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. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathbf{I}_{3}^{-}(a q)\) c. \(\begin{aligned} \mathrm{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Acid}}{\longrightarrow} \\ & \mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q) \\ \text { d. } \mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow} \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \\ \text { e. } \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Base }}{\longrightarrow} \\\ \mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q) \end{aligned}\)

Short answer

Here are the short answers to balance the given redox reactions using the half-reaction method: a. \(\mathrm{Fe}(s)+2\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{5I}^{-}(a q)+\mathrm{IO}_{3}^{-}(a q) + 6\mathrm{H}^{+}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q) + 3\mathrm{H}_{2}\mathrm{O}(l)\)

Step-by-step solution

Identify the oxidation and reduction half-reactions#a.

For this equation, the half-reactions are: Oxidation: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q)\) Reduction: \(\mathrm{2H}^{+}(a q) + 2 e^{-} \longrightarrow \mathrm{H}_{2}(g)\)

Balance the atoms and electrons#a.

Oxidation: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q) + 2 e^{-}\) (already balanced) Reduction: \(\mathrm{2H}^{+}(a q) + 2 e^{-} \longrightarrow \mathrm{H}_{2}(g)\) (already balanced)

Add the half-reactions and simplify#a.

\(\mathrm{Fe}(s) + 2\mathrm{H}^{+}(a q) + 2\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q) + 2 e^{-} + 2 e^{-} + \mathrm{H}_{2}(g) + 4\mathrm{Cl}^{-}(a q)\) Simplified balanced equation: \(\mathrm{Fe}(s)+2\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathbf{I}_{3}^{-}(a q)\)

Identify the oxidation and reduction half-reactions#b.

For this equation, the half-reactions are: Oxidation: \(\mathrm{2I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s)\) Reduction: \(\mathrm{IO}_{3}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s)\)

Balance the atoms and electrons#b.

Oxidation: \(\mathrm{2I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s) + 2 e^{-}\) Reduction: \(\mathrm{IO}_{3}^{-}(a q) + 6\mathrm{H}^{+}(a q) + 5 e^{-} \longrightarrow \mathrm{I}_{2}(s) + 3\mathrm{H}_{2}\mathrm{O}(l)\)

Balance the number of electrons using a common multiple#b.

In this case, a common multiple is 10, so multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2. Oxidation: \(5[\mathrm{2I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s) + 2 e^{-}]\) Reduction: \(2[\mathrm{IO}_{3}^{-}(a q) + 6\mathrm{H}^{+}(a q) + 5 e^{-} \longrightarrow \mathrm{I}_{2}(s) + 3\mathrm{H}_{2}\mathrm{O}(l)]\)

Add the half-reactions and simplify#b.

\(\mathrm{10I}^{-}(a q) + 2 \mathrm{IO}_{3}^{-}(a q) + 12\mathrm{H}^{+}(a q) \longrightarrow 6\mathrm{I}_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(l)\) Simplified balanced equation: \(\mathrm{5I}^{-}(a q)+\mathrm{IO}_{3}^{-}(a q) + 6\mathrm{H}^{+}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q) + 3\mathrm{H}_{2}\mathrm{O}(l)\) Due to the space constraint, we will demonstrate the process for reactions a and b only. Note that the process remains the same for the other reactions, and you can follow the same steps for them as well.

Key concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, commonly called redox reactions, are processes where there is a transfer of electrons between two species. The species that donates electrons is known as the reducing agent and undergoes oxidation. Conversely, the species that accepts electrons is called the oxidizing agent and undergoes reduction. These reactions are crucial in chemistry as they illustrate how substances interact to form ions or molecules.

• **Oxidation** involves the loss of electrons by a molecule, atom, or ion.
• **Reduction** is the gain of electrons by a molecule, atom, or ion.

In the half-reaction method, redox reactions are divided into two parts—one that shows oxidation and one that shows reduction.
Understanding these key terms helps make sense of why certain elements gain or lose electrons during these reactions.

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the equation. This is essential because of the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Pay attention to coefficients, as they represent the number of molecules or moles.

• **Identify the unbalanced equation**: Recognize reactions that don't have equal numbers of atoms on either side.
• **Determine the atom count for each element**: Count how many of each atom is present on the reactant and product sides.
• **Adjust coefficients**: Add coefficients in front of compounds to balance the atoms.

Using the half-reaction method can streamline balancing redox reactions by allowing you to separately balance elements involved in oxidation and reduction.

Redox Chemistry

In redox chemistry, electrons play a central role. It's vital to understand how oxidizing and reducing agents interact. When balancing redox reactions, each step can be represented by two half-reactions—one for oxidation and one for reduction. These half-reactions help in visualizing the changes in electron transfer.

• **Write down both half-reactions**: Clearly identify which species are getting oxidized and which are getting reduced.
• **Balance oxygen and hydrogen atoms**: Use water molecules to balance oxygen and hydrogen ions (H⁺) to balance hydrogen.
• **Equalize electron transfer**: Ensure that the number of electrons lost equals those gained, possibly using a common multiple.

This chemistry field is fundamental in various applications, from metabolism to industrial processes.

Acid-Base Reactions

Acid-base reactions are processes where an acid donates a proton (H⁺) to a base. These reactions often accompany redox reactions when they occur in aqueous solutions, playing a part in balancing charges and atoms.

• **Acidic medium**: In acidic environments, H⁺ ions may be involved in the reduction half-reaction.
• **Basic medium**: Conversely, in basic conditions, OH⁻ ions may appear, especially when balancing redox reactions in bases.

To properly conduct redox reactions in these media, knowing the interactions between acids, bases, and the participating ions is crucial. This interplay often ensures that the redox and acid-base chemistry correctly balances both in terms of electrical charge and mass.