Question

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) \(2 \mathrm{Fe}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

(a) In the reaction \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\), Hydrogen goes from 0 to +1, and Fluorine goes from 0 to -1. The total number of electrons transferred is 2. (b) In the reaction \(2 \mathrm{Fe}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\), Iron goes from +2 to +3, and Oxygen goes from -1 in \(\mathrm{H}_{2} \mathrm{O}_{2}\) to -2 in \(\mathrm{H}_{2} \mathrm{O}\). The total number of electrons transferred is 2. (c) In the reaction $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$, Carbon goes from -4 to +4, and Oxygen goes from 0 to -2. The total number of electrons transferred is 8.

Step-by-step solution

Assign oxidation numbers

Using the oxidation number rules, assign oxidation numbers to each element in the reactants and the products: - In \(\mathrm{H}_{2}(g)\), Hydrogen has an oxidation number of 0 since it's in its elemental form. - In \(\mathrm{F}_{2}(g)\), Fluorine has an oxidation number of 0 since it's in its elemental form. - In \(\mathrm{HF}(g)\), Hydrogen has an oxidation number of +1, and Fluorine has an oxidation number of -1.

Determine electrons transferred

To determine the number of electrons transferred in the reaction, find the change in oxidation numbers for the elements involved in the redox process: - Hydrogen: 0 (in \(\mathrm{H}_{2}\)) to +1 (in \(\mathrm{HF}\)), a change of +1 - Fluorine: 0 (in \(\mathrm{F}_{2}\)) to -1 (in \(\mathrm{HF}\)), a change of -1 Since there are 2 moles of \(\mathrm{HF}\) produced, two electrons are transferred in total. (b) $2 \mathrm{Fe}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$

Assign oxidation numbers

Assign oxidation numbers to each element in the reactants and the products: - In \(\mathrm{Fe}^{2+}\), Iron has an oxidation number of +2. - In \(\mathrm{H}_{2} \mathrm{O}_{2}\), Hydrogen has an oxidation number of +1, and Oxygen has an oxidation number of -1. - In \(\mathrm{H}^{+}\), Hydrogen has an oxidation number of +1. - In \(\mathrm{Fe}^{3+}\), Iron has an oxidation number of +3. - In \(\mathrm{H}_{2} \mathrm{O}\), Hydrogen has an oxidation number of +1, and Oxygen has an oxidation number of -2.

Determine electrons transferred

To determine the number of electrons transferred in the reaction, find the change in oxidation numbers for the elements involved in the redox process: - Iron: +2 (in \(\mathrm{Fe}^{2+}\)) to +3 (in \(\mathrm{Fe}^{3+}\)), a change of +1 - Oxygen: -1 (in \(\mathrm{H}_{2} \mathrm{O}_{2}\)) to -2 (in \(\mathrm{H}_{2} \mathrm{O}\)), a change of -1 Since there are 2 moles of \(\mathrm{Fe}^{2+}\) transformed to \(\mathrm{Fe}^{3+}\), and 1 mole of \(\mathrm{H}_{2} \mathrm{O}_{2}\) transformed to \(\mathrm{H}_{2} \mathrm{O}\), two electrons are transferred in total. (c) $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$

Assign oxidation numbers

Assign oxidation numbers to each element in the reactants and the products: - In \(\mathrm{CH}_{4}\), Carbon has an oxidation number of -4, and Hydrogen has an oxidation number of +1. - In \(\mathrm{O}_{2}(g)\), Oxygen has an oxidation number of 0 since it's in its elemental form. - In \(\mathrm{CO}_{2}\), Carbon has an oxidation number of +4, and Oxygen has an oxidation number of -2. - In \(\mathrm{H}_{2} \mathrm{O}\), Hydrogen has an oxidation number of +1, and Oxygen has an oxidation number of -2.

Determine electrons transferred

To determine the number of electrons transferred in the reaction, find the change in oxidation numbers for the elements involved in the redox process: - Carbon: -4 (in \(\mathrm{CH}_{4}\)) to +4 (in \(\mathrm{CO}_{2}\)), a change of +8 - Oxygen: 0 (in \(\mathrm{O}_{2}\)) to -2 (in \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\)), a change of -2 Since there are 2 moles of \(\mathrm{O}_{2}(g)\) transformed, oxygen transfers a total of 4 electrons. Hence, there is a transfer of 8 electrons in total in this reaction (4 from Carbon to Oxygen, and 4 from Hydrogen to Oxygen).

Key concepts

Oxidation Numbers

The concept of oxidation numbers helps us keep track of electron transfers during chemical reactions. They act as an indicator of the degree of oxidation (loss of electrons) or reduction (gain of electrons) for an element in a compound. To assign oxidation numbers, follow these rules:

• The oxidation number of an atom in its elemental form is always zero.
• For monoatomic ions, the oxidation number is equal to the charge of the ion.
• Hydrogen usually has an oxidation number of +1, while oxygen is typically -2.
• Total oxidation numbers in a neutral compound add up to zero, and for polyatomic ions, they equal the ion's charge.

By applying these rules, we can determine which elements undergo changes in oxidation state during a reaction, indicating an oxidation-reduction process has occurred.

Electron Transfer

Electron transfer is the fundamental mechanism in oxidation-reduction (redox) reactions. When one substance undergoes oxidation, it loses electrons, which are immediately gained by another substance undergoing reduction.

In a redox reaction, tracking the movement of electrons from one reactant to another provides insight into the changing oxidation states of the elements involved. Electron transfer is crucial, as it is the means by which energy is stored and released in many chemical processes, such as cellular respiration and photosynthesis.

In our reactions:

• For example, in reaction (a), electrons move from \(H_2 \) to \(F_2 \), resulting in the formation of \(HF \).
• Similarly, in reaction (b), electrons transfer from \(Fe^{2+} \) to \(H_2O_2 \).
• In reaction (c), the electrons from carbon in methane transfer to oxygen, forming carbon dioxide.

Understanding electron transfer helps us balance chemical equations and predict the products of a reaction.

Redox Reactions

Redox reactions are chemical reactions that involve the transfer of electrons between two species. These reactions are crucial as they drive numerous biological and industrial processes. Redox stands for reduction-oxidation.

• Reduction involves gaining electrons and reducing the oxidation state of a substance.
• Oxidation is the loss of electrons and an increase in the oxidation state.

In every redox process, there is always a substance that gets oxidized and another that gets reduced. Consider our exercises: - In reaction (a), hydrogen is oxidized by losing electrons to fluorine, which is reduced. - In reaction (b), iron ions are oxidized, while the peroxide is reduced. - Reaction (c) is more complex but involves carbon from methane being oxidized and oxygen molecules receiving electrons to form water and carbon dioxide. Understanding redox reactions gives valuable insight into the potential energy changes within a reaction.

Balanced Chemical Equations

Balanced chemical equations are essential for accurately representing a chemical reaction. They ensure that the same number of each type of atom exists on both sides of the equation. Balancing equations reflects the conservation of mass and the principle that matter cannot be created or destroyed.

In the context of redox reactions, balanced equations not only ensure that atoms are conserved but also that the charge is balanced. This becomes especially important when dealing with electron transfer.

In our examples:

• For \(H_2 + F_2 \rightarrow 2HF \), both atoms and charge need balance concurrently.
• In \(2Fe^{2+} + H_2O_2 + 2H^+ \rightarrow 2Fe^{3+} + H_2O \), differing oxidation numbers require attention to correctly reflect the electrons involved.
• The combustion of methane \(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \) must balance the carbon, hydrogen, and oxygen atoms on each side.

This comprehensive balance highlights the stoichiometry needed for accurate chemical analysis and ensures proper electron flow.