Question

For each of the following oxidation-reduction reactions, identify which element is oxidized and which is reduced. a. \(6 \mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow 2 \mathrm{Na}_{3} \mathrm{~N}(s)\) b. \(\operatorname{Mg}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{MgCl}_{2}(s)\) c. \(2 \mathrm{Al}(s)+3 \mathrm{Br}_{2}(l) \rightarrow 2 \mathrm{AlBr}_{3}(s)\) d. \(\mathrm{CuSO}_{4}(a q)+\mathrm{Mg}(s) \rightarrow \mathrm{MgSO}_{4}(a q)+\mathrm{Cu}(s)\)

Short answer

In the given reactions: a. Na is oxidized and N is reduced. b. Mg is oxidized and Cl is reduced. c. Al is oxidized and Br is reduced. d. Cu is reduced and Mg is oxidized.

Step-by-step solution

Determine the Oxidation Numbers of Elements

Start by determining the oxidation numbers of each element in the given reactions, both on the reactants' side and the products' side. Remember that for an element in its elemental state, the oxidation number is always 0. a. \(6 \mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow 2 \mathrm{Na}_{3}\mathrm{~N}(s) \) Before: Na: 0, N: 0 After: Na: +1, N: -3 b. \(\operatorname{Mg}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{MgCl}_{2}(s) \) Before: Mg: 0, Cl: 0 After: Mg: +2, Cl: -1 c. \(2 \mathrm{Al}(s)+3 \mathrm{Br}_{2}(l) \rightarrow 2 \mathrm{AlBr}_{3}(s) \) Before: Al: 0, Br: 0 After: Al: +3, Br: -1 d. \(\mathrm{CuSO}_{4}(a q)+\mathrm{Mg}(s) \rightarrow \mathrm{MgSO}_{4}(a q)+\mathrm{Cu}(s) \) Before: Cu: +2, S: +6, O: -2, Mg: 0 After: Cu: 0, S: +6, O: -2, Mg: +2

Identify the Oxidized and Reduced Elements

Now that the oxidation numbers of each element are determined, identify the changes in oxidation numbers for each reaction and determine which element is oxidized (increase in oxidation number) and which is reduced (decrease in oxidation number). a. Na: 0 -> +1 (oxidized), N: 0 -> -3 (reduced) b. Mg: 0 -> +2 (oxidized), Cl: 0 -> -1 (reduced) c. Al: 0 -> +3 (oxidized), Br: 0 -> -1 (reduced) d. Cu: +2 -> 0 (reduced), Mg: 0 -> +2 (oxidized)

Key concepts

Oxidation Numbers

Oxidation numbers are a fundamental concept in understanding redox (oxidation-reduction) reactions. These numbers are assigned to atoms to indicate their degree of oxidation or reduction. Essentially, an oxidation number is the charge that an atom would have if the compound was composed of ions.

An oxidation number can be a positive, negative, or zero value. In an elemental form, where it is not part of a compound or molecule, an atom's oxidation number is always zero. Some general rules that can be used to determine oxidation numbers include:

• For single elements, the oxidation number is zero (e.g., Na, Cl₂).
• The sum of oxidation numbers for all atoms in a neutral compound must be zero.
• Oxygen is usually assigned an oxidation number of -2, except in peroxides or when bonded to fluorine.
• Hydrogen is usually +1 unless it is bonded to metals where it can be -1.

Understanding and assigning correct oxidation numbers is the first step in identifying which element is oxidized and which is reduced in a redox reaction.

Identifying Oxidized Elements

In an oxidation-reduction reaction, an element that loses electrons is said to be oxidized. This process results in an increase in the oxidation number of that element. Visualize this as the element becoming more positive because it is losing negatively charged electrons.

To identify which element is oxidized in a reaction, look for an increase in the oxidation number from the reactants to the products. For example, in the reaction between magnesium and chlorine gas (Mg + Cl₂ → MgCl₂), magnesium has an oxidation number of 0 in the elemental state and changes to +2 in the compound MgCl₂. Therefore, magnesium is oxidized. It's like each magnesium atom is giving away its electrons to become more positively charged.

Identifying Reduced Elements

Conversely, an element that gains electrons during an oxidation-reduction reaction is said to be reduced. The element's oxidation number decreases, as it's effectively becoming more negative with the addition of electrons.

For identification, compare the oxidation numbers before and after the reaction. A reduced element will have a lower oxidation number in the product than in the reactant. Consider copper sulfate reacting with magnesium (CuSO₄ + Mg → MgSO₄ + Cu). Copper starts with an oxidation number of +2 in CuSO₄ but ends up in its elemental form with an oxidation number of 0 in Cu. Therefore, the copper is reduced. It's acquiring electrons and shedding its positive charge as it transitions to the elemental state.