Question

Balance each of the following oxidation-reduction reactions. In each, indicate which substance is being oxidized and which is being reduced. a. \(\mathrm{Na}(s)+\mathrm{S}(s) \rightarrow \mathrm{Na}_{2} \mathrm{S}(s)\) b. \(\operatorname{Mg}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MgO}(s)\) c. \(\mathrm{Ca}(s)+\mathrm{F}_{2}(g) \rightarrow \mathrm{CaF}_{2}(s)\) d. \(\operatorname{Fe}(s)+\mathrm{Cl}_{2}(g) \rightarrow \operatorname{Fe} \mathrm{Cl}_{3}(s)\)

Short answer

a. The balanced equation is \(2\,\mathrm{Na}(s) + \mathrm{S}(s) \rightarrow \mathrm{Na}_{2}\mathrm{S}(s)\). Na is being oxidized and S is being reduced. b. The balanced equation is \(2\,\mathrm{Mg}(s) + \mathrm{O}_{2}(g) \rightarrow 2\,\mathrm{MgO}(s)\). Mg is being oxidized and O is being reduced. c. The balanced equation is \(\mathrm{Ca}(s) + \mathrm{F}_{2}(g) \rightarrow \mathrm{CaF}_{2}(s)\). Ca is being oxidized and F is being reduced. d. The balanced equation is \(2\,\mathrm{Fe}(s) + 3\,\mathrm{Cl}_{2}(g) \rightarrow 2\,\mathrm{FeCl}_{3}(s)\). Fe is being oxidized and Cl is being reduced.

Step-by-step solution

1. Balance the reaction of Na and S to form Na2S.

To balance the reaction: \[\mathrm{Na}(s) + \mathrm{S}(s) \rightarrow \mathrm{Na}_{2}\mathrm{S}(s),\] we need 2 mol of Na to react with 1 mol of S to form 1 mol of Na2S. So, the balanced equation is: \[2\,\mathrm{Na}(s) + \mathrm{S}(s) \rightarrow \mathrm{Na}_{2}\mathrm{S}(s).\] As the oxidation state of Na increases from 0 to +1, Na is being oxidized. The oxidation state of S decreases from 0 to -2, making S the substance being reduced.

2. Balance the reaction of Mg and O2 to form MgO.

To balance the reaction: \[\mathrm{Mg}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{MgO}(s),\] we need 2 mol of Mg to react with 1 mol of O2 to form 2 mol of MgO. The balanced equation is: \[2\,\mathrm{Mg}(s) + \mathrm{O}_{2}(g) \rightarrow 2\,\mathrm{MgO}(s).\] Here, the oxidation state of Mg increases from 0 to +2, so Mg is being oxidized. The oxidation state of O decreases from 0 to -2, making O the substance being reduced.

3. Balance the reaction of Ca and F2 to form CaF2.

To balance the reaction: \[\mathrm{Ca}(s) + \mathrm{F}_{2}(g) \rightarrow \mathrm{CaF}_{2}(s),\] we need 1 mol of Ca to react with 1 mol of F2 to form 1 mol of CaF2. The balanced equation is: \[\mathrm{Ca}(s) + \mathrm{F}_{2}(g) \rightarrow \mathrm{CaF}_{2}(s).\] The oxidation state of Ca increases from 0 to +2, indicating that Ca is being oxidized. The oxidation state of F decreases from 0 to -1, so F is the substance being reduced.

4. Balance the reaction of Fe and Cl2 to form FeCl3.

To balance the reaction: \[\mathrm{Fe}(s) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{FeCl}_{3}(s),\] we need 2 mol of Fe to react with 3 mol of Cl2 to form 2 mol of FeCl3. The balanced equation is: \[2\,\mathrm{Fe}(s) + 3\,\mathrm{Cl}_{2}(g) \rightarrow 2\,\mathrm{FeCl}_{3}(s).\] In this case, the oxidation state of Fe increases from 0 to +3, meaning that Fe is being oxidized. The oxidation state of Cl decreases from 0 to -1, so Cl is the substance being reduced.

Key concepts

Chemical Balancing

Understanding how to balance chemical equations is a fundamental skill in chemistry. A balanced chemical equation ensures that the same number of atoms of each element is present on both sides of the equation, following the Law of Conservation of Mass.

Let's start with the basic concept of balancing equations. In a chemical reaction, you have reactants (substances that start the reaction) and products (substances that are produced by the reaction). To balance an equation, you adjust the coefficients (the numbers in front of the chemical formulas) to ensure that the number of atoms of each element is the same on both sides of the equation. For instance, in the reaction of sodium (Na) with sulfur (S) to form sodium sulfide (Na2S), it is balanced by ensuring that two sodium atoms react with one sulfur atom, which is explicitly shown in the balanced equation: \[2\,\mathrm{Na}(s) + \mathrm{S}(s) \rightarrow \mathrm{Na}_{2}\mathrm{S}(s).\]
This concept applies to all chemical reactions, and it is crucial for students to practice this skill to understand further chemical concepts.

Oxidation States

Oxidation states are important for understanding the transfer of electrons in chemical reactions, especially in oxidation-reduction reactions. An oxidation state, often known as an oxidation number, is a number assigned to an element in a compound that represents the number of electrons lost or gained by an atom of that element in the compound.

The oxidation state provides insight into the electron configuration of an element in a compound and helps us understand how electrons are transferred in a reaction. As an example, in the reaction of magnesium (Mg) with oxygen (O2) to produce magnesium oxide (MgO), the oxidation state of magnesium changes from 0 to +2, indicating it loses two electrons and is thereby oxidized. Similarly, oxygen gains those electrons, reducing its oxidation state from 0 to -2.

In simpler terms, the element with an increasing oxidation state is being oxidized (losing electrons), while the element with a decreasing oxidation state is being reduced (gaining electrons). Understanding how oxidation states change in a redox reaction is crucial for identifying what is being oxidized and what is being reduced.

Redox Reaction Identification

Redox reactions involve the transfer of electrons between two substances. Identifying redox reactions requires understanding which substance is oxidized and which is reduced. To do this, we look at the change in oxidation states of the elements involved.

Here's a breakdown for identifying a redox reaction: The substance undergoing oxidation shows an increase in oxidation state and is considered the reducing agent. Conversely, the substance undergoing reduction shows a decrease in oxidation state and is considered the oxidizing agent. For instance, in the reaction of calcium (Ca) with fluorine (F2) to form calcium fluoride (CaF2), calcium's oxidation state increases from 0 to +2, so it is the reducing agent and is being oxidized. Fluorine's oxidation state decreases from 0 to -1, making it the oxidizing agent and undergoing reduction.

Recognizing these changes in oxidation states and being able to determine which substances are oxidized and reduced is critical in the field of chemistry for interpreting and predicting the outcomes of redox reactions.