Question

For each of the following oxidation-reduction reactions of metals with nonmetals, identify which element is oxidized and which is reduced. a. \(4 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Na}_{2} \mathrm{O}(s)\) b. \(\operatorname{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{FeSO}_{4}(a q)+\mathrm{H}_{2}(g)\) c. \(2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \rightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g)\) d. \(3 \mathrm{Mg}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Mg}_{3} \mathrm{N}_{2}(s)\)

Short answer

In the given reactions: a. Na is oxidized and O is reduced. b. Fe is oxidized and H is reduced. c. Al is reduced and O is oxidized. d. Mg is oxidized and N is reduced.

Step-by-step solution

Determine oxidation numbers

To identify which element is oxidized and which is reduced, first determine the oxidation numbers of all atoms in the reactants and products. In this case, the oxidation number of Na in both Na(s) and Na2O(s) is +1, while the oxidation number of O in O2(g) is 0 and in Na2O(s) is -2.

Identify elements undergoing oxidation or reduction

Compare the oxidation numbers of each element in the reactants and products to find changes. As oxidation number of Na increases from 0 in Na(s) to +1 in Na2O(s), it is losing an electron. Hence, Na is oxidized. Oxidation number of O decreases from 0 in O2(g) to -2 in Na2O(s), meaning it gains electrons. Therefore, O is reduced. #b. Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)#

Determine oxidation numbers

Determine the oxidation numbers of all the atoms in the reactants and products. Oxidation numbers are: Fe: 0 (Fe(s)) and +2 (FeSO4(aq)); H: +1 (H2SO4(aq)) and 0 (H2(g)); S: +6 (H2SO4(aq)) and +6 (FeSO4(aq)); O: -2 (H2SO4(aq) and FeSO4(aq).

Identify elements undergoing oxidation or reduction

Fe's oxidation number increases from 0 in Fe(s) to +2 in FeSO4(aq), meaning it loses electrons and is oxidized. The oxidation number of H decreases from +1 in H2SO4(aq) to 0 in H2(g), so H is reduced. #c. 2Al2O3(s) → 4Al(s) + 3O2(g)#

Determine oxidation numbers

Determine the oxidation numbers of each species in reactants and products. Oxidation numbers are: Al: +3 (Al2O3(s)) and 0 (Al(s)); O: -2 (Al2O3(s)) and 0 (O2(g)).

Identify elements undergoing oxidation or reduction

Oxidation number of Al decreases from +3 in Al2O3(s) to 0 in Al(s), which means Al gains electrons and is reduced. Oxidation number of O increases from -2 in Al2O3(s) to 0 in O2(g), so O is oxidized. #d. 3Mg(s) + N2(g) → Mg3N2(s)#

Determine oxidation numbers

Determine the oxidation numbers of each species in reactants and products. Oxidation numbers are: Mg: 0 (Mg(s)) and +2 (Mg3N2(s)); N: 0 (N2(g)) and -3 (Mg3N2(s)).

Identify elements undergoing oxidation or reduction

Oxidation number of Mg increases from 0 in Mg(s) to +2 in Mg3N2(s), which means Mg loses electrons and is oxidized. Oxidation number of N decreases from 0 in N2(g) to -3 in Mg3N2(s), so N gains electrons and is reduced.

Key concepts

Oxidation Numbers

Oxidation numbers, also known as oxidation states, are vital for understanding redox reactions. An oxidation number represents the degree of oxidation of an atom in a chemical compound. It indicates how many electrons an atom has gained, lost, or shared when forming chemical bonds.
To assign oxidation numbers, you can follow these guidelines:

• The oxidation number of a free element (not bonded to anything) is zero, e.g., Na(s), O extsubscript{2}(g).
• For simple ions, the oxidation number is equal to the charge of the ion, e.g., Na extsuperscript{+} has an oxidation number of +1.
• In compounds, hydrogen generally has an oxidation number of +1, while oxygen usually has -2.
• The sum of oxidation numbers in a neutral compound must equal zero, and in polyatomic ions, it should equal the ion's charge.

By using these rules, we can determine the oxidation state of each atom in the compounds involved, which is essential for identifying the oxidation and reduction processes in reactions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are processes where one substance is oxidized (loses electrons) and another is reduced (gains electrons). Understanding which atoms are oxidized and which are reduced is crucial in redox reactions.
Here's how you can determine redox processes:

• Identify each element's oxidation number before and after the reaction.
• Observe the change: if an element's oxidation number increases, it loses electrons and is oxidized.
• If an element's oxidation number decreases, it gains electrons and is reduced.

For instance, in the equation involving sodium and oxygen: - Sodium ( ext{Na}) changes from an oxidation number of 0 in ext{Na}(s) to +1 in ext{Na}_2 ext{O}(s), indicating it is oxidized. - Oxygen ( ext{O}) shifts from an oxidation number of 0 in ext{O}_2(g) to -2 in ext{Na}_2 ext{O}(s), showing it is reduced. Understanding these changes helps in predicting how substances will interact during chemical reactions.

Oxidation States

The concept of oxidation states is fundamental to identifying what happens in redox reactions. An oxidation state, or oxidation number, represents how many electrons an atom effectively controls in a chemical substance. It is a hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic.
Oxidation states can:

• Help track electron flow in chemical reactions.
• Enable chemists to balance chemical equations easily.
• Provide insights into the reactivity of different materials.

When analyzing each reaction, such as in the case of magnesium and nitrogen: - Magnesium ( ext{Mg}) in ext{Mg}(s) has an oxidation state of 0 and transitions to +2 in ext{Mg}_3 ext{N}_2(s), showing oxidation. - Nitrogen ( ext{N}) starts with an oxidation state of 0 in ext{N}_2(g) and moves to -3 in ext{Mg}_3 ext{N}_2(s), indicating reduction. These shifts in oxidation states highlight the electron transfers, helping us recognize which elements are involved in oxidation or reduction.