Question

Each of the following atom/ion pairs undergoes the oxidation number change indicated below. For each pair, determine whether oxidation or reduction has occurred, and then write the electronic equation indicating the corresponding number of electrons lost or gained. a. \({K} \longrightarrow {K}^{+}\) e. \({H}_{2} \longrightarrow {H}^{+}\) b. \({S} \longrightarrow {S}^{2-}\) f. \({O}_{2} \longrightarrow {O}^{2-}\) c. \({Mg} \longrightarrow {Mg}^{2+}\) g. \({Fe}^{3+} \longrightarrow {Fe}^{2+}\) d. \({F}^{-} \longrightarrow {F}_{2}\) h. \({Mn}^{2+} \longrightarrow {MnO}_{4}^{-}\)

Short answer

(a): oxidation, \[ K \rightarrow K^+ + e^- \]. (b): reduction, \[ S + 2e^- \rightarrow S^{2-} \]. (c): oxidation, \[ Mg \rightarrow Mg^{2+} + 2e^- \]. (d): oxidation, \[ 2F^- \rightarrow F_2 + 2e^- \]. (e): oxidation, \[ H_2 \rightarrow 2H^+ + 2e^- \]. (f): reduction, \[ O_2 + 4e^- \rightarrow 2O^{2-} \]. (g): reduction, \[ Fe^{3+} + e^- \rightarrow Fe^{2+} \]. (h): oxidation, \[ Mn^{2+} \rightarrow MnO_4^- + 5e^- \].

Step-by-step solution

Identify the oxidation number change

For each pair, determine the change in oxidation state to understand if oxidation or reduction occurs.

Determine oxidation or reduction

Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons.

Write the electronic equation

For each pair, write the equation indicating the number of electrons lost or gained.

Analyze pair (a) \(K \rightarrow K^{+}\)

Potassium (K) has an initial oxidation number of 0 and changes to +1. It loses one electron (oxidation). \[ K \rightarrow K^+ + e^- \]

Analyze pair (b) \(S \rightarrow S^{2-}\)

Sulfur (S) has an initial oxidation number of 0 and changes to -2. It gains two electrons (reduction). \[ S + 2e^- \rightarrow S^{2-} \]

Analyze pair (c) \(Mg \rightarrow Mg^{2+}\)

Magnesium (Mg) has an initial oxidation number of 0 and changes to +2. It loses two electrons (oxidation). \[ Mg \rightarrow Mg^{2+} + 2e^- \]

Analyze pair (d) \(F^- \rightarrow F_2\)

Fluoride ion (F^-) has an initial oxidation number of -1 and changes to 0. Two fluoride ions lose two electrons to form fluorine gas (oxidation). \[ 2F^- \rightarrow F_2 + 2e^- \]

Analyze pair (e) \(H_2 \rightarrow H^+\)

Hydrogen gas (H_2) has an initial oxidation number of 0 and changes to +1. Each H atom loses one electron, so in total 2 electrons are lost (oxidation). \[ H_2 \rightarrow 2H^+ + 2e^- \]

Analyze pair (f) \(O_2 \rightarrow O^{2-}\)

Oxygen gas (O_2) has an initial oxidation number of 0 and changes to -2. Each O atom gains two electrons, so in total 4 electrons are gained (reduction). \[ O_2 + 4e^- \rightarrow 2O^{2-} \]

Analyze pair (g) \(Fe^{3+} \rightarrow Fe^{2+} \)

Ferric ion (Fe^{3+}) has an initial oxidation number of +3 and changes to +2. It gains one electron (reduction). \[ Fe^{3+} + e^- \rightarrow Fe^{2+} \]

Analyze pair (h) \(Mn^{2+} \rightarrow MnO_4^- \)

The manganese ion (Mn^{2+}) has an initial oxidation number of +2 and changes to +7 in MnO_4^-. It loses 5 electrons (oxidation). \[ Mn^{2+} \rightarrow MnO_4^- + 5e^- \]

Key concepts

Oxidation Number

Understanding the oxidation number is crucial in identifying oxidation and reduction reactions. The oxidation number, often referred to as the oxidation state, indicates the degree of oxidation of an atom in a chemical compound. It's a hypothetical charge that an atom would have if all its bonds were ionic.

• For example, in the problem above, we see potassium (K) changing from an oxidation state of 0 to +1.
• This change is identified as oxidation because potassium loses an electron.

A negative change would indicate a reduction, where electrons are gained. For sulfur (S), its oxidation state changes from 0 to -2, showing a gain of electrons and indicating a reduction.
Grasping these shifts in oxidation states helps us determine the number of electrons transferred during the reaction.

Electron Transfer

Electron transfer is at the heart of redox reactions. It involves the movement of electrons from one atom or molecule to another.

• In the example of magnesium (Mg), it changes to Mg²⁺, showing it has lost two electrons.
• This loss of electrons signifies oxidation. We write this as Mg → Mg²⁺ + 2e⁻.

For fluorine (F⁻) changing to fluorine gas (F₂), we notice two fluoride ions each losing one electron, resulting in F₂ + 2e⁻.
Conversely, gaining electrons, such as when O₂ becomes O²⁻, shows reduction. Here, each oxygen atom in O₂ gains two electrons, totaling four electrons for the diatomic molecule, ending with O₂ + 4e⁻ → 2O²⁻.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between two species. These are central to many chemical processes, including metabolism, combustion, and corrosion.

• Oxidation is the loss of electrons. For instance, H₂ transforming to 2H⁺ illustrates each hydrogen atom losing an electron (oxidation).
• Reduction is the gain of electrons, demonstrated by Fe³⁺ being reduced to Fe²⁺.

Both processes occur simultaneously in a redox reaction. In the equation Mn²⁺ converting to MnO₄⁻, manganese undergoes oxidation by losing five electrons, while another species in the reaction must gain these electrons (reduction).
Understanding these pairings ensures a balanced perspective on electron movement.

Chemical Equations

Writing chemical equations for redox reactions helps us visualize and balance these processes. The equations highlight the transfer of electrons, showing which species are oxidized and which are reduced.

• For instance, with Fe³⁺ converting to Fe²⁺, we write the reduction half-equation as Fe³⁺ + e⁻ → Fe²⁺.
• Similarly, for oxidation, like K to K⁺, the equation is K → K⁺ + e⁻.

These half-equations separately illustrate oxidation and reduction. One equation shows the species losing electrons, while the other shows the species gaining electrons. Balancing these charges and matter is essential for accuracy in chemical equations.
Utilizing these balanced reactions assists in forming a comprehensive view of the overall redox process.