Question

Balance the following equations for reactions occurring in an acidic solution. (a) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{OCl}^{-} \longrightarrow \mathrm{Cl}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (b) \(\mathrm{NO}_{3}^{-}+\mathrm{Cu} \longrightarrow \mathrm{NO}_{2}+\mathrm{Cu}^{2+}\) (c) \(\mathrm{IO}_{3}^{-}+\mathrm{H}_{3} \mathrm{AsO}_{3} \longrightarrow \mathrm{I}^{-}+\mathrm{H}_{3} \mathrm{AsO}_{4}\) (d) \(\mathrm{SO}_{4}^{2-}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{SO}_{2}\)

Short answer

(a) \(2 {S_2O_3}^{2-} + 8 {H^+} + {OCl^-} \rightarrow 2 {Cl^-} + {S_4O_6}^{2-} + 4 {H_2O} + 2e^-\), (b) \(3 {H^+} + {NO_3^-} + {Cu} \rightarrow {NO_2} + {Cu^{2+}} + 2e^-\), (c) \(6 {H^+} + {IO_3^-} + {H_3AsO_3} \rightarrow {I^-} + {H_3AsO_4} + 3 {H_2O} + 2e^-\), (d) \(8 {H^+} + {SO4}^{2-} + {Zn} \rightarrow {Zn}^{2+} + {SO_2} + 4 {H_2O} + 8e^-\).

Step-by-step solution

Balance atoms other than O and H

For each reaction, begin by balancing elements other than oxygen and hydrogen. For (a), balance sulfur by noticing that there are 2 sulfurs on the left and 4 on the right, so we need to double the thiosulfate ion. For (b), balance copper directly since it's already balanced. For (c), balance arsenic and iodine by making sure there is one of each on both sides, which they are initially. For (d), sulfur is already balanced.

Balance oxygen atoms

Balance oxygen by adding water molecules to the side that has fewer oxygen atoms. For reaction (a), we have 3 oxygen atoms on the left and 6 on the right, so we need to add 3 water molecules to the left. For (b), there are 3 oxygen atoms on the left and 1 on the right, so we add 2 water molecules to the right. For (c), add 3 water molecules to the right to have a total of 6 oxygen atoms on both sides. For (d), add 4 water molecules to the right to balance 4 oxygen atoms since there are none on the left.

Balance hydrogen atoms

Balance hydrogen by adding \(H^+\) ions to the opposite side of the water molecules added. For (a), add 6 \(H^+\) ions to the right to balance the 3 water molecules added to the left. For (b), add 4 \(H^+\) ions to the left to balance the 2 water molecules added to the right. For (c), add 6 \(H^+\) on the left to balance the 3 water molecules added to the right. For (d), add 8 \(H^+\) ions to the left to balance the 4 water molecules on the right.

Balance charge with electrons

Balance the charge by adding electrons to the side with the higher charge. For (a), the left has a charge of -1 and the right has 0 charge, so we add 1 electron to the right. For (b), there is a +1 charge on the left and a +2 charge on the right, so we add 1 electron to the right to balance the charge. For (c), the left side has a charge of -2 and the right has a charge of 0, so add 2 electrons to the right. For (d), the left side has a -2 charge and the right side has a +6 charge, so we add 8 electrons to the left.

Equalize the number of electrons in both half reactions

For redox reactions, ensure that the electrons lost in the oxidation half-reaction are equal to the electrons gained in the reduction half-reaction. If dealing with multiple half-reactions, multiply the half-reactions by appropriate factors to balance the electrons across. This step applies to any reactions that have separate reduction and oxidation half-reactions and may not be needed if the exercise consists of single-step reactions.

Verify the balance of the final equations

Check all elements and the overall charge to ensure each side of the equation is balanced. For each reaction, tally the atoms of each element and the total charge to confirm that the equation is balanced.

Key concepts

Redox Reactions

Redox reactions, or reduction-oxidation reactions, are processes where electrons are transferred between two species. These reactions are fundamental in chemistry as they are associated with energy changes and form the basis of many applications including batteries, corrosion, and cellular respiration. A redox reaction typically involves two half-reactions: reduction, where a molecule, atom, or ion gains electrons, and oxidation, where they lose electrons. Identifying redox reactions is critical, and this can be done by observing changes in oxidation states of the elements involved in the reaction. Recognizing which species undergoes oxidation and which undergoes reduction is essential to balance redox reactions correctly.

In our textbook exercise examples, when balancing redox reactions, we observe the transfer of electrons between different species in acidic solutions, an environment often encountered in biological and industrial settings. By balancing the equations, we can understand stoichiometry, or the quantitative relationships between reactants and products in a chemical reaction, which ultimately informs both theoretical calculations and practical applications like the formulation of reactions in a lab.

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions. This method involves breaking down a complex redox reaction into two simpler half-reactions: one for oxidation and one for reduction. By focusing on each half-reaction separately, balancing becomes more manageable. Each half-reaction is balanced for mass and charge independently by adding electrons, water molecules, and hydrogen ions as necessary, particularly in an acidic solution.

Once the individual half-reactions are balanced, they are then recombined to form the balanced overall equation. Ensuring that the number of electrons lost in oxidation equals the number of electrons gained in reduction is key to maintaining the balance. Additional adjustments may be needed if multiple half-reactions are involved to equalize electron transfer between them.

Oxidation State

The oxidation state, also known as oxidation number, is a concept that helps in understanding the electron transfer processes in redox reactions. It is a measure of the degree of oxidation of an atom in a substance. Oxidation state can be thought of as the hypothetical charge that an atom would have if all its bonds to atoms of different elements were completely ionic. Oxidation states are essential for balancing redox reactions because they allow us to determine which atoms are oxidized and which are reduced.

Changes in oxidation state through a reaction indicate the loss or gain of electrons by different elements. For instance, an increase in oxidation state signifies oxidation while a decrease points to reduction. By following the changes in oxidation states, you can effectively apply the half-reaction method to balance equations. Moreover, they play a crucial role in identifying the species that must be balanced for charge using electrons during the balancing process.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the relative quantities of reactants and products in chemical reactions. It is based on the conservation of mass and the principle that each element must have the same number of atoms on both sides of a chemical equation. Stoichiometry involves calculations that use the relationships between reactants and products in a balanced chemical equation.

In the context of redox reactions, stoichiometry is crucial because it requires the number of atoms and the charge to be balanced on both sides of the equation. This means that along with the species' changes in oxidation states, it's equally important to ensure that the physical quantities (the stoichiometric coefficients) of all reactants and products are accounted for. Correct stoichiometry is vital for preparing reactions in the lab, where precise amounts of substances are crucial, and in industry for scaling up reactions for manufacturing processes.