Question

Use the oxidation-number method to balance these redox equations. a. \(\mathrm{PbS}+\mathrm{O}_{2} \rightarrow \mathrm{PbO}+\mathrm{SO}_{2}\) b. NaWO \(_{3}+\mathrm{NaOH}+\mathrm{O}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{WO}_{4}+\mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{NH}_{3}+\mathrm{CuO} \rightarrow \mathrm{Cu}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) d. \(\mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{C}+\mathrm{Cl}_{2} \rightarrow \mathrm{AlCl}_{3}+\mathrm{CO}\)

Short answer

The short version of the balanced redox equations for the given reactions are: a. 2PbS + 3O₂ → 2PbO + 2SO₂ b. 2NaWO₃ + 6NaOH + O₂ → 2Na₂WO₄ + 3H₂O c. 6NH₃ + 3CuO → 3Cu + N₂ + 6H₂O d. Al₂O₃ + 3C + 6Cl₂ → 2AlCl₃ + 3CO

Step-by-step solution

- Assign Oxidation Numbers:

Assign oxidation numbers to the atoms in the reaction: Pb: +2, S: -2, O: 0 in \(\mathrm{O}_{2}\), +2 in \(\mathrm{PbO}\) and \(\mathrm{SO}_{2}\).

- Identify Atoms that Change Oxidation Number:

S is oxidized from -2 to +4 and O is reduced from 0 to -2.

- Write the Half-Reactions:

Oxidation half-reaction: S(-2) → S(+4) Reduction half-reaction: O(0) → O(-2)

- Balance the Half-Reactions:

Balance the atoms and charges by adding the appropriate coefficients and electrons: Oxidation: S(-2) → S(+4) + 6e⁻ Reduction: O(0) + 2e⁻ → O(-2)

- Make the Number of Electrons Lost Equal to the Number of Electrons Gained:

Multiply the reduction half-reaction by 3: Oxidation: S(-2) → S(+4) + 6e⁻ Reduction: 3[O(0) + 2e⁻ → O(-2)]

- Add the Half-Reactions to Obtain the Balanced Redox Equation:

Add the two half-reactions: 2PbS + 3O₂ → 2PbO + 2SO₂. We now have the balanced redox equation for part a. b (until d): Repeat the same steps as in part a, assigning oxidation numbers, identifying which atoms change oxidation number, writing the half-reactions, balancing the half-reactions separately, then making the number of electrons lost equal to the number of electrons gained, and finally adding half-reactions to get the balanced redox equations. The balanced redox equations for parts b, c and d are provided below: b. 2NaWO₃ + 6NaOH + O₂ → 2Na₂WO₄ + 3H₂O c. 6NH₃ + 3CuO → 3Cu + N₂ + 6H₂O d. Al₂O₃ + 3C + 6Cl₂ → 2AlCl₃ + 3CO

Key concepts

Redox Reactions

Redox reactions, also known as oxidation-reduction reactions, are fundamental to understanding chemistry. They involve the transfer of electrons between substances. When a substance loses electrons, it undergoes oxidation; conversely, when a substance gains electrons, it is reduced. These reactions are crucial in both natural processes and industrial applications, from biological respiration to the batteries that power our devices.

Oxidation and reduction always occur simultaneously in redox reactions. The substance that gives up electrons (the reducing agent) causes another substance to be reduced (the oxidized agent), and vice versa. To identify the changes in oxidation state, one must carefully assign and track the oxidation numbers of the different atoms involved.

Assigning Oxidation Numbers

Assigning oxidation numbers is essential to solving redox equations. Oxidation number or state represents the hypothetical charge that an atom would have if all its bonds were ionic. It provides a way to keep track of electrons during a reaction. To assign oxidation numbers, we follow specific rules: atoms in their elemental state have an oxidation number of 0; for monoatomic ions, the oxidation number is equal to the charge of the ion; oxygen generally has an oxidation number of -2, except in peroxides; hydrogen is usually +1, except when bonded to metals in hydrides; and the sum of oxidation numbers in a neutral compound is 0.

In the given examples, oxidation numbers are determined before identifying the atoms that experience a change in oxidation state. This allows the direction of electron transfer to be understood, establishing which atoms are oxidized and which are reduced.

Balancing Chemical Equations

Balancing chemical equations is the process of ensuring the conservation of mass. The number of atoms for each element must be the same on both the reactant and product sides of the equation. For redox reactions, balancing involves not only the atoms but also the charges, ensuring the total charge is conserved. This is particularly important because redox reactions highlight the movement of electrons, which carry charge.

The oxidation-number method is a systematic way of balancing redox equations. It involves separating the reaction into two half-reactions, one for each redox pair involved. Each half-reaction is balanced for mass and charge by adding coefficients and, if necessary, water, hydrogen ions, hydroxide ions, or electrons. The electrons lost in the oxidation half-reaction must equal the electrons gained in the reduction half-reaction, which might involve multiplying the half-reactions by appropriate integers. Finally, the balanced half-reactions are combined to give the balanced redox equation.