Question

Is the reaction $$ 2 \mathrm{~K}(\mathrm{~s})+\mathrm{Br}_{2}(\ell) \rightarrow 2 \mathrm{KBr}(\mathrm{s}) $$ an oxidation-reduction reaction? Explain your answer.

Short answer

Yes, it is an oxidation-reduction reaction because there is a transfer of electrons.

Step-by-step solution

Identify oxidation states of reactants

For potassium (K) in its solid state, the oxidation state is 0. For bromine (Br2) in its liquid state, the oxidation state is also 0 because it is in its elemental form.

Determine oxidation states of products

In the product 2 KBr, potassium (K) has an oxidation state of +1 and bromine (Br) has an oxidation state of -1 when they form the ionic compound KBr.

Analyze changes in oxidation states

Potassium changes from an oxidation state of 0 in K(s) to +1 in KBr, indicating it has lost electrons. Bromine changes from 0 in Br2(l) to -1 in KBr, indicating it has gained electrons.

Conclude type of reaction

Since there is a transfer of electrons with potassium losing electrons (oxidation) and bromine gaining electrons (reduction), this indicates an oxidation-reduction reaction.

Key concepts

Oxidation States

In chemistry, the oxidation state of an element describes its degree of oxidation, which is the loss of electrons. This concept helps us understand how electrons are distributed in a chemical reaction. Oxidation states are often represented by integers and can be positive, negative, or zero.

• If an element's oxidation state increases, it has lost electrons and is said to be oxidized.
• Conversely, if the oxidation state decreases, it has gained electrons and is considered reduced.
• When an element is in its pure or elemental form, its oxidation state is zero.

In the given reaction of potassium (\(K\) ) changing to \(KBr\) , potassium goes from an oxidation state of 0 to +1, indicating it loses an electron. For bromine, \(Br_2\) , its oxidation state also starts at 0, changing to -1 when forming \(KBr\) , showing a gain of electrons. These changes in oxidation states are key to identifying the electron transfer occurring in oxidation-reduction reactions.

Electron Transfer

Electron transfer is a fundamental concept in chemical reactions, especially in oxidation-reduction (redox) reactions. This process involves electrons moving from one atom or molecule to another. Such transfers result in changes in the oxidation states of the involved species.

• When an atom loses electrons, it undergoes oxidation.
• When an atom gains electrons, it undergoes reduction.

In the exercise, potassium atoms lose one electron each, moving from an oxidation state of 0 in \(K\) to +1 in \(KBr\) . This is oxidation. Bromine gains these electrons, moving from 0 in \(Br_2\) to -1 in \(KBr\) , and hence is reduced. The presence of both oxidation and reduction indicates a redox reaction, confirming the role of electron transfer.

Ionic Compounds

Ionic compounds are formed when atoms transfer electrons to achieve stable electron configurations. This results in atoms having fully occupied outer electron shells, akin to noble gas configuration.

• In ionic compounds, metals typically lose electrons to form positively charged ions, called cations.
• Non-metals gain these electrons, forming negatively charged ions, or anions.

In the given example, potassium donates electrons to bromine. Potassium becomes a \(K^+\) ion, while bromine forms two \(Br^-\) ions to produce \(KBr\) , an ionic compound. The electrostatic attraction between these oppositely charged ions forms the crystal lattice structure typical of solid ionic compounds. Understanding ionic compounds is crucial for visualizing how atoms interact, leading to the formation of compounds through electron transfer in redox reactions.