" system to help us determine how many electrons are gained or lost.

• In the exercise, selenium undergoes a change in oxidation state from +6 in \(\mathrm{H}_2 \mathrm{SeO}_4\) to 0 in elemental selenium \((Se)\). This indicates a reduction, as it gains electrons.
• Sulfur changes from +4 in \(\mathrm{SO}_2\) to a higher state, meaning it loses electrons and undergoes oxidation.

Understanding oxidation states is crucial for analyzing and balancing redox reactions, as it helps identify which elements are oxidized and which are reduced.

Reduction

Reduction is the gain of electrons by an atom, ion, or molecule. This concept is often paired with oxidation, as redox reactions involve the transfer of electrons.
In the context of our reaction, selenium is reduced as it gains electrons, moving from an oxidation state of +6 to 0 in elemental selenium (Se). This results in the formation of selenium from \(\mathrm{H}_2 \mathrm{SeO}_4\).
By identifying which elements are reduced in a reaction, we can better understand the overall chemical changes taking place and balance the equation accurately.

Sulfur Compounds

Sulfur compounds feature prominently in many chemical reactions, noteworthy for their distinct oxidation states and roles in redox reactions.
In the reaction provided, \(\mathrm{SO}_2\) (sulfur dioxide) participates as a reactant and undergoes oxidation, transitioning from an oxidation state of +4 to +6.
This shift facilitates the reduction of selenium from \(\mathrm{H}_2 \mathrm{SeO}_4\), leading to a balanced redox equation.

• Understanding how sulfur compounds behave in reactions aids comprehension of the essential steps in balancing redox reactions.
• They often provide a framework for assessing other elements' behaviors within the reaction.

Familiarity with sulfur compounds and their chemical behavior is vital in mastering the intricacies of redox reactions.