Question

For each of the following oxidation-reduction reactions, identify which element is being oxidized and which is being reduced. a. \(\mathrm{Cl}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{ICl}(g)\) b. \(\mathrm{Cl}_{2}(g)+2 \mathrm{Li}(s) \rightarrow 2 \mathrm{LiCl}(s)\) c. \(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\) d. \(\mathrm{Cl}_{2}(g)+2 \mathrm{NaBr}(a q) \rightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(l)\)

Short answer

In the given reactions: a. Cl is being reduced, and I is being oxidized. b. Cl is being reduced, and Li is being oxidized. c. H is being reduced, and Na is being oxidized. d. Cl is being reduced, and Br is being oxidized.

Step-by-step solution

Reaction a

For the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{ICl}(g)\), we assign oxidation numbers: Cl: 0 (in \(\mathrm{Cl}_{2}\)) -> -1 (in \(\mathrm{ICl}\)) I: 0 (in \(\mathrm{I}_{2}\)) -> +1 (in \(\mathrm{ICl}\)) Cl is being reduced and I is being oxidized.

Reaction b

For the reaction \(\mathrm{Cl}_{2}(g)+2 \mathrm{Li}(s) \rightarrow 2 \mathrm{LiCl}(s)\), we assign oxidation numbers: Cl: 0 (in \(\mathrm{Cl}_{2}\)) -> -1 (in \(\mathrm{LiCl}\)) Li: 0 (in elemental form) -> +1 (in \(\mathrm{LiCl}\)) Cl is being reduced and Li is being oxidized.

Reaction c

For the reaction \(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2\mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\), we assign oxidation numbers: Na: 0 (in elemental form) -> +1 (in \(\mathrm{NaOH}\)) O: -2 (in \(\mathrm{H}_{2}\mathrm{O}\)) -> -2 (in \(\mathrm{NaOH}\)) H: +1 (in \(\mathrm{H}_{2}\mathrm{O}\)) -> 0 (in \(\mathrm{H}_{2}\)) H is being reduced and Na is being oxidized.

Reaction d

For the reaction \(\mathrm{Cl}_{2}(g)+2 \mathrm{NaBr}(a q) \rightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(l)\), we assign oxidation numbers: Cl: 0 (in \(\mathrm{Cl}_{2}\)) -> -1 (in \(\mathrm{NaCl}\)) Na: +1 (in \(\mathrm{NaBr}\)) -> +1 (in \(\mathrm{NaCl}\)) Br: -1 (in \(\mathrm{NaBr}\)) -> 0 (in \(\mathrm{Br}_{2}\)) Cl is being reduced and Br is being oxidized.

Key concepts

Oxidation Numbers

Oxidation numbers are a handy tool in redox chemistry that help us understand how electrons are transferred between atoms in a chemical reaction. Think of the oxidation number as a bookkeeping method that gives us insight into how the electrons are distributed.
These numbers make it easier to see which atoms are gaining or losing electrons during a reaction.
To assign oxidation numbers:

• Pure elements in their standard states (e.g., \(\mathrm{O}_2\), \(\mathrm{N}_2\), or \(\mathrm{Cl}_2\)) have an oxidation number of 0.
• For ions and compounds, the sum of oxidation numbers is equal to the charge of the molecule or ion. For example, in \(\mathrm{NaCl}\), Na has an oxidation number of +1 and Cl has -1. Combined, they sum to 0.
• Oxygen usually has an oxidation number of -2, and hydrogen is generally +1 unless specified otherwise.

Once we have the oxidation numbers, we can identify which element is oxidized or reduced based on whether its oxidation number increases or decreases:

• Oxidation involves an increase in the oxidation number.
• Reduction means a decrease in the oxidation number.

This allows us to see the specific role each element plays in a redox reaction.

Redox Chemistry

Redox chemistry, or oxidation-reduction reactions, are fascinating because they show us how atoms share electrons. An easy way to remember what happens in a redox reaction is through the acronym OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).
In a redox reaction, two things happen side by side:

• One species (the reducing agent) gives away electrons, becoming oxidized in the process.
• Another species (the oxidizing agent) receives those electrons, becoming reduced.

Therefore, oxidation and reduction always occur together. The total number of electrons lost in the oxidation half must be the same number gained in the reduction half. This balance maintains charge and mass continuity in chemical reactions.
In our original examples:

• In reaction \(a\), iodine is oxidized from 0 to +1, and chlorine is reduced from 0 to -1.
• In reaction \(b\), lithium is oxidized from 0 to +1, while chlorine is reduced from 0 to -1.
• In reaction \(c\), sodium is oxidized from 0 to +1, and hydrogen is reduced from +1 in water to 0 in hydrogen gas.
• In reaction \(d\), bromine is oxidized from -1 to 0, and chlorine is reduced from 0 to -1.

Understanding these roles enhances our insight into the workings of chemical processes.

Chemical Reactions Analysis

Analyzing chemical reactions, particularly redox reactions, involves understanding the processes at a molecular level. By examining the changes in oxidation numbers, we can deduce the details of electron transfer and gain insights into the interaction between reactants.
Here's how to perform a structured analysis:

• First, write down all the reactants and products of the reaction.
• Use rules for assigning oxidation numbers (as discussed earlier) to all elements involved in the reaction.
• Identify the changes in oxidation numbers for each element.
• Determine which element is oxidized and which is reduced.
• Calculate the number of electrons transferred in the redox process.

This systematic approach allows chemists to understand not just which elements undergo oxidation or reduction, but also how the overall reaction pathway occurs.
Such comprehension is crucial for tasks ranging from balancing chemical equations to industrial chemical applications and environmental chemistry. It is this interplay of elements that power reactions and innovations in technology and nature.