Question

For each of the following oxidation-reduction reactions, identify which element is being oxidized and which is being reduced. a. \(2 \mathrm{Fe}(s)+3 \mathrm{F}_{2}(g) \rightarrow 2 \mathrm{FeF}_{3}(s)\) b. \(\mathrm{O}_{2}(g)+2 \mathrm{Cu}(s) \rightarrow 2 \mathrm{CuO}(s)\) c. \(\mathrm{F}_{2}(g)+2 \mathrm{KI}(a q) \rightarrow 2 \mathrm{KF}(a q)+\mathrm{I}_{2}(s)\) d. \(2 \mathrm{Al}(s)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{AlH}_{3}(s)\)

Short answer

a. In the reaction, Fe is being oxidized (from 0 to +3) and F2 is being reduced (from 0 to -1). b. In the reaction, Cu is being oxidized (from 0 to +2) and O2 is being reduced (from 0 to -2). c. In the reaction, I is being oxidized (from -1 to 0) and F2 is being reduced (from 0 to -1). d. In the reaction, Al is being oxidized (from 0 to +3) and H2 is being reduced (from 0 to -1).

Step-by-step solution

Determine the oxidation numbers of each element

Before the reaction, Fe has an oxidation number of 0, since it is in its elemental form. F2 also has an oxidation number of 0, as it is in its molecular form. In FeF3, Fe has an oxidation number of +3, and F has an oxidation number of -1.

Identify which element is being oxidized and which is being reduced

Since Fe gains a positive oxidation number (+3), it is being oxidized. F2 loses its oxidation number (from 0 to -1) and is therefore being reduced. b. \(\mathrm{O}_{2}(g)+2 \mathrm{Cu}(s) \rightarrow 2 \mathrm{CuO}(s)\)

Determine the oxidation numbers of each element

Before the reaction, O2 has an oxidation number of 0, since it is in its molecular form. Cu has an oxidation number of 0, as it is in its elemental form. In CuO, Cu has an oxidation number of +2, and O has an oxidation number of -2.

Identify which element is being oxidized and which is being reduced

Since Cu gains a positive oxidation number (+2), it is being oxidized. O2 loses its oxidation number (from 0 to -2) and is therefore being reduced. c. \(\mathrm{F}_{2}(g)+2 \mathrm{KI}(a q) \rightarrow 2 \mathrm{KF}(a q)+\mathrm{I}_{2}(s)\)

Determine the oxidation numbers of each element

Before the reaction, F2 has an oxidation number of 0, as it is in its molecular form. In KI, K has an oxidation number of +1, and I has an oxidation number of -1. In KF, K has an oxidation number of +1, and F has an oxidation number of -1. I2 has an oxidation number of 0 because it is in its molecular form.

Identify which element is being oxidized and which is being reduced

Since I loses its negative oxidation number (from -1 to 0), it is being oxidized. F2 loses its oxidation number (from 0 to -1) and is therefore being reduced. d. \(2 \mathrm{Al}(s)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{AlH}_{3}(s)\)

Determine the oxidation numbers of each element

Before the reaction, Al has an oxidation number of 0, as it is in its elemental form. H2 also has an oxidation number of 0, as it is in its molecular form. In AlH3, Al has an oxidation number of +3, and H has an oxidation number of -1.

Identify which element is being oxidized and which is being reduced

Since Al gains a positive oxidation number (+3), it is being oxidized. H2 loses its oxidation number (from 0 to -1) and is therefore being reduced.

Key concepts

Oxidation Numbers

Understanding oxidation numbers is essential in redox reactions. They help track the transfer of electrons.

• An oxidation number is a theoretical charge that an atom would have if all bonds to other atoms were completely ionic.
• For atoms in their pure elemental form, the oxidation number is always 0. For example, both Fe and F in their elemental forms have oxidation numbers of 0.
• When forming compounds, elements usually gain or lose electrons, changing their oxidation states. For example, in the compound FeF₃, the oxidation number of Fe is +3, while for F, it's -1.

By determining which element's oxidation number increases or decreases, you can identify if the element underwent oxidation or reduction.

Oxidation

Oxidation refers to the loss of electrons or an increase in oxidation number. Oxidation is happening if you notice that the oxidation number of an atom increases during a chemical reaction.

• For example, in the reaction involving Cu and O₂, Cu goes from an oxidation number of 0 to +2, indicating it has lost electrons. Hence, Cu is oxidized.
• Oxidation is often remembered using the phrase 'OIL RIG'—Oxidation Is Loss (of electrons).

Understanding oxidation is crucial for balancing redox reactions, as it shows where electrons are being lost.

Reduction

Reduction is the gain of electrons or a decrease in oxidation number. An element is reduced if its oxidation number becomes more negative or decreases through the reaction.

• Take the example of O₂ reacting with Cu. O decreases its oxidation state from 0 to -2. This shows O has gained electrons, meaning O is reduced in this reaction.
• The 'OIL RIG' acronym helps remember this—Reduct Simultaneously, the other element must be oxidized.

ion Is Gain (of electrons). When an element is reduced, an electron is added, balancing the loss from the oxidized element.

Chemical Equations

Chemical equations represent the transformation of substances through redox reactions. They are essential for understanding how elements swap electrons and undergo changes in state.

• A balanced equation must have equal numbers of each type of atom on both sides. Use oxidation numbers to ensure this balance, especially in redox reactions.
• In \[ 2 \text{Fe}(s) + 3 \text{F}_{2}(g) \rightarrow 2 \text{FeF}_{3}(s) \]Fe is oxidized and F₂ is reduced, showing the transfer of electrons.
• Writing balanced chemical equations requires careful observation of changes in oxidation numbers to depict a realistic and correct redox process.

Overall, understanding chemical equations allows you to predict product formation and the dynamics of chemical reactions.