Question

Which element is oxidized and which is reduced in the following reactions? (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\) (c) \(\mathrm{CuCl}_{2}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{Cu}(s)\) (d) \(2 \mathrm{NaCl}(a q)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NaF}(a q)+\mathrm{Cl}_{2}(g)\)

Short answer

In (a), S is oxidized and O is reduced; in (b), Na is oxidized and Cl is reduced; in (c), Zn is oxidized and Cu is reduced; in (d), Cl is oxidized and F is reduced.

Step-by-step solution

Identify Changes in Oxidation Numbers for Reaction (a)

For the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\), determine the oxidation numbers of sulfur in \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\):- Sulfur in \(\mathrm{SO}_{2}\) is +4.- Sulfur in \(\mathrm{SO}_{3}\) is +6.Oxygen is an element in both \(\mathrm{O}_{2}\) and is -2 in \(\mathrm{SO}_{3}\). **Conclusion:** Sulfur is oxidized from +4 to +6; oxygen is reduced as it goes from 0 to -2 in forming \(\mathrm{SO}_{3}\).

Identify Changes in Oxidation Numbers for Reaction (b)

For the reaction \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\), determine the changes in oxidation numbers:- \(\mathrm{Na}\) goes from 0 in \(\mathrm{Na}(s)\) to +1 in \(\mathrm{NaCl}\).- \(\mathrm{Cl}_{2}\) goes from 0 to -1 in \(\mathrm{NaCl}\).**Conclusion:** \(\mathrm{Na}\) is oxidized from 0 to +1; \(\mathrm{Cl}_{2}\) is reduced from 0 to -1.

Identify Changes in Oxidation Numbers for Reaction (c)

For the reaction \(\mathrm{CuCl}_{2}(aq)+\mathrm{Zn}(s) \longrightarrow \mathrm{ZnCl}_{2}(aq)+\mathrm{Cu}(s)\), check the oxidation states:- \(\mathrm{Cu}\) in \(\mathrm{CuCl}_{2}\) is +2 and becomes 0 in \(\mathrm{Cu}(s)\).- \(\mathrm{Zn}\) is 0 in \(\mathrm{Zn}(s)\) and becomes +2 in \(\mathrm{ZnCl}_{2}\).**Conclusion:** \(\mathrm{Cu}\) is reduced from +2 to 0; \(\mathrm{Zn}\) is oxidized from 0 to +2.

Identify Changes in Oxidation Numbers for Reaction (d)

For the reaction \(2 \mathrm{NaCl}(aq)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NaF}(aq)+\mathrm{Cl}_{2}(g)\):- Chlorine in \(\mathrm{NaCl}\) is -1 and becomes 0 in \(\mathrm{Cl}_{2}\).- Fluorine in \(\mathrm{F}_{2}\) is 0 and becomes -1 in \(\mathrm{NaF}\).**Conclusion:** \(\mathrm{Cl}\) in \(\mathrm{NaCl}\) is oxidized from -1 to 0; \(\mathrm{F}\) in \(\mathrm{F}_{2}\) is reduced from 0 to -1.

Key concepts

Understanding Oxidation States

Oxidation states, also known as oxidation numbers, play a crucial role in chemical reactions, particularly in redox reactions. They are used to keep track of the electrons in an atom that occur during these reactions. The concept of oxidation states helps us determine whether a chemical species has been oxidized or reduced.

Here are some basic rules for determining oxidation states:

• The oxidation state of an element in its standard state (like O₂ or Na) is always 0.
• For monoatomic ions, the oxidation state is equal to the ion charge.
• Oxygen almost always has an oxidation state of -2 in compounds (except in peroxides).
• Hydrogen is typically +1 when bonded with non-metals and -1 with metals.

By assigning oxidation states to atoms before and after the reaction, we can determine how they change. An increase in oxidation state indicates oxidation (loss of electrons), while a decrease indicates reduction (gain of electrons). Understanding these shifts is key for analyzing oxidation-reduction reactions.

Exploring Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are a type of chemical reaction where the oxidation states of molecules are changed. These reactions involve the transfer of electrons between two species. The species that loses electrons is said to be oxidized, while the species that gains electrons is reduced.

To remember these processes, you can use the mnemonic "OIL RIG" which stands for "Oxidation Is Loss, Reduction Is Gain." In this context, it refers to the electrons.

In the reactions we analyzed:

• Sulfur in reaction (a) is oxidized as it goes from an oxidation state of +4 to +6.
• Oxygen in reaction (a) is reduced as it reduces its state from 0 to -2.
• Sodium in reaction (b) loses electrons, increasing its oxidation state from 0 to +1, and is oxidized.
• Chlorine in reaction (b) gains electrons by changing from 0 to -1, and is reduced.

Understanding the particular behaviors of oxidation and reduction in redox reactions is fundamental for deciphering chemical processes in broader chemical equations.

The Role of Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They depict the reactants and products, their physical states (solid, liquid, gas, aqueous), and the stoichiometric coefficients that indicate the ratio in which they participate in a reaction. In oxidation-reduction reactions, chemical equations are essential for illustrating the electron transfers taking place.

In a typical redox reaction, the balanced chemical equation ensures that the number of electrons lost in oxidation matches the number of electrons gained in reduction. For example, in reaction (c):

• Zinc goes from an oxidation state of 0 to +2 as it forms ZnCl₂. This change reflects the oxidation of zinc.
• Copper changes from +2 in CuCl₂ to 0 in Cu, highlighting its reduction.

Balancing chemical equations is vital not only for determining the conservation of mass but also for ensuring the proper accounting of electrons in redox reactions. These balanced equations provide insights into how atoms rearrange and how electrons are transferred, enhancing our understanding of the underlying chemical processes.