Question

Balance the following equations by the half-reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\) c. \(\mathrm{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Acid }}{\longrightarrow}\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}{ }^{2-}(a q)\) d. \(\operatorname{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow}\) \(\operatorname{CrO}_{4}{ }^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) e. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Base }}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)

Short answer

a. 2 Fe(s) + 6 HCl(aq) → 2 HFeCl4(aq) + 3 H2(g) b. IO3^- + 6H^+ + 5I^- → 3I^- + 3H2O

Step-by-step solution

Determine the oxidation states of species

For this reaction, we first find the oxidation states of the species involved: Fe: 0 (solid state) H in HCl: +1 Cl: -1 H in HFeCl4: +1 Fe in HFeCl4: +3 Cl in HFeCl4: -1 H in H2: +1

Separate the half-reactions

In this reaction, Fe is oxidized (from 0 to +3) while H in HCl is reduced (from +1 to 0 in H2). Separate the half-reactions: 1. Oxidation: Fe → Fe^3+ + 3e^- 2. Reduction: 2H^+ + 2e^- → H2 (from HCl)

Balance the half-reactions

1. Oxidation half-reaction is already balanced. 2. Reduction half-reaction: Add 6 Cl^- to the right side 2H^+ + 2e^- + 6 Cl^- → H2 + 6 Cl^-

Combine the half-reactions

To combine the half-reactions, we must first make sure that the number of electrons lost in the oxidation reaction is equal to the number of electrons gained in the reduction reaction. To do that, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3, and then add them together. 2(Fe → Fe^3+ + 3e^-) + 3(2H^+ + 2e^- + 6 Cl^- → H2 + 6 Cl^-) Resulting in: 2 Fe(s) + 6 HCl(aq) → 2 HFeCl4(aq) + 3 H2(g) b. Balance \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\)

Determine the oxidation states of species

For this reaction: I in IO3^-: +5 I in I^-: -1 I in I3^-: 0

Separate the half-reactions

In this reaction, I in I^- is oxidized (from -1 to 0) while I in IO3^- is reduced (from +5 to 0). Separate the half-reactions: 1. Oxidation: 2I^- → I2 + 2e^- 2. Reduction: IO3^- + 2e^- → I-

Balance the half-reactions

1. Oxidation half-reaction is already balanced. 2. Reduction half-reaction: Add 6 H2O on the right side and then 6H^+ on the left side IO3^- + 2e^- +6H^+ → I- + 3H2O

Combine the half-reactions

To combine the half-reactions, we must first make sure that the number of electrons lost in the oxidation reaction is equal to the number of electrons gained in the reduction reaction. We get: (IO3^- + 2e^- +6H^+ → I- + 3H2O) + (2I^- → I2 + 2e^-) Resulting in: IO3^- + 6H^+ + 5I^- → 3I^- + 3H2O Therefore, the balanced equation is: IO3^- + 6H^+ + 5I^- → 3I^- + 3H2O Please, continue these steps to balance the remaining reactions.

Key concepts

Balancing Redox Equations

Balancing redox equations requires an understanding of the movement of electrons during chemical reactions. Redox reactions involve both oxidation (loss of electrons) and reduction (gain of electrons). The half-reaction method is a systematic way of balancing these equations.

Let's break down the method used in our exercise. First, split the overall reaction into two half-reactions, one for oxidation and one for reduction. Each half-reaction is balanced separately for mass and charge. For the mass, make sure atoms other than oxygen and hydrogen are balanced. Balance oxygen by adding water molecules, and hydrogen by adding hydrogen ions. For the charge, add electrons to either side of the half-reaction as needed.

Finally, we combine the two half-reactions by scaling them so that the number of electrons lost in one half equals the number of electrons gained in the other half. The electrons should cancel out, giving a balanced overall reaction. Throughout this process, ensure that the conditions (acidic or basic) are kept in mind, as they affect the balancing of hydrogen and oxygen.

For students who find it challenging, remember to meticulously check each step and practice with a variety of equations. Each redox reaction may introduce new nuances to the balancing process, but the fundamental steps remain the same.

Oxidation States

Understanding oxidation states is pivotal to grasping redox reactions. The oxidation state, or oxidation number, indicates the degree of oxidation of an atom in a compound. It is a hypothetical charge that an atom would have if all its bonds to different atoms were completely ionic.

To determine oxidation states, we follow a set of rules such as: hydrogen usually has an oxidation state of +1, oxygen usually has -2, and alkali metals in compounds have +1. The sum of oxidation states for all atoms in a molecule or ion must equal the charge of the molecule or ion. For example, in water (H₂O), hydrogen has an oxidation state of +1 and oxygen has -2, giving a total charge of zero, which corresponds to the neutral charge of a water molecule.

In our exercise, recognizing changes in oxidation states helps to identify which species are oxidized and reduced. These changes signal the transfer of electrons that dictate how we balance the reactions. Encourage students to practice determining oxidation states as it is fundamental to solving redox equations.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are a type of chemical reaction that involves a transfer of electrons between two species. An increase in oxidation state corresponds to oxidation, while a decrease signifies reduction.

Understanding the concept of redox reactions extends beyond the scope of simple electron transfer; it's central to many biological processes, industrial applications, and environmental systems. For instance, cellular respiration and photosynthesis are redox reactions crucial for life. Corrosion and battery operation also fall under redox chemistry.

To fully grasp redox reactions, it's beneficial to recognize the signs of oxidation and reduction, study how electrons move during the reaction, and observe how redox reactions impact everyday life. These reactions are not isolated; they are always coupled, as one species gains the electrons that another species loses. Emphasizing the interdependent nature of oxidation and reduction can aid students in comprehending more complex concepts in chemistry.

Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. It is based on the law of conservation of mass where the total mass of the reactants equals the total mass of the products. Stoichiometry relies on balanced chemical equations to determine the proportions of chemicals involved.

When balancing redox equations, stoichiometry plays a crucial role. Once you correctly balance the half-reactions for mass and charge, stoichiometry can be used to find the coefficients that balance the overall equation. These coefficients correspond to the number of moles of each substance involved in the reaction.

Encourage students to familiarize themselves with the concept of the mole, molar masses, and avogadro's number for stoichiometric calculations. Many students struggle with stoichiometry because it requires careful attention to detail and robust problem-solving skills, but with practice, it becomes an invaluable tool for predicting the outcomes of chemical reactions and understanding the relationships within them.