Question

Balance each of the following equations that describe oxidation-reduction reactions. a. \(\mathrm{Al}(s)+\mathrm{Br}_{2}(l) \rightarrow \operatorname{AlBr}_{3}(s)\) b. \(\operatorname{Zn}(s)+\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{Zn}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2}(g)\) c. \(\mathrm{Na}(s)+\mathrm{P}(s) \rightarrow \mathrm{Na}_{3} \mathrm{P}(s)\) d. \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)\) e. \(\mathrm{Cu}(s)+\mathrm{AgNO}_{3}(a q) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Ag}(s)\)

Short answer

a. \(2 \mathrm{Al}(s) + 3 \mathrm{Br}_{2}(l) \rightarrow 2 \operatorname{AlBr}_{3}(s)\) b. \( \mathrm{Zn}(s) + 2 \mathrm{HClO}_{4}(a q) \rightarrow \mathrm{Zn}\left(\mathrm{ClO}_{4}\right)_{2}(a q) + \mathrm{H}_{2}(g)\) c. \(3\mathrm{Na}(s) + \mathrm{P}(s) \rightarrow \mathrm{Na}_{3} \mathrm{P}(s)\) d. \(\mathrm{CH}_{4}(g) + 4\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(l) + 4\mathrm{HCl}(g)\) e. \( \mathrm{Cu}(s) + 2\mathrm{AgNO}_{3}(a q) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) + 2\mathrm{Ag}(s)\)

Step-by-step solution

Balancing Equation a

To balance the equation, we compare the number of atoms on both sides: Al(s) + Br2(l) → AlBr3(s) First, we balance the Al atoms: 1 Al (on reactants side) → 1 Al (on product side) (Already balanced) Then, we balance the Br atoms: 2 Br (on reactants side) → 3 Br (on product side) To balance the Br atoms, we can multiply 2 (from Br2) by 3, and then divide the product by the number of Br atoms on the product side (3). Therefore, 6 / 3 = 2, we need to multiply AlBr₃ by 2 to get 6 Br atoms: Balanced Equation: \(2 \mathrm{Al}(s) + 3 \mathrm{Br}_{2}(l) \rightarrow 2 \operatorname{AlBr}_{3}(s)\)

Balancing Equation b

To balance the equation, we compare the number of atoms on both sides: Zn(s) + HClO₄(aq) → Zn(ClO₄)₂(aq) + H₂(g) First, we balance the Zn atoms: 1 Zn (on reactants side) → 1 Zn (on product side) (Already balanced) Next, we balance the H atoms: 1 H (on reactants side) → 2 H (on product side) To balance the H atoms, we multiply HClO₄ by 2: Zn(s) + 2 HClO₄(aq) → Zn(ClO₄)₂(aq) + H₂(g) Now we balance the Cl atoms: 2 Cl (from 2 HClO₄) → 2 Cl (from Zn(ClO₄)₂) (Already balanced) The final balanced equation is: \( \mathrm{Zn}(s) + 2 \mathrm{HClO}_{4}(a q) \rightarrow \mathrm{Zn}\left(\mathrm{ClO}_{4}\right)_{2}(a q) + \mathrm{H}_{2}(g)\)

Balancing Equation c

To balance the equation, we compare the number of atoms on both sides: Na(s) + P(s) → Na₃P(s) First, we balance the Na atoms: 1 Na (on reactants side) → 3 Na (on product side) To balance the Na atoms, we multiply Na(s) by 3: 3 Na(s) + P(s) → Na₃P(s) Now, the equation is balanced. Balanced Equation: \(3\mathrm{Na}(s) + \mathrm{P}(s) \rightarrow \mathrm{Na}_{3} \mathrm{P}(s)\)

Balancing Equation d

To balance the equation, we compare the number of atoms on both sides: CH₄(g) + Cl₂(g) → CCl₄(l) + HCl(g) First, we balance the C atoms: 1 C (on reactants side) → 1 C (on product side) (Already balanced) Next, we balance the H atoms: 4 H (on reactants side) → 1 H (on product side) To balance the H atoms, we multiply HCl by 4: CH₄(g) + Cl₂(g) → CCl₄(l) + 4 HCl(g) Now we balance the Cl atoms: 2 Cl (from Cl₂) → 4 Cl (from CCl₄) + 4 Cl (from 4 HCl) To balance the Cl atoms, we multiply Cl₂ by 4: Balanced Equation: \(\mathrm{CH}_{4}(g) + 4\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(l) + 4\mathrm{HCl}(g)\)

Balancing Equation e

To balance the equation, we compare the number of atoms on both sides: Cu(s) + AgNO₃(aq) → Cu(NO₃)₂(aq) + Ag(s) First, we balance the Cu atoms: 1 Cu (on reactants side) → 1 Cu (on product side) (Already balanced) Next, we balance the Ag atoms: 1 Ag (from AgNO₃) → 1 Ag (on product side) (Already balanced) Now we balance the N atoms: 1 N (from AgNO₃) → 2 N (from Cu(NO₃)₂) To balance the N atoms, multiply AgNO₃ by 2: Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2 Ag(s) Now the equation is balanced. Balanced Equation: \( \mathrm{Cu}(s) + 2\mathrm{AgNO}_{3}(a q) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) + 2\mathrm{Ag}(s)\)

Key concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, commonly referred to as redox reactions, involve the transfer of electrons between two substances. In these processes, one element gains electrons (reduction), while another loses electrons (oxidation). These reactions are fundamental because they underpin many physical and biological processes. For instance, when aluminum (\(\mathrm{Al}(s)\)) reacts with bromine (\(\mathrm{Br}_{2}(l)\)) to form aluminum bromide (\(\operatorname{AlBr}_{3}(s)\)), the aluminum atoms undergo oxidation by losing electrons, while the bromine molecules are reduced as they gain electrons.

• Oxidation: Loss of electrons
• Reduction: Gain of electrons

Each redox reaction component must balance in terms of atoms and charge, leading us to the balancing procedure where stoichiometry plays a critical role.

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. It enables us to understand the proportions in which chemicals react and are produced. When balancing chemical equations, stoichiometry ensures that the number of atoms of each element is the same on both sides of the equation.
For example, when balancing the reaction between zinc (\(\mathrm{Zn}(s)\)) and perchloric acid (\(\mathrm{HClO}_{4}(a q)\)), stoichiometry helps determine how many molecules react to produce the intended products. This means if you need two molecules of perchloric acid to completely react with one atom of zinc and produce hydrogen gas (\(\mathrm{H}_{2}(g)\)), stoichiometry guides the proportions:

• Zinc contributes to the product side by maintaining the same number of atoms.
• Perchloric acid must be doubled to produce the proper number of hydrogen molecules.

Knowing these calculations allows chemists to predict reaction behavior and product amounts precisely.

Chemical Reactions

Chemical reactions represent the change of substances into new ones, involving rearrangement of atoms. In a balanced chemical reaction, like the combination of sodium (\(\mathrm{Na}(s)\)) and phosphorus (\(\mathrm{P}(s)\)), products must have atoms exactly equal to those of reactants.
Balancing is crucial as it respects the law of conservation of mass. Every atomic inventory must be tallied up correctly, so no atoms are lost or gained in reactions. For example, when 3 sodium atoms react with a phosphorus atom forming sodium phosphide (\(\mathrm{Na}_{3} \mathrm{P}(s)\)), stoichiometry ensures each and every atom is properly accounted for at the start and end of the reaction.
Understanding this concept helps grasp more complex interactions, such as predicting whether reactions will continue or what catalysts are needed. It forms the backbone of chemical computations and real-world chemical applications.